Physics engineering-first-year notes, books, e book pdf download
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The document provides information about Studynama.com, an online education hub that provides free study materials for students from classes 9-12, engineers, managers, lawyers, and doctors. It includes lecture notes, project reports, and solved papers for various subjects. Disclaimers are provided noting that Studynama.com does not own the copyright for this content. The remainder of the document is a study material on elastodynamics that covers topics like simple harmonic motion, displacement and area vectors, Coulomb's law, and more.
![Unit-1 Elasto-dynamics
Page 20
̅ − ̅ = Δ
Hence the contribution to the line integral ∮ .� ⃗⃗⃗⃗ from the path and
= Δ Δ …………………………………………… (2)
Similarly by the path and
= Δ Δ …………………………………………… (3)
Therefore the line integral along the whole rectangular form (2) and (3) is given by-
= ∮� +� .� ⃗⃗⃗⃗
= ∮� .� ⃗⃗⃗⃗
= − Δ Δ ……………………………… (4)
Now = lim
Δ Δ →
=
lim
Δ Δ →
( − )� Δ Δ
Δ Δ
= − ……………………………………. (5)
Similarly
=
( − ) ……………………………………. (6)
and = − ……………………………………. (7)
Summing up the results given in (5), (6) and (7) we get
= ̂ +� ̂ +� ̂
= ̂� ( − )� +� ̂� (− )� +�̂ ( − )
=
[
̂ ̂ ̂
]
= ∇⃗⃗ ×�
Note: The curl of a vector field is sometime called circulation or rotation or simply .
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![Unit-1 Elasto-dynamics
Page 27
⃗ =
We know that ⃗ =� −
So
− =
⇒
−∇⃗⃗ .� ∇⃗⃗ =
⇒
∇ = −
⇒
+ + = −
Gauss law in Presence of dielectrics :
The Gauss’ la elates the ele t i flu a d ha ge. The theo e states that the et ele t i flu a oss a
arbitrary closed surface drown in an electric field is equal to times the total charge enclosed by the
surface. Now we want to extend this theorem for a region containing free charge embedded in dielectric.
In figure the dotted surface in an imaginary closed surface drown in a dielectric medium. There is certain
amount of free charge in the volume bounded by surface. Let us assume that free charge exists on the
surface of three conductors in amount ,� ,� …� ..In a dielectric there also exits certain amount of
polarisation (bound) charge .
He e Gauss’ theo e
∬� ⃗ .� ⃗⃗⃗⃗
= ( ′
+� ) ……………….
Where � =� +� +� is the total free charge and is the polarisation (bound) charge by
= ∬ ⃗ .� ⃗⃗⃗⃗
+ +
+� ∭ −
�
……………….
Here is the volume of the dielectric enclosed by . As there is no boundary of the dielectric at ,
therefore the surface integral in equation (2) does not contain a contribution from . If we transform
volume integral in (2) into surface integral by means of Gauss divergence theorem, we must include
contribution from all surface bounding , namely ,� ,� and i..e.
∫ −
�
= [ ∬ ⃗ .� ⃗⃗⃗⃗
+ +
+� ∭ −
�
]
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![Unit-II (LASER)
Figure(1):Induced absorption
This may also be shown by the equation
+ ℎ� ⟶∗
[* represents the exited state ]
The probability of absorption transition is given by
∝ �
= �
Where � is the energy state density
And the number of absorption transition in material is equal to the product of number of atoms at � and
absoption transition is given as
=
= �
Where is the number of atoms in ground state �
2. Spontaneous Emission: When an atom at lower energy level is exited to the high energy level, it cannot stay
in the exited state for relatively longer time. In a time of about −8
��, the atom reverts to the lower energy
state by releasing a photon of energy ℎ� = �− �. This emission of photon by an atom without any external
input is called spontaneous emission.
Figure(2): Spontaneous emission
We may write the transition as
∗
⟶ + ℎ�
Probability of spontaneous emission depends only on the properties of energy states and is depends on the
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![Unit-II (LASER)
Einstein’s coefficient:
Let there is a lasering medium in which the number of atoms in the ground state are and number of atoms
in the excited state are , � is the energy density of radiation for frequency �.
The rate of self-emission
∝
=
And the rate of stimulated emission
∝
And ∝ �
∴ ∝ �
= �
The rate of absorption
∝
And ∝ �
= �
Here, coefficient and and are respectively called the Ei stei ’s and coefficients. It is clear that
the rate of stimulated emission and rate of absorption determined by the same coefficient . This is why
simulated emission is also called the inverse absorption.
Relation between Einstein’s and coefficient:
Let there be an assembly of atoms in thermal equilibrium at temperature T with radiation frequency �.
Since the rate of absoption of radiation i.e. transition from state � ⟶ �is proportional to the energy desity
of radiation � . The number of transition per unit time per unit volume from � → �is given by
= �
Where is the number of atoms in energy state � and is the probability of the transition from � → �
Similarly the number of transition per unit time per unit volume from state � → �may be given as
= { + � }
Where is the number of atpms in energy state � and is the probability of the transition from � → �
In equilibrium state
=
� = { + � }
� = + �
� − � =
[ − ] � =
� =
�
� −�
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![Unit-II (LASER)
� = [
�
�
�
�
−�
]
� = [�
�
�
�
−
�
�
]
� = [�
�
�
�
−
]
By Boltzmann distribution law
= �− � / �
= �− � / �
Where is the total number of atoms
�
�
= � � −� / �
Then
� = [
{
� −�
�� }
�
�
−
]
But � − �= ℎ�
So
� = [
{
ℎ�
�� }
�
�
−
]
But a o di g to the Pla k’s the e e gy de sity of the adiatio of f e ue y � at temperature T is given by
� =
8�ℎ�
. [
( ℎ�/��)−
]
On comparing
=
�ℎ�
�
and =
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![Unit-3 (Fiber Optics)
5. Temperature Changes: A temperature change from 0 to ℃ could add as much as to the cable
losses. Stress (Strain and tension) could add another .
6. Attenuation Losses: Attenuation losses of an optical fiber is defined as the ratio of optical output
power � from a fiber of length � to the output power �� . In symbol � is expressed attenuation in
� / .
� =
�
log [
��
�
]
In case a fiber is an ideal when �� = � , therefore � = which means that there will no
attenuation loss. In actual practice, a low loss fiber may have � = /.
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![Unit IV- Quantum Mechanics
Page 3
Emissivity:
The emissivity of a material is the irradiative power of its surface to emit heat by radiation, usually it is shown
by or . It is the ratio of energy radiated by a material to the energy radiated by the black body.
True black body has maximum emissivity � =� (highly polished silver has an emissivity for about . at
least.)
Plank’s Quantum Hypothesis:
Plank assumes that the atoms of the wall of blackbody behave as an oscillator and each has a characteristic
frequency of oscillation. He made the following assumption-
1) An oscillator can have any arbitrary value of energy but can have only discrete energies as per the
following relation
� =� ℎ�
Where � =� , , , � …� ..and � and ℎ a e k o as f e ue a d Pla k’s o sta t.
2) The oscillator can absorb or emit energy only in the form of packets of energy ℎ� but not
continuously.
� =� ℎ�
Average energy of Plank’s Oscillators:
If be the total number of oscillations and as the total energy of these oscillators, then average energy will
be given by the relation.
̅ = ……………………………………………………………………………. (1)
Now consider ,� ,� …� …� …� …� .� .�as the number of oscillators having the energy values
,� ℎ�,� ℎ�� …� .� .� ℎ�espe ti el . The the Ma ell’s dist i utio fo ula
= +� +� +� ⋯� …� …� …� …� ..
= � +�−
ℎ�
�� +� −
ℎ�
�� +� ⋯� …� …� .�
=
( � −�−
ℎ�
��)
.............................................................. (2)
And the total energy
= ×� + ×� ℎ�+ ×� ℎ�+� ⋯� ….
= ×� +� ( −
ℎ�
�� � ×� ℎ�)� +� (−
ℎ�
�� � ×� ℎ�)� +� ⋯
= −
ℎ�
�� � ×� ℎ�[ � +� −
ℎ�
�� +� −
ℎ�
�� +� ⋯� …� .� .]
=
−
ℎ�
��
ℎ�
( � −�−
ℎ�
��)
………………………………………….. (3)
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![Unit IV- Quantum Mechanics
Page 7
Matter wave:
According to Louis de-Broglie every moving matter particle is surrounded by a wave whose wavelength
depends up on the mass of the particle and its velocity. These waves are known as matter wave or de-
Broglie waves.
Wavelength of the de-Broglie wave:
Consider a photon whose energy is given by � =� ℎ�� =
ℎ
�
[ � � =� ��……………………………………… (1)
Where ℎ is Pla k’s o sta t. � ×� −
, � is the frequency and � is the wavelength of photon.
No Ei stei ’s ass e e g elatio
= ……………………………………… (2)
By equation (1) and (2)
=
ℎ
�
� =
ℎ
� =
ℎ
Where � =�
In place of the photon a material particle of mass is moving with velocity then
� =
ℎ
…………………………….…… (3)
(i)
Now we know that the kinetic energy of the material particle of mass moving with velocity � is given by-
=
= �
= [ � � =� ]
= √
So by equation (3)
� =
ℎ
√
(ii)
According to kinetic theory of gasses the average kinetic energy of the material particle is given by � =
� where � =� . � ×�−
/ i.e. Boltzmann constant
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![Unit IV- Quantum Mechanics
Page 8
= �
= �
= � � � =
= √ �
So by equation (3)
� =
ℎ
√ �
……………………………… (4)
Group or Envelope of the wave:
When a mass particle moves with some velocity than it emits the matter waves, those waves interacts each
other and where there they interfere constructively they form an envelope around the particle which is
known as wave group or simply envelope.
Figure(2): Formation of the wave packet
Group velocity:
Group velocity of a wave is the velocity with which the overall shape of the a e’s amplitudes (modulation
or envelope) of the wave propagates through space. It is denoted by �.
Phase velocity:
The phase velocity of a wave is the rate at which the phase or the wave propagates in the space. It is
denoted by �.
Expression for Group velocity and phase velocity:
Let us suppose that the wave group arises from the combination of two waves that have some amplitude
but differ by an amount ∆ in angular frequency and an mount ∆ in wave number.
= � cos � −� ………………………………
= � cos[ � +� ∆� − � +� ∆] ………………………………
By the principle of superposition
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![Unit IV- Quantum Mechanics
Page 9
= +� …………….…………………………… (3)
= [cos � −� +� cos{ � +� ∆� − � +� ∆}]
Using the identity
cos� � +� cos� � =� � cos� (
� +�
)� cos� (
� −�
)
And cos −� =� cos� �
� +� � =
� −� +� { � +� ∆ � −� � −� ∆ }
� +� � =
� −� � +� ∆ � −� ∆
� +� � =
�+∆ �− � −∆�
� +� � =
+∆ �− �+∆�
� −� � =
� −� −� { � +� ∆ � −� � −� ∆
� −� � =
� −� � −� � −� ∆ � +� � +
� −� � =
−∆ � +� ∆
� −� � =� −
∆ � −� ∆
� =� � [ � cos{
� +� ∆� − � +� ∆
} .� cos{
∆ � −� ∆
}]
Let � +� ∆ � =�and � +� ∆ � =�
So we have
= [cos� (
� −�
) .� cos� (
∆ � −� ∆
)]
⟹ = [cos � −� .� cos�
∆
� −
∆�
]…………………… (4)
This is the resultant wave equation of superposition of two waves having the amplitude
� cos�
∆
� −
∆�
and phase cos � −� where and are mean values of angular frequency and
prapogation constant of the wave.
Phase velocity:
Since phase � −� =
Differentiating with respect to we get
� −� =
⟹
�
= �
But � =
�
phase
velocity
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![Unit IV- Quantum Mechanics
Page 12
⟹
=
� �
ℎ√ −
�
�
...................................................(9) � � =
√ −
�
�
Now phase velocity � =
�
So
� = (
��
ℎ
.
√ −
�
� )
(
� �
ℎ√ −
�
� )
� =
�
ℎ
.
√ −
�
�
� �
ℎ√ −
�
�
� �
� =
�
………………………………………………………………… (10)
Now group velocity � =
�
The expression can be written as
� = ……………………………………………………………… (11)
In order to find the value of � we have to solve the following terms-
⟹
=
[
�
ℎ
.
√ � − ]
[By equation (8)]
⟹
= �
ℎ
√ � −
−
⟹
= �
ℎ
(− ) � −
−
.� (− )
⟹
= �
ℎ
� −
−
……………………………………… (12)
Again
⟹ =
[
�
ℎ√ � − ]
⟹ =
�
ℎ
[
√ � − ]
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![Unit IV- Quantum Mechanics
Page 13
⟹ =
�
ℎ
.
[
√ � − .� � −� .( � − )
−
.� −
( � − )
]
⟹ =
�
ℎ
.
[
√ � − � +� .
√ � −
( � − )
]
⟹ =
�
ℎ
.
[
� − +
(
� −�
)� .√ � − ]
⟹ = [
�
ℎ
� −�
−
] …………………………… (13)
Putting the value from (13) and (14) into (11) we get
⟹ � =
[
{
�
ℎ
� ( � −)
−
}
{
�
ℎ
( � − )
−
}
]
Group velocity � = ………………………………………………………………………… (14)
By equation (10) we have
�.� =
By equation (14) i.e. � =�
So �.� � =
3. Non-Relativistic free Particle:
According to the de-Broglie hypothesis
� =
ℎ
��
Now the total energy
= � ……………………………………………………………………… (1)
And = ℎ� ……………………………………………………………………… (2)
By equation (1) and (2)
ℎ� = �
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![Unit IV- Quantum Mechanics
Page 15
Uncertainty Principle:
It is impossible to determine the exact position and momentum of a particle simultaneously.
Let us consider a particle surrounded by a wave group of de-Broglie wave as shown in the figure
Figure(3): particle surrounded by a wave packet
Let us consider two such waves of angular frequency and and prapogation constant and
traviling along the same direction are-
= � sin � −� …………………………………………… (1)
= � sin � −� …………………………………………… (2)
According to the principal of superposition
= +�
= � sin � −� � +� � sin� −�
= [sin � −� +� sin � −� ]
= [ � sin� (
� −� � +� � −�
)� .� cos� (
� −� � −� � +�
)]
� sin� � +� sin� � =� � sin� (
� +�
)� .� cos� (
� −�
)
= [sin� (
+�
� −
+�
)� .� cos� (
−�
� −
−�
)]
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![Unit IV- Quantum Mechanics
Page 16
Let � =
+�
(3)
And � =
+�
…………………………………….........................
And ∆ � =� −�
And ∆ � =� −�
So we have
= [sin � −� .� cos�
∆
� −
∆�
] ………………………………………………………… (4)
The resultant wave is plotted in the figure (4). The position of the particle cannot be given with certainty it is
somewhere between the one node and the next node. So the error in the measurement of the position of
the particle is therefore equal to the distance between these two nodes.
Figure(4): The envelope created by the superposition of two waves.
The node is formed when
cos�
∆
� −
∆�
� =� � cos� � =� � ⟹� � =� +�
�
⟹
∆
� −
∆
= � +�
�
Thus and represents the positions of two successive nodes, then at any instant , we get-
∆
� −
∆�
= � +�
�
………………………………………………… (5)
∆
� −
∆�
= � +�
�
………………………………………………… (6)
Now on subtracting (5) from (6) we get
∆
� −
∆�
−
∆
� +
∆�
= .
�
+
�
−� .
�
−
�
∆�
−� = �
∆ = � ………………………………………………… (7)
But � =
�
�
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![Unit IV- Quantum Mechanics
Page 20
The momentum of the scattered photon is
ℎ
�
then the momentum along the x-axis is –� � sin� �and
+ � sin� �. But when the photon of wavelength �′
collied to the electron then this photon recoile the
electron by giving some momentum to it.
Now the uncertainty in the momentum transfer to the electron will be � =[ � +� ] − [ � −� ]
=
ℎ
�′ +
ℎ
�
sin� � � −�
ℎ
�′ −
ℎ
�
sin� �
=
ℎ
�
sin� � ................................................................. (2)
By equation (1) and (2)
.� =
�
� si � �
×
ℎ� si � �
�
.� = ℎ
.�
ℏ
Where ℏ� =
ℎ
�
Diffraction of electron beam by a single slit:
Suppose a narrow beam of electron passes through a narrow single slit and produces a diffraction on the
screen as shown in figure.
Figure(7): Diffraction pattern of electron beam by single slit
But the theo of F au hofe ’s diffraction at a single slit � sin� �� =� ± �the first minima is given by
� sin� �= � …………………………………….……………………………………… (1)
In producing the diffraction pattern on the screen, all the electrons have passed through the slits but we
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![Unit IV- Quantum Mechanics
Page 21
can-not say definitely at what place of the slit. Hence the uncertainty in determining the position of electron
is equal to the width of the slit then by equation (1)
=
�
� si � �
…………………………………………….…………………… (2)
Initially the electrons are moving x-axis and hence they have no component of momentum along y-axis.
After diffraction on the slit, they are deviated from their initial path to form the pattern. Now they have a
component � sin� �.As y component of momentum may be anywhere between � sin� �and –� � sin� �.
Hance the uncertainty in component is
= � sin� �� −� − � sin� �
= � sin� �
= �
ℎ
�
� sin� � [ � � =
ℎ
�
] .............................. (3)
By equation (2) and (3)
.� y =
�
� si � �
×
ℎ
�
sin� �
.� y ℎ
.� y
ℏ
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![Unit IV- Quantum Mechanics
Page 23
we get
In x-direction
� � = � �
ℎ�� +� =
ℎ�′
cos� �� +� � cos� �…………………………………. (1)
In y-direction
� +� =
ℎ�′
sin� �� −� � sin� �…………………………………. (2)
By equation (1)
� cos� �= ℎ�� −� ℎ�′
cos� � …………………………………. (3)
By equation (2)
� sin� �= ℎ�′
sin� � …………………………………. (4)
Squaring equation (3) and (4) then adding we get
sin �� +� cos� = ℎ�� −� ℎ�′
cos� � + ℎ�′
sin� �
= ℎ � +� ℎ�′
cos �� −� ℎ��′
cos� �� +� ℎ�′
sin �
= ℎ � +� ℎ�′
cos �� +� sin� −� ℎ��′
cos� �
= ℎ � +� ℎ�′
−� ℎ��′
cos� �
ℎ � +� ℎ�′
−� ℎ��′
cos� � = …………………………………. (5)
Now by conservation of energy
� � = � �
ℎ�� +� = ℎ ′
+� …………………………………. (6)
ℎ�� +� −� ℎ′ =
ℎ�� −� ℎ�′
+� =
Squaring both side
[ℎ�� −� ℎ�′
+� ] =
As we know that � +� � +�=� +� +� +� � +� � +�
so � −� � +�=� +� +� −� � −� � +�
ℎ � +� ℎ�′
+� −� ℎ��′
−� ℎ�′
+� ℎ� =
ℎ � +� ℎ�′
−� ℎ��′
+� +� ℎ�� −� �′
.� = …………………………………. (7)
Subtracting equation (5) from (7) we get
ℎ � +� ℎ�′
−� ℎ��′
+� +� ℎ�� −� �′
.� −� {ℎ� +� ℎ�′
−
ℎ ��′
cos� �}
= −�
ℎ � +� ℎ�′
−� ℎ��′
+� +� ℎ�� −� �′
.� −� ℎ� −� ℎ�′
+� ℎ��′
cos� �
= −�
−ℎ .� ��′
� −� cos� �+� ℎ�� −� �′
+� = −�
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![Unit IV- Quantum Mechanics
Page 26
tan� � =
[
�
� +
ℎ�
sin
�
] .� sin� �
�� −[
�
� +
ℎ�
sin
�
] .� cos� �
Let
ℎ�
=� �
then
tan� � =
[
�� sin� �
� +� �� sin
� ]
�� −[
�� cos� �
� +� �� sin
� ]
tan� � =
[
�� sin� �
� +� �� sin
� ]
� [ � −
cos� �
� +� �� sin
� ]
tan� � =
[
sin� �
� +� �� sin
� ]
[
� +� �� sin
�
� −� cos� �
� +� �� sin
� ]
tan� � =
sin� �
� +� �� sin
�
� −� cos� �
tan� � =
sin� �
� −� cos� � � +� �� sin
�
tan� � =
.� sin�
�
� .� cos�
�
.� sin
�
� +� �� sin
�
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![Unit IV- Quantum Mechanics
Page 27
tan� � =
.� sin�
�
� .� cos�
�
.� sin
�
� � +� �
tan� � =
cot�
�
[ � +� �]
tan� � =
cot�
�
[ � +
ℎ�
]
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![Unit IV- Quantum Mechanics
Page 33
�ℏ = (
ℏ
�
) � +� �
�ℏ = −
ℏ
+� � ………..…………………….………… (4)
S h odi ge ’s ti e i depe de t a e e uatio i 3-dimentional form:
�ℏ
�
=
ℏ
+ + � � +� �
But the Laplacian operator is given as ∇ =� + +
So the above equation can be written as-
�ℏ
�
=
ℏ
∇ � +� � ………………………….…………… (5)
Schrodinger’s time independent wave equation in 1-dimention:
Again from wave function-
= −
�
ℏ
��−�
=
−
�
ℏ
��
.�
�
ℏ
�
=
�
ℏ
�
.� −
�
ℏ
��
=
−
�
ℏ
��
Where =�
�
ℏ
�
Now differentiating partially with respect to we get
�
= −
��
ℏ
−
�
ℏ
�� …………………………………………… (7)
Now differentiating partially with respect to we get
=
−
�
ℏ
��
Again =
−
�
ℏ …………………………………………… (8)
Putting the value from (7) and (8) into equation (5) we get i.e. �ℏ � =
ℏ
∇ � +� �
�ℏ [−
�
ℏ
−
�
ℏ ] = −
ℏ
[ −
�
ℏ ] +� �
−
�
ℏ
−
�
ℏ =
−
�
ℏ
��
[
ℏ
+� � ]
=
ℏ
+� �
− � =
ℏ
� −� =
ℏ
ℏ
� −� =
− ℏ
� −� = ………………………………………………………. (10)
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![Unit IV- Quantum Mechanics
Page 36
=
ℎ
�
......................................... (6)
It is clear from expression (6) that inside an infinitely deep potential well, the particle can have only discrete
set of energy i.e. the energy of the particle is quantised. The discreet energies are given by
=
ℎ
�
=
ℎ
�
=�
=
ℎ
�
=�
=
ℎ
�
=�
.. = .. .. ..
.. = .. .. ..
The constant A of equation (4) can be obtained by applying the normalization condition i.e.
∫ | |
=�
= =
∫ | � sin� |
�
=
∫ sin �
�
=
∫
−c s� �
�
�
=
∫ � −� cos�
�
=
[{ }�
− {
si � �
�
}
�
] =
[ � −� � −sin� � −� sin�] =
[ � − sin
� � −� sin�] = � � =
�
[ � − � −� ] = � sin� �� =�
=
= √
�
…………………........................................... (7)
Now the wave function will be given by
= √
�
sin
�
�
� =� , , ,� …� …� .�
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![Unit 5: Wave Optics
Page 5
Mathematical treatment of interference:
Let two waves of amplitude and and angular frequency � super imposes and re-unit at a point after
traveling different path and ,let the phase difference of these two waves is �
If and are two waves then
= sin � …………
= sin � + � ………… (2)
By the principle of superposition of waves , the resultant waves will be
= +
= sin � +sin � + �
sin + = sin cos + cos sin
= sin � +[sin � cos � + cos � sin �]
= sin � +sin � cos � +cos � sin �
= sin � [+ cos �] + cos � [sin �] ………… (3)
Let cos �= + cos � …………
sin �= sin � …………
by the equation , and we get-
= sin � . cos � + cos � . A sin �
= [sin � cos � + cos � sin �]
= sin � + � …………
Here and � are constant and can be given by equation and as
+ cos � + sin � = cos � + sin �
+ cos � + cos � +sin � = [cos � + sin�]
+ cos � + sin� + cos � = [cos � + sin�]
+ + cos � =
cos� + sin� =
= + + cos �
Now the resultant intensity at any point is given as � ∝ for simplicity let
� =
So � = + + cos �
Condition for maxima:
For maximum intensity
cos � =
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![Unit 5: Wave Optics
Page 7
�� =
∫ �
�
∫ �
�
�� =
∫ + + cos �
�
∫ �
�
�� =
[ � + � + sin �] �
[�] �
�� =
+ . �
�
�� = +
�� = � + �
The average intensity is the average of the maximum and minimum intensities. It can be given by-
Now if = = then,
�� =
The average intensity is equal to the sum of the separate intensities. Whatever the intensity disappears at the
minima is actually appears at the maxima. Thus there is no violation of the law of conservation of energy in
the phenomena of interference.
Condition for the sustained interference of light.
1. Two sources of light must be coherent.
2. Difference in the amplitudes of the two waves must be small.
3. Sources should be narrow or point source.
4. The separation between two sources should as small as possible.
5. If the interfering waves are polarised then the plane of polarisation must be same.
6. The sources should be monochromatic.
7. Interfering waves from two coherent sources of light should travel in the same direction.S
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![Unit 5: Wave Optics
Page 8
Fringe width:
Consider a narrow monochromatic source and two parallel narrow slits and very close together and
equidistance from . Let be the distance between two slits and and be the distance of screen from
coherent source. The path difference between the rays reaching from and to is zero so the point
has maximum intensity.
Considering a point at a distance from . The wave reaches at the point from and hence =
− and = +
Figure(13): Measurement of fringe width
− = [ + ( +) ] − [+ ( −) ]
+ − = [ + + + . − { + + − .}]
+ − = + + + . − − − + .
+ − =
− = +
Now from the figure
If the point is very close to point
so − = Δand ≈ =
Δ =
+
Δ =
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![Unit 5: Wave Optics
Page 9
Δ =
1. Bright Fringes: For bright fringes the path difference is the integer multiple of the i.e.
= .
=
=
This equation gives the distance of the bright fringes from the point . Hence for the ℎ
bright fringe
(replacing by )
=
For next bright fringe
+ =
+
Therefor the distance between any two consecutive bright fringes
+ − =
+
−
� =
2. Dark Fringes: For dark fringes the path difference is an odd multiple of
So = +
=
+
Hence the ℎ
dark fringe (replacing by )
=
+
And for the + ℎ
dark fringe
+ =
[ + + ]
+ =
+
Therefore the distance between two consecutive dark fringes
+ − =
+
−
+
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![Unit 5: Wave Optics
Page 10
+ − = [ + − −]
+ − = .
�′
=
As the distance between two consecutive bright or dark fringes is same and is called fringe width and denoted
by �.
� =
i. The fringe width is directly proportional to the wavelength of the light used i.e. � ∝
ii. The fringe width is directly proportional to the distance of the slits from the screen i.e. � ∝
iii. The fringe width is inversely proportional to the distance between the slits i.e. � ∝
Shape of the interference fringes:
Actually these interfering fringes are hyperbolic in shape, but the eccentricity of fringes is quite large and
hence these hyperbolic fringes appear more or less strength lines.
Angular Fringe Width:
The angular fringe width is defined as the angular
separation between consecutive or dark fringes and
is denoted by �.
Figure (14): Angular fringe width
=
�
� = � + − �
� =
+
−
� =
+ −
� =
β
But � =
so � =
λ
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![Unit 5: Wave Optics
Page 13
Determination of the thickness of a thin sheet of
transparent material:
Figure(17): Shift in fringes on introducing the thin film
Distance travelled by the light in air = −(by the velocity c)
Distance travelled by the light in film= (by the velocity )
Time taken by the light to cover this distance
=
−
+
But =
�
ℎ =
So we have
=
−
+
=
−
+
=
− +
=
+ −
Thus the path to �. . is eqivelent to an air path + −
Now the path difference at
Δ = ℎ � � − ℎ � �
Δ = − [ + − ]
Δ = − − −
But − =
�
�
So we have ∆ = − − … …
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![Unit 5: Wave Optics
Page 14
but for ℎ
maxima,
Δ = … …
So by equation (1) and (2)
− − =
= [ + − ] … …
Where is the position of the ℎ
maxima
Now in absence of the plate �. . =
The ℎ
maxima ′
=[
�
] (in the absence of ) …………..
1. Displacement of the fringes:
If denotes the displacement of the ℎ
maxima by introducing the mica sheet, then
− ′
= [ + − ] −
=
�
+
�
− −
�
`
= −
This equation is free from so the displacement of each maxima will be same.
2. Thickness of mica sheet:
The displacement of any maxima by introducing a mica sheet of thickness is given by
=
×
−
3. Refractive index of the material of prism:
Once if we know the displacement of the fringes and thickness of the film we can calculate the
refractive index of the material of the film as-
− =
×
=
×
+
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![Unit 5: Wave Optics
Page 31
Now to find the intensity at point there are following two methods are available
1. Phase diagram Method.
2. Integral Method.
Phase diagram method:
In the figure, draw vectors , , … … … … ..such that the magnitude of each vector is and angle
between the two consecutive vectors is � . The vector gives the resultant vector. Let the magnitude of
the resultant vector is �. If is the centre of the polygon formed by the vector then by the simple
geometry we can see that each vector substance � at the centre and the angle substaended by the
resultant vector at the centre is �.
Let and are the normal drawn of first vector and resultant vector from centre .
Figure(28): Phase diagram
from right angle triangle Δ
= sin (
�
)
= sin (
�
)
But = = [ =]
= sin (
�
) … … …
Similarly from Δ
= sin (
�
)
= sin (
�
)
But =
��
so
��
= sin
� … … …
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![Unit 5: Wave Optics
Page 32
By (3) and (4) ��
�
=
sin
�
sin
�
� =
sin
�
sin
�
… … …
Now putting the value of � from the equation no. (2) we get
� =
sin
�
sin �
sin
�
sin �
… … …
Let
� si �
=
Then � =
sin
sin
Now is very small so sin ≈
Then
� =
sin
So
� =
sin
� =
sin
Where =
Now the intensity
� ∝ �
� = � [Where is a constant
� = [
sin
]
� = [
�
]
� = � [
sin
] … . … . . Where � =
Conditions for maxima and minima:
From the equation � = �[
si
] it is clear that the resultant intensity � at point on the screen depends on
the angle of diffraction � or on . For maxima, the derivation of � with respect to must be zero. �. .
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![Unit 5: Wave Optics
Page 33
{� [
sin
] } =
� . . (
sin
) (
cos − sin
) = ……
Condition for Minima:
For the minima, the first term in the above equation (8) should be zero i.e.
sin
=
sin =
= ± �
Putting the value of we get-
� sin �
= ± �
sin �= ±
Where = , , , , … … … ….
Condition for maxima:
Now the second term of equation (8) will show the condition for maxima-
i.e.
cos − sin
=
cos = sin
=
sin
cos
= tan
The condition for maxima is = tan
To find the value of for which the above condition may hold, we draw two curves
= … … …(9)
= tan … … …(10)
On the same graph as shown
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![Unit 5: Wave Optics
Page 35
first subsidiary maxima � = � [
sin
�
� ] =
�
Second subsidiary maxima � = � [
sin
�
� ] =
�
Third subsidiary maxima � = � [
sin
�
� ] =
�
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![Unit 5: Wave Optics
Page 40
Fraunhofer’s Diffraction of � Parallel slit:
Diffraction Grating:
It is an arrangement consisting of several parallel and equidistant slits each of equal width.
It is constructed by drawing the several equidistance parallel lines on an optically plane glass plate with a
pointed diamond. The distance between two consecutive slits is + = which is called the grating
element. Genrally the value of for the grating to be used with the visual light is of the order of
−
�. . lines drawn on length of the grating.
Figure(32): Diffraction Grating Figure(33): Diffraction Grating
Theory:
In figure, is a grating of parallel and equidistance slits , , … … … …the width of each slit is ′ ′ and
width of opaque space between the two consecutive slits is ′ ′. �. .the grating element = +.
Let a plane wavefront of wavelength is incidents normally on the grating. Then diffracted by it is focused on
a screen by means of a convergent lens on screen.
Intensity distribution: It is clear from the figure that
diffracted waves do not reach a point on the screen in
the same phase since their optical paths are not equal.
The path difference between the two consecutive wave
is Δ = sin �
Therefor the phase diff.
� =
�
× ∆
Wave diffracted at an angle � from each slit is � =
si
where =
� e si �
[by single slits diffraction]
Figure(34): Phase Diagram
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![Unit 5: Wave Optics
Page 41
Now we can find the resultant amplitude due to the superposition of such waves by phase diagram method.
In figure we draw vector , , … ..such that magnitude of each vector is � and the angle between
the consecutive vector is �. The vector which joints the initial points of first vector and final point of last
vector is and this vector sustained an angle �. and are the normal plotted from the centre of
polygon on first and resultant vectors.
From the figure in Δ
= sin
�
⇒ = sin
�
But = =
��
�
= sin
�
… … … .
Similarly in Δ
⇒ = sin (
�
)
⇒ = sin (
�
)
But = =
= sin (
�
) …………. (2)
Dividing equation (2) by (1)
�
�� =
sin
��
sin
�
�
= [
sin
��
sin
�
]
= � [
sin
��
sin
�
]
On substituting the value of � we get
= (
sin
) [
sin
��
sin
�
]
Multiplying and dividing by we get
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![Unit 5: Wave Optics
Page 42
= (
sin
) [
sin
��
sin
�
]
So the resultant intensity will be given as
� ∝
� =
Where is the proportionality constant
Putting the value of from (3) we get
� = (
sin
) [
sin
��
sin
�
]
� = � (
sin
) [
sin
��
sin
�
] … … …(3)
Where � =
In this expression the term
si
represents the intensity due to diffraction due to a single slit, while second
term [
si
��
2
� si
�
2
] represents the intensity due to interference of wave obtained from slits.
Condition for Principle Maxima:
For principal maxima the path difference will be zero so the phase diff then sin
�
= ⇒
�
= ± �where
= , , … ..then sin
��
is also zero and in the limit when sin
�
→ the value of term [
si
��
2
si
�
2
] will be
. Hence from equation the resultant intensity will be maximum.
i.e.
lim�
2
→
[
sin
��
sin
�
] =
So we have � = � (
sin
) ……..
Which is the intensity at principle maxima i.e. similar to the intensity by a single slit.
Condition for Minima:
From the equation (3) it is clear that the intensity will be minimum when sin
��
= but sin
�
≠
i.e. sin (
�
) =
⇒ � = ± �
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![Unit 5: Wave Optics
Page 43
Where = , , , … …
But � =
� si �
So
� sin �
= � … … …(5)
sin �=
This is the condition for the minimum intensity for N-slit diffraction.
Condition for Secondary maxima:
Condition for maxima is
�
�
=
So from equation (3)
�
[� (
sin
) {
sin
��
sin
�
} ] =
�
(
sin
)
�
{
sin
��
sin
�
} =
�
(
sin
) [{
sin
��
sin
�
}
sin
�
.
�
cos
��
− sin
��
.cos
�
{sin
�
}
] =
sin (
�
) . cos (
�
) − sin (
�
) . cos (
�
) =
sin (
�
) . cos (
�
) = sin
��
. cos
�
……… 6)
⇒ tan (
�
) = tan (
�
) … … …
again by equation
tan (
�
) cos (
�
) = sin (
�
)
⇒ sin (
�
) = tan (
�
) cos (
�
)
sin (
�
) =
tan
�
sec
��
sin (
�
) =
tan
�
√ + tan
��
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![Unit 5: Wave Optics
Page 44
sin (
�
) =
tan
�
√ + tan
�
By (7)
On squaring both sides we get
sin (
�
) =
si
�
2
c s
�
2
+ tan
�
sin
��
sin
� =
[ + tan
�
] . cos
�
sin
��
sin
� =
[cos
�
+ tan
�
. cos
�
]
sin
��
sin
� =
cos
�
+
si 2 �
2
c s2 �
2
cos
�
sin
��
sin
� =
[cos
�
+ . sin
�
]
sin
��
sin
� = − sin�
+ . sin
�
[
sin
��
sin
�
] =
+ − sin
�
[
sin
��
sin
�
] =
+ − sin
� ……… (7)
So by equation (3) and (7)
� = � (
sin
) ×
+ − sin
�
This is the expression for the intensity at the subsidiary maxima in N-slit diffraction.
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![Unit –V: Semiconductors
Page 10
= ∫
�
ℎ
−
+
(
�−�
��
)
∞
��
=
�
ℎ
∫
−
+
(
�−�
��
)
∞
��
As ≫ and − ≫ so +
�−�
�� ≈
�−�
��
So
=
�
ℎ
∫
−
(
�−�
��
)
∞
��
=
�
ℎ
∫ −
(
� −�
��
)
∞
��
=
�
ℎ
∫ −
(
� −��+��−�
��
)
∞
��
=
�
ℎ
∫ − −
�−��
��
(
� −��
��
)
∞
��
=
�
ℎ
� −��
�� ∫ −
−
�−��
��
∞
��
Let
�−��
��
= so that = .
Limits:-
As → ⟹ =
And → ∞ ⟹ = ∞
so
=
�
ℎ
(
� −��
��
)
∫ −�
.
∞
=
�
ℎ
(
� −��
��
)
∫ −�
∞
But ∫ / −�∞
=
√�
=
�
ℎ
�
(
� −��
��
)
[
√�
]
= (
�
ℎ
)
(
� −��
��
) … . .
This is the density or concentration of electron in conduction band in intrinsic semiconductor.
Hole concentration in valance band:
Since holes are created by removal of an electron so Fermi function will be −
So let us calculate the Fermi-Dirac distribution for holes as
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![Unit –V: Semiconductors
Page 11
− = −
+
(
�−�
��
)
− = − [ +
(
�−�
��
)
]
−
− = − [ −
(
�−�
��
)
]
− = − +
(
�−�
��
)
− = (
�−�
��
) … . .
And for the top of the valance band the density of the states will be given as
=
�
ℎ ℎ � −
Here ℎ is the effective mass of the hole near the top of the balance band.
So the hole concentration will be given as
ℎ =
�
ℎ ℎ � −
(
�−�
��
)
On integrating we get
ℎ = ∫
�
ℎ ℎ � −
(
�−�
��
)
��
−∞
ℎ =
�
ℎ ℎ ∫ � −
(
�−�
��
)
��
−∞
ℎ =
�
ℎ ℎ ∫ � −
(
�−��+��−�
��
)
��
−∞
ℎ =
�
ℎ ℎ ∫ � −
�−��
��
(
��−�
��
)
��
−∞
ℎ =
�
ℎ ℎ
(
��−�
��
)
∫ � −
−
��−�
��
��
−∞
Let
��−�
��
= ⟹ − =
And limits
As ⟶ −∞ ⟹ ⟶ ∞
And ⟶� ⟹ ⟶
ℎ =
�
ℎ ℎ
(
��−�
��
)
∫ −�
−
∞
ℎ = −
�
ℎ ℎ
(
��−�
��
)
∫ −�
∞
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![Unit –V: Semiconductors
Page 13
=
+ �
+ ln
ℎ
Now if the effective mass of the electrons and holes are same then
=
+ �
This shows that the Fermi level lies exactly in the middle of the forbidden energy gap as depicted in
figure. The Fermi level can also be defined as the energy level at which there is a . probability of finding
an electron. It depends on the distribution of energy level and number of electron available.
(B) Fermi level in extrinsic semiconductor:
(i) n-type extrinsic semiconductor:
At usual temperature all the donor level will be fully activated and the donor atoms will be ionised. It
means the density of electrons will be increase. It means the density of electrons in the conduction band
will be approximately equal to the density of donor atoms, i.e. = (density of donor atoms)
Then = = [
�
ℎ
] .
(
� −��
��
)
Let [
�
ℎ
] = =
Then = .
(
� −��
��
)
= (
� −��
��
)
= −(
� −��
��
)
Taking log on both sides
ln [ ] = − (
−
)
− = − ln[ ]
= − ln[ ]
It shows that the Fermi level lies below the bottom of the conduction band, as shown in the figure.
In intrinsic semiconductor, Fermi level lies in the middle of the forbidden energy indicating equal
concentrations of free electrons and holes. When a donor type impurity is added to the crystal, then if we
assume that all the donor atoms are ionised, the donor electrons will occupy the states near the bottom
of the conduction band. Hence it will be more difficult for the electrons from the valance band to cross the
energy gap by thermal agitation. Consequently, the number of holes of the valance band is decreased.
Since Fermi level is a measure of the probability of occupancy of the allowed energy states, for n-type
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![Unit –V: Semiconductors
Page 14
semiconductors must move closer to the conduction band, as shown in the figure.
(ii) P-Type extrinsic semiconductor:
When an acceptor type impurity is added it also modifies the energy level diagram of semiconductor and
makes the conduction easier. The presence of impurity creates new energy levels which are in the gap in
the neighbourhood of the top of valence band of energies as shown in figure. Ambient temperature
results in ionisation of most acceptor atoms and thus an apparent movement of holes takes place from
the acceptor level to the valance band. The energies for holes are highest near the valance band decreases
vertically upward in the energy level diagram. Alternatively, one may say that electrons are accepted by
the acceptors and these electrons are supplied form the valance band, thus leaving a preponderance of
holes in the valance band.
If we assume that there are only acceptor atoms present and that these are all ionised, we have
ℎ = = [
� ℎ
ℎ
] .
(
��−�
��
)
=
�
(
��−�
��
)
Where � = [
��ℎ��
ℎ
]
/
=
�
= (
��−�
��
)
�
= −(
��−�
��
)
Taking log on both side,
ln [
�
] = − � − /
= � + ln[
�
]
It shows that the Fermi level lies above the top of the valance band.
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![Unit-7 Nuclear Physics
Page 6
radioactive decay , , � �, successfully. Moreover this model can also explain the nuclear
quadruple moment.
Un-success of the liquid drop model:
This model could not explain the reason for most stability of the lighter nuclei such as , , etc.
which have equal number of protons and neutrons. Similarly this model could not be successful to explain
the magic numbers.
Mass Defect:
The actual mass of the nucleus is less than the total
mass of protons and nucleons present inside it,
because some mass is lost in the form the binding
energy. This defect is called the mass defect and
denoted by Δ .
Δ = [ . � + − . �] −
Where
= Mass number of nuclei.
= Number of protons.
− = Number of neutrons
� = Mass of protons.
� = Mass of neutrons.
= Measured mass.
Curve between the mass defect and nuclear number:
Figure shows the variation of mass defect per nucleons with atomic numbers.
The mass defect per mass number in the nuclear is known as packing fraction and denoted by =
Δ�
�
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![Unit-7 Nuclear Physics
Page 7
Figure(2): Graph between packing fraction and atomic number
From the graph, we can say
1. Mass defect is positive for elements having mass number less than 20.3
2. Mass defect/ packing fraction is negative for elements having mass number between .
3. The mass defect is again positive for elements having mass number above 200.
The significance of packing fraction is that it is directly related to the availability and its stability of
nuclear.
Binding Energy:
The energy required to remove any nucleon (neutron or proton) from the nucleus is called binding energy
of nucleus.
The binding energy is the energy equivalent to the mass defect.
We know that the mass defect is given by
Δ = [ . � + − . �] −
Where all the symbols have their usual meaning.
Now the binding energy
= Δ
= { . � + − . � − }
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![Unit-7 Nuclear Physics
Page 8
Figure(3): Curve plotted between Binding energy per nucleon and mass number
The average binding energy per nucleon is found to be about for all elements.
Some important features from binding energy curves-
1. Binding energy of very light nuclei like is very small. The curve rise sharply with increase in
mass number and reaches a maximum value of . for . This makes iron most stable.
2. Binding energy decreases for element having > . It is found to be decreases to . for
uranium.
Semi-empirical mass formula:
The actual mass is slightly less than the calculated mass of the nucleus, because some mass is lost in the
form of the binding energy
This mass defect is given by
Δ =[ . � + − . �] −
Where the symbols have their usual meanings.
Now the binding energy
= Δ
= { . � + − . � − }
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![Unit-7 Nuclear Physics
Page 9
=
[ . � + − . �] −
On the basis of the liquid drop model the binding energy is considered to be the sum of independent
different energy terms and is expressed as
= + + + + … … … … .
1. Volume Energy:
For most of the nuclei, the binding energy per nucleon is constant
So
∝
∝
= � … … … …
2. Surface Energy:
∝ −
∝ −
= − ( )
= − … … … …
Negative sign shows that the binding energy decreases due to surface energy.
3. Coulo b’s e ergy:
The electrical repulsion between each pair of protons in the nucleus also contributes in decreasing its
binding energy. The coulomb energy is proportional to the number of proton pair [
−
!
] In the nucleus,
]and inversely proportional to the nuclear radius ∝ so the coulomb energy � = − [
−
] the
coulomb energy is negative, because if arises from an effect that opposes nuclear stability.
4. Asymmetric energy:
We know that the light nuclei having same number of neutron and proton are more stable but as mass
number increases, the number of neutron increases more as compare to the protons, due to which
the stability of nucleus decreases. Thus it is concluded that if the number of neutrons increases as
compare to protons the binding energy will decrease.
= −
= − −
= −
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![Unit-7 Nuclear Physics
Page 10
So = − .
−
…………….
5. Energy due to even odd effect:
The nucleus having even number of neutron and proton are more stable as compared to the nucleus
having odd number of neutron and proton while the nucleon having even number of proton and odd
number of neutrons or even number of protons and odd number of neutrons are found to be either less
stable or more stable. So
= ± �
− / …………… 6
Here (+) sign is for the even number of neutron and protons
(-) sign is for odd number of neutron and protons
And � = for even number of neutron and odd number of protons and vice-versa.
On combining all the above terms we get
= � + − � − [ � − − � −
−
+ ]
Isomerism:
The existance of the atomic nuclei those have same atomoic number and the same mass numbers but
different energy states.
Magic numbers:
The nuclei which have number of protons or number of nutrons equal to
, , , , , are relatively stable. These numbers are called the magic numbers.
The existence of magic numbers is established by the following facts.
i) There is abundance of nuclei in nature which have number of nucleons equal to magic no.
ii) From the binding energy curve it is clear that the nuclei = − =and
= − =are especially stable.
iii) The nuclei of even atomic number or atomic number more than 28 have isotopic abundance
more than % are only = − =, = − =and
= − =
iv) There are not more than 5 isotones of all the nuclei except for = and = .
v) = has ten stable isotopes and = has six stable isotopes.
vi) After = ,the ∝ − energy shows the discontinious behabiour.
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![Unit-7 Nuclear Physics
Page 13
nucleons thereby changing their identity equally fast and are responsible for keeping them bound
together.
Thus nuclear force between a proton and a neutron is the result of the exchange of charged mesons (�+
and �−
) between them. When a �+
meson jumps from a proton to a neutron, the proton is converted
into a neutron and vice versa.
− �+
→ or + �+
→
Similarly when a �−
meson jumps form a neutron to a proton, it is converted to a proton, and vice versa.
− �−
→ or + �−
→
In the same way, the forces between two protons and those between two neutrons arise due to the
exchange of neutral mesons between them. In this way nucleus is an ever-changing structure. It should be
remembered that numbers of protons and neutrons remains the same in the nucleus.
Nuclear Fission:
The phenomenon of braking of a heavy nucleus in to two or more light nuclei of almost equal masses
together with the release of a huge amount of energy is known as nuclear fission. The process of nuclear
fission was first discovered by the German scientists, Otto Hahn and Strassman, in the year 1939. In this
process, when uranium nucleus ( ) was bombarded with slow neutrons, this nucleus was found to split
up in to two radioactive nuclei which were identified as isotopes of Barium ( ) and Krypton ( ). It
is given by the following nuclear reaction-
+ → [ ] → + + +
.
Figure (5): Liquid drop model for nuclear fission
It is not that Barium and Krypton are only isotopes to be obtained by the fission of . Actually, this is a
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Physics engineering-first-year notes, books, e book pdf download
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- 3. Unit-1 Elasto-dynamics Page 1 Unit-1 Elasto-dynamics Syllabus: Simple Harmonic Motion, Electric Flux, displacement vector, Columb law, Gradient, Divergence, Curl, Gauss Theorem, Stokes theorem, Gauss law in dielectrics, Maxwell’s equation: Integral & Differential form in free space, isotropic dielectric medium. S t u d y n a m a .c o m
- 4. Unit-1 Elasto-dynamics Page 2 Periodic motion: If an object repeats its motion on a definite path after a regular time interval then such type of motion is called periodic motion. 1) Vibratory motion or oscillatory motion 2) Uniform circular motion 3) Simple harmonic motion Vibratory motion: If a body in periodic motion moves to and fro about a definite point on a single path, the motion of the body is said to be vibratory or oscillatory motion. Mean or equilibrium position: The point on either side of which the body vibrates is called the mean position or equilibrium position of the motion. Time period: The definite time after which the object repeats its motion, is called time period and it is denoted by . Frequency: The number of complete oscillation in one second is called the frequency of that body, it is represented by the letter or or � its unit is . Uniform circular motion: Figure(1): Uniform circular motion Let an object is moving on a circular path of radius with uniform angular velocity � = . S t u d y n a m a .c o m
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- 6. Unit-1 Elasto-dynamics Page 3 In right angle triangle Δ ∠ = � +� � = cos � +� � = cos � +� � = .� cos � +� � But � = so = .� cos� � +� � Similarly = sin � +� � = sin � +� � = .� sin � +� � = r.� sin� � +� � Both equation (1) and (2) represents the uniform circular motion. Simple (armonic Motion S(M : When a body moves periodically on a straight line on either side of a point, the motion is called the simple harmonic motion. S t u d y n a m a .c o m
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- 8. Unit-1 Elasto-dynamics Page 4 Graphical representation of SHM Figure(2): Graphical representation of SHM Displacement in SHM: Let a particle is moving on a circular path with uniform angular velocity " " and the radius of the circular path is " "; then movement of the point on their axis i.e. and is the SHM about the mean position Figure(3): SHM Let at time � =� the particle is on point and after time the position of the particle is then In Δ = sin S t u d y n a m a .c o m
- 9. Unit-1 Elasto-dynamics Page 5 = .� sin …………………………………………… (1) = .� sin� This equation represents the displacement of foot dropped from the position of particle on � −� . Velocity in SHM: Differentiating equation (1) with respect to we get- = .� sin� = rω cos� ……………. = √ � −� sin = √ −� sin = √ −� Using (1) (i) In equilibrium condition � =� So = √ −� = (ii) In the position of maximum displacement i.e. � =� So = √ −� = Acceleration: Again differentiating equation (2) we get- = � = rω cos� = −rω sin� S t u d y n a m a .c o m
- 10. Unit-1 Elasto-dynamics Page 6 = − ………………………………………………………… (3) +� = This is a second order differential equation which denotes the equation of SHM in the differential form Again by equation (3) = − Multiplying by i.e. the mass of the particle executing SHM then = − = − Here negative sing shows that the direction of displacement and acceleration are opposite to one another So ∝ − � =� Time period and frequency: = ⇒ = √ � = √ � = √ � = √ � And = √� Question: A uniform circular motion is given by the equation � =� � sin � +� ., find 1) Amplitude 2) Angular frequency 3) Time period 4) Phase Sol: Given: � =� � sin� +� . Comparing the given equation with the standard equation of uniform circular motion i.e. � = � sin � +� � S t u d y n a m a .c o m
- 12. Unit-1 Elasto-dynamics Page 7 We get � =� � =� � � = = =� . � = = =� . Question: A particle is moving with SHM in a straight line. When the displacement of the particle from equilibrium position has values and , the corresponding position has valocities and show that the time period of oscillation is given by � =� √ −� −� Sol: In the SHM the velocity is given by- = √ −� …………………………………… (1) At velocity is So = √ −� Squaring both sides = −� …………………………. (2) Again at the velocity is So = −� …………………………. (3) By equation (2) and (3) −� = −� � −� −� −� = −� −� −� = S t u d y n a m a .c o m
- 13. Unit-1 Elasto-dynamics Page 8 = −� −� = √ −� −� …………………………. (4) Now � = So = √ −� −� Question: If the earth were a homogeneous sphere and a straight hole was bored in it through the centre, then a body dropped in the hole, execute SHM. Calculate the time period of its vibration. Radius of the earth is . � ×� 6 and � =� . − Solution: The time period of oscillation executed by the body dropped in the hole along the diameter of earth � =� √=� √ . � ×� 6 . =� . Energy of a particle executing SHM: A particle executing SHM possess potential energy on the account of its position and kinetic energy on account of motion. Potential energy: We know that the acceleration in a simple harmonic motion is directly proportional to the displacement and its direction is towards the mean position = − Let is the mass of particle executing SHM then the force acting on the particle will be- = .� = − If the particle undergoes an infinitesimal displacement against the restoring force, then the small amount of work done against the restoring force is given by = − .� Here negative sign shows that the restoring force is acting the displacement than S t u d y n a m a .c o m
- 14. Unit-1 Elasto-dynamics Page 9 = So the total amount of work done = ∫� = This work done is equal to the potential energy of the particle at displacement i.e. = Kinetic energy: If is the velocity of the particle executing SHM, when the displacement is then kinetic energy = But for SHM � =� √ −� Where is the amplitude of SHM So = � √ −� ⇒ = −� ……………………………. (2) Total energy: Now the total energy = � +� ⇒ = + −� ⇒ = + − ⇒ = Thus we find that the total energy: 1) � ∝� 2) � ∝� of SHM 3) � ∝� of SHM Graphical representation of total energy of SHM S t u d y n a m a .c o m
- 16. Unit-1 Elasto-dynamics Page 10 Figure(4): Total energy of SHM Position vector: A position vector expresses the position of a point P in space in terms of a displacement from an arbitrary reference point O (typically the origin of a coordinate system). Namely, it indicates both the distance and direction of an imaginary motion along a straight line from the reference position to the actual position of the point. Displacement Vector: A displacement is the shortest distance from the initial to the final position of a point P. Thus, it is the length of an imaginary straight path, typically distinct from the path actually travelled by particle or object. A displacement vector represents the length and direction of this imaginary straight path. Figure(5): Displacement vector Area Vector: S t u d y n a m a .c o m
- 17. Unit-1 Elasto-dynamics Page 11 In many problems the area is treated as a vector, an area element is represented by ⃗⃗⃗⃗ , such that the area representing the area vector ⃗⃗⃗⃗ is perpendicular to the area element. The length of the vector ⃗⃗⃗⃗ represents the magnitude of the area element Figure(6): Area vector Coulomb’s Law: According to it the force of attraction or repulsion between the two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. If two charges and are separated at a distance form one another then the force between these charges will be- Figure(7): Two electric charges separated a distance r i) Force is proportional to the product of the magnitude of the charges i.e. � ∝� .� ii) The force is inversely proportional to the distance between the charges i.e. � ∝ So � ∝ .� � =� .� Where is a proportionality called electrostatic force constant, its value depends on the nature of the medium in which the two charges are located and also the system of units adopted to measure ,� and . So � =� . .� S t u d y n a m a .c o m
- 18. Unit-1 Elasto-dynamics Page 12 Case 1:(when the medium between the charges is air or vacuum ) As we know that the force between the charges is given as- � =� . .� If we put =� =� and � =� then � =� So is the force feels by two charges of placed apart from one another in vacuum or free space. Its value is � =� � ×�9 � ×� ×� Case 2:(When the medium between the charges is other than the vacuum) If the changes are located in any other medium then � = . =� � ×�9 . Where is the dielectric constant of relative permittivity. Putting this value in equation (1) we get ′ = . .� Where ′ is the force in the medium ′ = . .� Where � =� is called the relative permittivity of the medium. Vector form of the Coulomb’s Law Consider two like charges and present at and in vacuum at a distance apart. The two charges will exert equal repulsive force on each other, Let be the force on charge due to the charge and be the force on charge due to charge . According to the Coulo s’ la , the ag itude of fo e o ha ge and is given by | |. | | = . ………………………… (1) Let ̂ and ̂ are the unit vectors in the direction from to and vice versa. So the force is along the direction of unit vector ̂ , we have ⃗ = . .� ̂ S t u d y n a m a .c o m
- 19. Unit-1 Elasto-dynamics Page 13 And ⃗ = . .� ̂ These two equations show the Coulo s’ la i e to fo . Electric flux: Number of electric lines of forces passing normally through the surface, when held in the electric field. It is denoted by � . There are two types of electric flux- 1. Positive electric flux: When electric lines of forces leave any body through its surface it is considered as positive electric flux. 2. Negative electric flux: When lines of forces enter through any surface, it is considered as the negative electric flux. Measurement: Let us consider a small area ⃗⃗⃗⃗ of a closed surface . The electric field ⃗ produced due to the charge will be radially outwards which will be along ̂. Now the normal to the surface area is ⃗⃗⃗⃗ as shown in the figure, hence the angle between ⃗⃗⃗⃗ and ̂ is � So the electric lines of forces from the surface area will be given as- � = ⃗ .� ⃗⃗⃗⃗ � = � cos� �…………. Figure(8): Electric flux Where � cos� �is the component of electric field ⃗ along ⃗⃗⃗⃗ . Hence the electric flux through a small elementary surface area is equal to the product of the small area and normal component of ⃗⃗ along the direction of the elementary area⃗⃗⃗⃗⃗ . Over the hole surface, � = ∮ � cos� � � = ∮ ⃗ .� ⃗⃗⃗⃗ ………………………… (2) S t u d y n a m a .c o m
- 20. Unit-1 Elasto-dynamics Page 14 Gradient of a scalar field: The gradient of a scalar function � is a vector whose magnitude is equal to maximum rate of chcnge of scalar function � with respect to the space variable ∇⃗⃗ and has direction along that change. �� = � ̂ In the scalar field let there be two level surfaces and close together characterised by the scalar function � and �� +� � respectively. Consider point and on the level surfaces and respectively. Let and � +�⃗⃗⃗⃗ be the position vector of and . Then ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗ =� ̂ � +� ̂ � +�̂ Now as � is a function of ,� ,� i.e. �� =� � ,� ,� Then the total differentiation of this function can be given as Figure(9): Gradient of a scalar field � = � � + � � + � � = ( ̂ � +� ̂ � +� ̂ � )� .� ( ̂ � +� ̂ � +�̂ ) � = ∇⃗⃗ � ⃗⃗⃗⃗ …………………………………………………… (1) Agian if represents the distance along the normal from point to the surface to point , then = In the ∆ = cos� � = cos� � Now if we consider a unit vector along as ̂ then = ⃗⃗⃗⃗ .� ̂ …………………………………………………… (2) If we proceed form to then value of scalar function � increases by an amount � � = � � = � � ⃗⃗⃗⃗ .� ̂ [Usi g ……………………………. (3) S t u d y n a m a .c o m
- 21. Unit-1 Elasto-dynamics Page 15 By equation (1) and (2) (∇⃗⃗ .� �).� ⃗⃗⃗⃗ = � � ⃗⃗⃗⃗ .� ̂ (∇⃗⃗ .� �) = � � ̂ � = � � ̂ Note: ∇⃗⃗ =� ̂ +� ̂ +� ̂ is called del or Nabla operator. Note: � = ∇⃗⃗ .� � � = ( ̂ +� ̂ +� ̂ )� .� � � = ( ̂ � +� ̂ � +� ̂ � ) Note: The gradient of a scalar field has great significant in physics. The negative gradient of electric potential of electric field at a point represents the electric field at that point. i.e. ⃗ =� − Note: The gradient of a scalar field is a vector quantity. S t u d y n a m a .c o m
- 22. Unit-1 Elasto-dynamics Page 16 Divergence of a vector field: The divergence of a vector field at a certain point ,� ,� is defined as the outward flux of the vector field per unit volume enclosed through an infinitesimal closed surface surrounding the point " ". =� lim �→ .� ⃗⃗⃗⃗ � =� lim �→ � � Consider a infinitesimal rectangular box with sides Δ ,� Δ ,� Δand one corner at the point ,� ,� in the region of any vector function with rectangular faces perpendicular to co-ordinates axis. Figure(10): divergence of a vector field The flux emerging outwards from surface i.� e.� surface, � = ∬ ̅ .� � = ∬ ( ̂ ̅ +� ̂̅ +� ̂ ̅ ).� ̂Δ ,� Δ Where ̅̅̅ is the average of the vector function over thesurface i.e. surface � = ∬ ̅̅̅̅̅.� Δ .� Δ…………………………………………. (1) Similarly The flux emerging out from the surface i.e. surface ,� � = ∬ ̅ .� � = ∬ ( ̂ ̅ +� ̂̅ +� ̂ ̅ ).� − ̂Δ ,� Δ � = ∬ − ̅ .� Δ .� Δ………………………………………. (2) Thus net outwards flux of vector through the two faces perpendicular to � −axis, S t u d y n a m a .c o m
- 23. Unit-1 Elasto-dynamics Page 17 � = � +� � � = ∬ ̅ −� ̅ Δ .� Δ ……………….. (3) But ̅̅̅̅̅ −� ̅̅̅̅̅ = � +� Δ ,� ,�−� ,� ,� ̅̅̅̅̅ −� ̅̅̅̅̅ = Δ …………………………………………………… (4) Where �� is the variation of with distance along � −axis by equation (2) and (3) Thus net outward flux of vector function through the two faces perpendicular to � −axis � = �� Δ Δ ,� Δ [ Using equation (3) Similarly perpendicular to � −axis � = Δ Δ Δ Similarly perpendicular to � −axis � = Δ Δ Δ Therefore whole outward flux through infinitesimal box � = � +� �+� � � = + + Δ Δ Δ Now at any point, which is the flux enclosed per unit infinitesimal volume surrounding that point is given by- = lim Δ Δ Δ → � Δ Δ Δ = lim Δ Δ Δ → ( + + )� Δ Δ Δ Δ Δ Δ = + + = ( ̂ +� ̂ +� ̂ ) ( ̂ +� ̂ +� ̂ ) = ∇⃗⃗ .� Note: Divergence of a vector field is a scalar quantity. Note: If =� + S t u d y n a m a .c o m
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- 25. Unit-1 Elasto-dynamics Page 18 it indicates the existence of the source of fluid at that point. Note: If =� − It means fluid is flowing towards the point and thus there exist a sink for the fluid. Note: If =� It means the fluid is flowing continuously from that point. In other words this means that the flux of the vector function entering and leaving this region is equal. This condition is called solenoidal vector. S t u d y n a m a .c o m
- 26. Unit-1 Elasto-dynamics Page 19 Curl of a vector field: If is any vector field at any point and an infinitesimal test area at point then =� lim → ∮ .� ⃗⃗⃗⃗ ̂ Let us consider an infinitesimal rectangular area with sides Δ and Δ parallel to � −� plane in the region of vector function ⃗⃗ . Let the coordinate of be ,� ,� . If ,� ,� are the Cartesian components of at then ⃗⃗ =� ̂ +� ̂ +� ̂ Figure(11): Curl of a vector field Now the line integral of vector field along the path = ∫� .� ⃗⃗⃗⃗ = (̂̅ +� ̂̅ +� ̂̅ ).� ̂ Δ = ̅ Δ Where ̅ is the average value of � −component of the vector function over the path Similarly for the Path = ∫� .� ⃗⃗⃗⃗ = (̂̅ +� ̂̅ +� ̂̅ ).� − ̂ Δ = − ̅ Δ Where ̅ is the average value of � −component of vector function over the path . Hence the contribution to line integral ∮ ⃗⃗ . ⃗⃗⃗⃗ form two path and parallel to � −axis is = − = − − Δ As the rectangle is infinitesimal the difference between the average of .� .̅ − ̅ along these two paths may be approximated to the difference between the values of at and Thus- ̅ − ̅ = −� ̅ − ̅ = ,� � +� Δ ,�−� ,� ,� S t u d y n a m a .c o m
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- 28. Unit-1 Elasto-dynamics Page 20 ̅ − ̅ = Δ Hence the contribution to the line integral ∮ .� ⃗⃗⃗⃗ from the path and = Δ Δ …………………………………………… (2) Similarly by the path and = Δ Δ …………………………………………… (3) Therefore the line integral along the whole rectangular form (2) and (3) is given by- = ∮� +� .� ⃗⃗⃗⃗ = ∮� .� ⃗⃗⃗⃗ = − Δ Δ ……………………………… (4) Now = lim Δ Δ → = lim Δ Δ → ( − )� Δ Δ Δ Δ = − ……………………………………. (5) Similarly = ( − ) ……………………………………. (6) and = − ……………………………………. (7) Summing up the results given in (5), (6) and (7) we get = ̂ +� ̂ +� ̂ = ̂� ( − )� +� ̂� (− )� +�̂ ( − ) = [ ̂ ̂ ̂ ] = ∇⃗⃗ ×� Note: The curl of a vector field is sometime called circulation or rotation or simply . S t u d y n a m a .c o m
- 29. Unit-1 Elasto-dynamics Page 21 Note: If =� then vector field is called Lamellar field. S t u d y n a m a .c o m
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- 31. Unit-1 Elasto-dynamics Page 22 Gauss’ Divergence Theorem: According to this theorem the volume integral of divergence of a vector field over a volume is equal to the surface integral of that vector field taken over the surface which enclosed that volume . i.e. ∭( ) � =� ∬�.� ⃗⃗⃗⃗ � Consider a volume enclosed by a surface this volume can be divided into small elements of volumes ,� …� …�� which are enclosed by the elementary surface ,� …� …� …� …�� respectively. By definition the flux of a vector field diverging out of the ℎ element is Figure(12): Gauss’ Di e ge e tho e ( )� = .� � ⃗⃗⃗⃗⃗⃗ � � ( )� .� � = ∬� .� ⃗⃗⃗⃗ � ………………………………………………… (1) On LHS of equation we add the quantity ( )� .� � for each element ,� …� …�� ∑( )� .� � � �= = ∭( ) � On RHS of equation (1) if we add the quantity .� ⃗⃗⃗⃗ � for each ,� …� …� …� …�� we get the terms only on the outer surface Sum comes out to be ∑� ∬�.� � ⃗⃗⃗⃗⃗⃗ � � �= = ∬� .� ⃗⃗⃗⃗ So putting these values in equation (1) we get So ∭( ) � = ∬� .� ⃗⃗⃗⃗ This is the Gauss’ di e ge e theo e . S t u d y n a m a .c o m
- 32. Unit-1 Elasto-dynamics Page 23 Stokes theorem: According to this theorem, the line integral of a vector field along the boundary of a closed curve is equal to the surface integral of curl of that vector field when the surface integration is done over a surface enclosed by the boundary i.e. ∮� .� ⃗⃗⃗ =� ∬� .� ⃗⃗⃗⃗ Figure(13): Stokes theorem Consider a vector which is a function of position. We are to find the line integral ∮ .� ⃗⃗⃗ along the boundary of a closed curve . If we divide the area enclosed by the curve in two parts by a line , we get two closed curve and . The line integral of vector along the boundary of will be equal to the sum of integral of along and ∮� .� ⃗⃗⃗ = ∮� .� ⃗⃗⃗ +� ∮�.� ⃗⃗⃗ Similarly if we divide the area enclosed by the curve in small element of area …� …� …� …by the curve ,� …� …� …� ….As shown in the figure. Then the sum of line integrals along the boundary of these curves ,� …� …� …� ..(taken anticlockwise) will be ∮� .� ⃗⃗⃗ = ∑� ∮�.� ⃗⃗⃗ By the definition of curl, we have S t u d y n a m a .c o m
- 33. Unit-1 Elasto-dynamics Page 24 = ∮ .� ⃗⃗⃗ � .� � = ∮� .� ⃗⃗⃗ ∮� .� ⃗⃗⃗ = ∑� .� � ⃗⃗⃗⃗⃗⃗⃗ =� ∬� .� ⃗⃗⃗⃗ ∮� .� ⃗⃗⃗ = ∬� .� ⃗⃗⃗⃗ Gauss Law According to this law, the net electric flux through any closed surface is times of the total charge present inside it. � = ………………………… (1) But by the definition of electric flux ⇒ � = ∬� ⃗ .� ⃗⃗⃗⃗ …� …� …� …� …� …� …� …� …� …(2) So by equation (1) and (2) so ∬� ⃗ .� ⃗⃗⃗⃗ = This is the i teg al fo of Gauss’ la . Proof: Case1: When the charge lies inside the arbitrary closed surface. Let charge lies inside the arbitrary surface at point Now let us consider an infinitesimal area ⃗⃗⃗⃗ on this surface which contain the point , the direction of the area vector ⃗⃗⃗⃗ is perpendicular to the surface and electric field Figure(14): Gauss Law S t u d y n a m a .c o m
- 34. Studynama’s MBA Community is one of India’s Largest Community of MBA & PGDM Students. About 29,882 Indian Management students are members of this community and share FREE study material, notes, eBooks, projects, case studies exam papers etc. to enable each other to do well in their semester exams. Links to Popular Study Material for Management (MBA/PGDM) students: MBA General MBA Marketing MBA Finance MBA HR MBA Operations MBA/PGDM first year notes Enterprise resource planning (ERP) pdf Security analysis & portfolio Mgmt. Business environment Notes MBA Operations Notes MBA/PGDM second year notes Marketing management pdf International Fin. management eBook Human resource management pdf MIS pdf Principles of management pdf International marketing eBook Advanced financial management eBook Compensation management Notes Industrial safety eBook Managerial economics eBook Retail marketing eBook Corporate taxation pdf eBook International human resource management pdf Import export mgmt. lecture notes Organizational behavior pdf Sales management eBook Financial management eBook Human resource development eBook TQM eBook Operation research lecture notes Brand management pdf Management accounting notes Organization & management process pdf Management control systems And 12,998 more free downloads for MBA & PGDM Students. Other Popular Links for MBA Study Material: • MBA/PGDM Core (General) Notes, eBooks, Projects, Cases & Reports FREE PDF Download • MBA/PGDM Marketing Notes, eBooks, Projects, Cases & Reports FREE PDF Download • MBA/PGDM Finance Notes, eBooks, Projects, Cases & Reports FREE PDF Download • MBA/PGDM HR Notes, eBooks, Projects, Cases & Reports FREE PDF Download • MBA/PGDM Operations Notes, eBooks, Projects, Cases & Reports FREE PDF Download
- 35. Unit-1 Elasto-dynamics Page 25 ⃗ makes an angle � with ⃗⃗⃗⃗ then electric field will be given as- ⃗ = ……………………………………. (3) Now the flux emerging out of the surface area ⃗⃗⃗⃗ will be � = ⃗ .� ⃗⃗⃗⃗ ⇒ � = � cos� � Where � is the angle between ⃗ and ⃗⃗⃗⃗ So putting the value of ⃗ we get � = � cos� � ⇒ � = � cos� � But � cos� � =� i.e. solid angle � = Now total flux � = ∫ ⇒ � = ∫� But � =� � = ⇒ � = Case 2: When the charge lies outside the closed surface then the flux entering and leaving the surface S t u d y n a m a .c o m
- 36. Unit-1 Elasto-dynamics Page 26 area will be equal and opposite then �� =� Gauss law in the differential form (Poisson’s equation and Laplace’s equation If the charge is continuous distributed over the volume and charge density is then = ∭� � Now by Gauss theorem the flux emerging out of this surface which enclosed volume ∬� ⃗ .� ⃗⃗⃗⃗ = ∭� � …� …� …� …� …� …� …� …� …� …(1) By Gauss divergence theorem ∬� ⃗ .� ⃗⃗⃗⃗ = ∭� ⃗ � …� …� …� …� …� …� …� …� …� …(2) By equation (1) and (2) ⇒ ∭� ⃗ � = ∭� � ⇒ ∭� ( ⃗ − ) � = But as we know that � ≠� So ⃗ − = ⇒ ⃗ = …� …� …� …� …� …� …� …� …� …(3) This is the diffe e tial fo of Gauss’ la a d also k o as Poisso ’s e uatio Now if we consider the charge less volume then � =� So ⃗ = …� …� …� …� …� …� …� …� …� …(4) This equation is Laplace equation. Again by equation (3) S t u d y n a m a .c o m
- 37. Unit-1 Elasto-dynamics Page 27 ⃗ = We know that ⃗ =� − So − = ⇒ −∇⃗⃗ .� ∇⃗⃗ = ⇒ ∇ = − ⇒ + + = − Gauss law in Presence of dielectrics : The Gauss’ la elates the ele t i flu a d ha ge. The theo e states that the et ele t i flu a oss a arbitrary closed surface drown in an electric field is equal to times the total charge enclosed by the surface. Now we want to extend this theorem for a region containing free charge embedded in dielectric. In figure the dotted surface in an imaginary closed surface drown in a dielectric medium. There is certain amount of free charge in the volume bounded by surface. Let us assume that free charge exists on the surface of three conductors in amount ,� ,� …� ..In a dielectric there also exits certain amount of polarisation (bound) charge . He e Gauss’ theo e ∬� ⃗ .� ⃗⃗⃗⃗ = ( ′ +� ) ………………. Where � =� +� +� is the total free charge and is the polarisation (bound) charge by = ∬ ⃗ .� ⃗⃗⃗⃗ + + +� ∭ − � ………………. Here is the volume of the dielectric enclosed by . As there is no boundary of the dielectric at , therefore the surface integral in equation (2) does not contain a contribution from . If we transform volume integral in (2) into surface integral by means of Gauss divergence theorem, we must include contribution from all surface bounding , namely ,� ,� and i..e. ∫ − � = [ ∬ ⃗ .� ⃗⃗⃗⃗ + + +� ∭ − � ] S t u d y n a m a .c o m
- 38. Unit-1 Elasto-dynamics Page 28 Using above equation, we note that = ∬ ⃗ .� ⃗⃗⃗⃗ + + ………………. Substituting this value in (1) We get ∬� ⃗ .� ⃗⃗⃗⃗ = � − ∬� ⃗ .� ⃗⃗⃗⃗ ∬ ⃗ + ⃗ .� ⃗⃗⃗⃗ = Multiplying through by ∬( ⃗ +� ⃗ ).� ⃗⃗⃗⃗ = ………………. This equation states that the flux of the vector ⃗ +� ⃗ through a closed surface is equal to the total free charge enclosed by the surface. This vector quantity is named as electric displacement ⃗⃗ i.e. ⃗⃗ = ⃗ +� ⃗ …………………….. Evidently electric displacement ⃗⃗ has the same unit as ⃗ . i.e. charge per unit area. In terms of electric displacement vector ⃗⃗ , equation (4) becomes ∬� ⃗⃗ .� ⃗⃗⃗⃗ = …………………….. i.e. the flux of electric displacement vector across an arbitrary closed surface is equal to the total free charge enclosed by the surface. This esult is usuall efe ed to as Gauss’ theo e i diele t i . If e o side i to a la ge u e of i fi itesi al olu e ele e ts, the Gauss’ theo e a expressed as ∬� ⃗⃗ .� ⃗⃗⃗⃗ = ∭� � …………………….. Where is the charge density at a point within volume element such that � →�. ∭� ⃗⃗ .� � = ∭� � ∭ ⃗⃗ −� .� � = Volume is arbitrary, therefor we get ⃗⃗ −� = S t u d y n a m a .c o m
- 39. Unit-1 Elasto-dynamics Page 29 ⃗⃗ = This result is called differential fo of Gauss’ theo e i a diele t i . The main advantage of this method is that the total electrostatic field at each point in the dielectric medium may be expressed as the sum of parts ,� ,� = ⃗⃗ ,� ,� − ,� ,� …………………..…. Where the first term ⃗⃗ is related to free charge density through the divergence and the second theorem is proportional to the polarisation of the medium. In vacuum � =� so ⃗ = ⃗⃗ Electric Polarization � When a dielectric is placed in any external electric field then the dielectric gets polarized and induced electric dipole moment is produced which is proportional to the external applied electric field. Now if there are number of dipoles induced in per unit volume of dielectric then total polarization will be- ⃗ = �� ⃗⃗⃗⃗⃗ …� …� …� …� …� …� …� …� …� …(1) But �� ⃗⃗⃗⃗⃗ ∝ ⃗⃗⃗⃗ So �� ⃗⃗⃗⃗⃗ = ��� ⃗⃗⃗⃗ Putting this value in equation (1) we get ⇒ ⃗ = ��� ⃗⃗⃗⃗ It is clear from the above equation that the direction of polarization is in the direction of the applied external electric field. And the unit is / Electric displacement We know that the value of electric field depends on the nature of the material, so to study the dielectric we need such a quantity which does not depends on the nature of the material and this quantity is known as electric displacement vector ⃗⃗ . S t u d y n a m a .c o m
- 40. Unit-1 Elasto-dynamics Page 30 Both ⃗ and ⃗⃗ are same except that we define ⃗ by a force in a charge placed at a point while the displacement vector is measure by the displacement flux per unit area at that point. ∭� ⃗⃗ .� ⃗⃗⃗⃗ = Or = ⇒ = � Where � is the surface charge density. Relation between ⃗⃗ and ⃗⃗ We know that the Gauss law is given as- ∬� ⃗ .� ⃗⃗⃗⃗ = Where is the permittivity of the dielectric medium ⇒ ⃗ = . But � =� so we have ⃗ = ⃗⃗ ⇒� ⃗⃗ =� ⃗ ⇒� ⃗⃗ =� ⃗⃗ � � =� Where is the permittivity of the free space Current: Current for study current � = S t u d y n a m a .c o m
- 41. Unit-1 Elasto-dynamics Page 31 If the charge passing per unit time is not constant, then the current at any instant will be given as � = Current density: = ⃗⃗⃗⃗ = .� ⃗⃗⃗⃗ = ∫� .� ⃗⃗⃗⃗ = From the above equation we can see that the current is the flux of current density as �� =� ∫�⃗ ⃗⃗⃗⃗ Its SI unit is S t u d y n a m a .c o m
- 42. Unit-1 Elasto-dynamics Page 32 Equation of continuity: The law of conservation of charge is called the equation of the continuity. � =� ∬�.� ⃗⃗⃗⃗ For steady current charge does not stay at any place, so the current will be constant for all the places. Figure(17): Flux of current ⇒ = ∬� .� ⃗⃗⃗⃗ =� By divergence theorem ⇒ ∬� .� ⃗⃗⃗⃗ = ∭� .� � So ∭� .� � = On differentiating we get = This is the equation of continuity for study current. Now if current is not stationary i.e. if the current is the function of the time and position then = ∬� .� ⃗⃗⃗⃗ =� − Here negative sign shows that the charge is reduced with respect to time. But if is the charge per unit volume then- = ∭� .� � So ∬� .� ⃗⃗⃗⃗ = − ∭� .� � ⇒ ∭� .� � = − ∭� .� � S t u d y n a m a .c o m
- 43. Unit-1 Elasto-dynamics Page 33 ⇒ ∭� ( + )� � = ⇒ = − This is the equation of continuity for time varying current. Maxwell’s equations James Clerk Maxwell took a set of known experimental laws (Faraday's Law, Ampere's Law) and unified them into a symmetric coherent set of Equations known as Maxwell's Equations. These equations are nothing but the relation between electric field and magnetic field in terms of divergence and curl. S.N. Name Integral form Differential form 1 Gauss’ La fo electricity ∬� ⃗ .� ⃗⃗⃗⃗ = ∭� � ⃗ = 2 Gauss’ la fo magnetism ∬� ⃗ .� ⃗⃗⃗⃗ =� ⃗ =� 3 Fa ada ’s La of induction ∮� ⃗ .� ⃗⃗⃗ = ∬� ⃗ .� ⃗⃗⃗⃗ ⃗ =� − ⃗⃗ 4 A pe e’s la ∮� ⃗ .� ⃗⃗⃗ =� ∬�⃗⃗⃗ .� ⃗⃗⃗⃗ + ∬� ⃗⃗ .� ⃗⃗⃗⃗ ⃗ =� � +� ⃗ Maxwell’s first equation Gauss’ law in electric): Let us consider a volume which is enclosed in a surface , the Gauss’ la the ele t i flu is given as ∬� ⃗ .� ⃗⃗⃗⃗ = � …� …� …� …� …� …� …� …� …� …(1) Where is the totat charge enclosed in the volume Now if is the volume charge density then = ∭� � …� …� …� …� …� …� …� …� …� … (2) By equation (1) and (2) S t u d y n a m a .c o m
- 44. Unit-1 Elasto-dynamics Page 34 ⇒ ∬� ⃗ .� ⃗⃗⃗⃗ = ∭� � This is the i teg al fo of Ma ell’s e uatio . B Gauss’ di e ge e theo e ⇒ ∬� ⃗ .� ⃗⃗⃗⃗ = ∭� ⃗ � So by applying this on above equation we get ⇒ ∭� ⃗ � = ∭� � ⇒ ∭� ( ⃗ − )� � = But � ≠� so ⇒ ⃗ − = ⇒ ⃗ = ⇒ ⃗ = ⇒ ⃗⃗ = [ � ⃗⃗ =� ⃗ Maxwell’s second equation Gauss’ law in magnetism : Since the magnetic lines of forces are closed curves so the magnetic flux entering any orbitri surface should be equal to leaving it mathematically ⇒ ∬� ⃗ .� ⃗⃗⃗⃗ = � …� …� …� …� …� …� …� …� …�(1) This is i teg al fo of Ma ell’s se o d e uatio . No Gauss’ di e ge e theo e ⇒ ∬� ⃗ .� ⃗⃗⃗⃗ = ∭� ⃗ � So equation (1) can be written as- ⇒ ∭� ⃗ � = S t u d y n a m a .c o m
- 45. Unit-1 Elasto-dynamics Page 35 As � ≠� so ⇒ ⃗ = Maxwell’s third equation Faraday’s law : A o di g to Fa ada ’s la of ele t o ag eti i du tio if the ag eti flu li ked ith a losed circuit changes with time then a is induced in the close circuit which is known as induced the direction of the induced will be such as it oppose the change in the magnetic flux. It is given as ⇒ = − � …� …� …� …� …� …� …� …� …� …(1) But Gauss’ theo e e k o that ⇒ � = ∬� ⃗ . ⃗⃗⃗ So = − ∬� ⃗ ⃗⃗⃗⃗ Now if ⃗ is the electric field produced due to the change in the magnetic flux then the induced will be equal to the line integral of ⃗ along the circuit. i.e. ⇒ = ∮ ⃗ .� ⃗⃗⃗ …� …� …� …� …� …� …� …� …� …(2) By equation (1) and (2) ⇒ ∮� ⃗ .� ⃗⃗⃗ = − ∬� ⃗ ⃗⃗⃗⃗ ⇒ ∮� ⃗ .� ⃗⃗⃗ = −� ∬ ⃗ ⃗⃗⃗⃗ …� …� …� …� …� …� …� …� …� …(3) No Stokes’ theo e ⇒ ∮� ⃗ .� ⃗⃗⃗ = ∬� ⃗ .� ⃗⃗⃗⃗ S t u d y n a m a .c o m
- 46. Unit-1 Elasto-dynamics Page 36 Applying this to the above equation, we get ∬� ⃗ .� ⃗⃗⃗⃗ = −� ∬ ⃗ ⃗⃗⃗⃗ ⇒ ∬� ⃗ + ⃗ � ⃗⃗⃗⃗ = As ⃗⃗⃗⃗ ≠� So ⃗ + ⃗ = ⇒ ⃗ = − ⃗ Maxwell’s fourth equation Maxwell’s correction in Ampere’s law A pe e’s La is gi e as ⇒ ⃗ = This equation is true only for time independent electric field and to correct this equation for time varying field a term must be added ⇒ ⃗ = ( +� ) …� …� …� …� …� …� …� …� …� …(1) Taking of both side and for simplicity writing as ⇒ ( ⃗ ) = ( +� ) But divergence of curl of any quantity is always zero so ( ⃗ )� =� Then ( +� ) = ………………………………………. (2) ⇒ = − …� …� …� …� …� …� …� …� …� …(3) But by the equation of continuity ⇒ = − …� …� …� …� …� …� …� …� …� …(4) A d Ma ell’s fi st e uatio S t u d y n a m a .c o m
- 47. Unit-1 Elasto-dynamics Page 37 ⃗ = ⇒ = ⃗ …� …� …� …� …� …� …� …� …� …(5) By (4) and (5) ⇒ = − ( ⃗ ) ⇒ = − ( ⃗⃗ ) …� …� …� …� …� …� …� …� …� …(6) Again by (3) and (6) ⇒ − = − ( ⃗⃗ ) ⇒ = ( ⃗⃗ ) ⇒ = � ( ⃗⃗ ) ⇒ = ⃗⃗ Putti g this alue i A pe e’s la e get ⃗ =� + ⃗⃗ This is Ma ell’s fou th e uatio . For vacuum ⃗ =� and � =� So ⃗⃗ = +� ⃗ ⇒ ⃗⃗ = +� ⃗S t u d y n a m a .c o m
- 49. Unit-II (LASER) Unit-2 LASER Syllabus: Properties of lasers, types of lasers, derivation of Einstein A & B Coefficients, Working He-Ne and Ruby lasers. S t u d y n a m a .c o m
- 50. Unit-II (LASER) LASER: The word LASER is acronym for light amplification by stimulated emission of radiation. Laser source produces coherent, monochromatic, least divergent, unidirectional and high intense beam. Einstein gave the theoretical description of stimulated emission in 1917. In 1954 G.H. Towne developed microwave amplifier MASER using Einstein’s theo y a d put forward to light and first Laser was developed. Characteristics of Laser beam: i) Coherent: The Laser light is coherent. A Laser emits the light waves of same wavelength and in same phase. ii) Monochromatic: If the light coming from a source has only one frequency or single wavelength is called monochromatic source and the light is called monochromatic light. In case of Laser beam it has the wavelength confirmed to very narrow range of a few angstroms. iii) Divergence: Divergence is the measure of its spread with distance. The angular spread in ordinary light is very high because of its propagation in the form of a spherical wave front. The divergence in the Laser beam is negligible. A very small divergence is due to the diffraction of Laser light when it emerges out from the partially silvered mirror. iv) Directionality: An ordinary source of light emits light in all directions. In case of Laser the photons of a particular direction are only allowed to escape. Thus the Laser beam is highly directional. v) Intensity: The intensity of ordinary light decreases as it travels in the space. This is because of its spreading. The Laser does not spread with distance. It propagates in the space in the form of narrow beam and its intensity remains almost constant over long distance. Three Quantum Process: 1. Induced absorption: When an atom gains some energy by any mean in the ground state, the electrons of the atoms absorbs some energy and are excited to high energy level. Let us consider two energy levels � and � of an atom. Suppose this atom is expose to light radiation it can excite the atom from ground state � to the high energy state � by absorbing a photon of frequency �. The frequency � is given as � = � − � ℎ This process is called the induced absorption. Pictorially it is represented as in figure(1) S t u d y n a m a .c o m
- 51. Unit-II (LASER) Figure(1):Induced absorption This may also be shown by the equation + ℎ� ⟶∗ [* represents the exited state ] The probability of absorption transition is given by ∝ � = � Where � is the energy state density And the number of absorption transition in material is equal to the product of number of atoms at � and absoption transition is given as = = � Where is the number of atoms in ground state � 2. Spontaneous Emission: When an atom at lower energy level is exited to the high energy level, it cannot stay in the exited state for relatively longer time. In a time of about −8 ��, the atom reverts to the lower energy state by releasing a photon of energy ℎ� = �− �. This emission of photon by an atom without any external input is called spontaneous emission. Figure(2): Spontaneous emission We may write the transition as ∗ ⟶ + ℎ� Probability of spontaneous emission depends only on the properties of energy states and is depends on the S t u d y n a m a .c o m
- 52. Unit-II (LASER) photon density. It is equal to the life time of � = is Ei stei ’s oeffi ie t fo spo te ious e issio . Number of spontaneous transition depends only on number of atoms in the excited state � . Thus = Process of spontaneous emission cannot be controlled from outside and photon are emitted in random order so light is non-directional, non-monochromatic, incoherent and no amplification of light takes place. 3. Stimulated Emission: In 1916 Einstein predicted the existence of stimulated emission. A photon of appropriate energy when incidents to an atom which is in the exited state, then it may causes the de- excitation by the emission of an additional photon of same frequency as that of incident one, now the two photons of same frequency moves together. This process is called the stimulated or induced emission. The emitted photon have same direction, phase, energy and state of polarization as that of incident photon we can rewrite the transition as ∗ + ℎ� ⟶ + ℎ� The probability of stimulated emission is given by � ∝ � � = � is Ei stei ’s oeffi ie t of sti ulated e issio . The number of stimulated transition in a material is given by � = � Where is the number of atoms in excited state � The light produced by this process is essentially directional, monochromatic, and coherent, the outstanding feature of this process is the multiplication of photons i.e. if there are exited atoms, photons will be produced. Figure (3): Multiplication of stimulated photons into an avalanche Population Inversion: In the thermal equilibrium number of atoms in higher energy levels is less than population of lower energy S t u d y n a m a .c o m
- 53. Unit-II (LASER) level . Then if � and � are two energy levels with population and then by Boltzmann relatin. = × ( � − � ) Since � > �⟹ < . In this situation the system absorbes electromagnetic radiation incident on it for laser action to take place, the higher energy level should be more populated as compared to the lower energy state i.e. > . Thus the process by which the population of a particular high energy state is made more than that of a specified lower energy state is called population inversion. Figure(4): Population inversion Meta stable States: An atom in the exited state has very short life time which is of the order of −8 ��. Therefore even if continuous energy is given to the atoms in ground state to transfer them to exited state they immediately comes back to the ground state. Thus population inversion cannot be achieved. To achieve population inversion, we Figure(5): must have energy states which has a longer lifetime. The life time of meta stable state is − to − �� which is time of exited states thus allows accumulation of large number of excited atoms and result in population inversion. Components of Laser the essential components of Laser are- S t u d y n a m a .c o m
- 54. Unit-II (LASER) Active Medium: The active medium is the material in which the laser action takes place. It may be solid, liquid, or gas. The medium determines the wavelength of the laser radiation. Atoms are characterized by the large number of the energy levels, but all types of atoms are not suitable for Laser operation. Even in a medium consisting of different species of atoms, only a fraction of atoms of particular type have energy level system suitable for achieving population inversion. Such atoms can produce more stimulated emission than spontaneous emission causes amplification of light. These atoms are called active center. The rest of the medium acts as the host medium and supports the active medium. Thus the active medium is the one which when excites, reaches the state of population inversion and promotes stimulated emission leading to light amplification. Figure(6): Component of LASER Optical Resonator: It is specially designed cylindrical tube having two opposite optically plane mirrors with active medium filled between them, one mirror is fully silvered and other is partially silvered and are normal to the light intensity by multiple reflection. Science active medium is maintained in population inversion state photon produced through spontaneous emission produces the stimulated emission in every direction. The photons having parallel to the axis of the resonators are only augmented while other are reflected trough the walls of resonator. If these unidirectional photons reach fully reflecting mirror they reflects and while transverse through the medium they produce the stimulated emission in other atoms thus increased stimulated photons reaches partially silvered mirror. At this end some photons are transmitted and other are reflects back in the medium. This process repeats itself again and again. Working of optical resonator: a) Non-exited medium before pumping. b) Optical pumping. S t u d y n a m a .c o m
- 55. Unit-II (LASER) c) Spontaneous/stimulated emission. d) Optical feedback. e) Light amplification. f) Light oscillation and laser output. Pumping: The process by which we can achieve the population inversion is called the pumping. Pumping Schemes: Figure(7): Pumping scheme Two level pumping scheme: A two level pumping scheme is not suitable for obtaining population inversion. The time span ∆ , for which atom has to stay at upper level � , must be longer for achieving population inversion condition. As according to the Heise e g’s uncertainty principle ∆�. ∆ ≥ ℏ Figure (8): Two level pumping scheme. ∆ will be longer if ∆� is smaller i.e. � is narrow. If ∆� is smaller, the pumping efficiency is smaller as a consequence of which less number of atoms are exited and population inversion is not achieved. Three level pumping: Let an atomic system has three energy levels, the state � is the ground state, � is the metastable state and � is the energy excited state. When light is incident, the atom are rapidly exited to upper most state � . They Pumping Scheme Two Level Pumping Scheme Three Level Pumping Scheme Four Level Pumping Scheme S t u d y n a m a .c o m
- 56. Unit-II (LASER) comes back in the lower energy level. The atom does not stay at the � level for long time. The probability of spontaneous transition � ⟶ �is comparable to � ⟶ �, � is the metastable state. Science probability of � ⟶ � transition is extremely small when the medium is expose to a large number of photons a large number of atoms will be exited to the higher energy level � . Some of these atoms make spontaneous transition to the � state trough the radiative transition. As the spontaneous transition from � to � occurs rarely. The atoms get trapped into the state � . This process continues when more than half of the ground state atoms accumulate at � , the population inversion is achieved between � and � . Figure(9): Three Level Pumping In this scheme a very useful-pumping process is required because to achieve population inversion more than half of ground state atoms must be pumped to the upper state. Four Level Pumping: In four level pumping process the active medium are pumped from ground state � to the uppermost level � from where they rapidly fall to intermediate � level i.e. meta stable state, leaving level � empty. Now � is populated inversely with respect to � . If a triggering incident beam has frequency � the transition � ⟶ �is the stimulated transition. It could be the atom from � may go to � sponteniously. This transition � ⟶ �is non radiative transition. S t u d y n a m a .c o m
- 57. Unit-II (LASER) Figure(10): Four level pumping S t u d y n a m a .c o m
- 58. Unit-II (LASER) Einstein’s coefficient: Let there is a lasering medium in which the number of atoms in the ground state are and number of atoms in the excited state are , � is the energy density of radiation for frequency �. The rate of self-emission ∝ = And the rate of stimulated emission ∝ And ∝ � ∴ ∝ � = � The rate of absorption ∝ And ∝ � = � Here, coefficient and and are respectively called the Ei stei ’s and coefficients. It is clear that the rate of stimulated emission and rate of absorption determined by the same coefficient . This is why simulated emission is also called the inverse absorption. Relation between Einstein’s and coefficient: Let there be an assembly of atoms in thermal equilibrium at temperature T with radiation frequency �. Since the rate of absoption of radiation i.e. transition from state � ⟶ �is proportional to the energy desity of radiation � . The number of transition per unit time per unit volume from � → �is given by = � Where is the number of atoms in energy state � and is the probability of the transition from � → � Similarly the number of transition per unit time per unit volume from state � → �may be given as = { + � } Where is the number of atpms in energy state � and is the probability of the transition from � → � In equilibrium state = � = { + � } � = + � � − � = [ − ] � = � = � � −� S t u d y n a m a .c o m
- 59. Unit-II (LASER) � = [ � � � � −� ] � = [� � � � − � � ] � = [� � � � − ] By Boltzmann distribution law = �− � / � = �− � / � Where is the total number of atoms � � = � � −� / � Then � = [ { � −� �� } � � − ] But � − �= ℎ� So � = [ { ℎ� �� } � � − ] But a o di g to the Pla k’s the e e gy de sity of the adiatio of f e ue y � at temperature T is given by � = 8�ℎ� . [ ( ℎ�/��)− ] On comparing = �ℎ� � and = S t u d y n a m a .c o m
- 60. Unit-II (LASER) Ruby Laser: Solid state laser is the first laser operated successfully. It was fabricated by Mainman in 1960. Ruby is the lasing medium consist of the crystal of mixture and the . Here some aluminum atoms are replaced by the . % cromiume atoms. Construction: Chromium atoms doped into the aluminum atoms. The active medium in ruby with which main laser action takes place is + ions. Length of the cylindrical rod lies in between 2 to 20cm and the diameter of the rod is about 0.1 to 2cm. The end faces of the rod are polished flat and parallel. In this one face is partially silvered and other face is fully silvered. Ruby rod is surrounded by the helical Xenon photo flash lamp which provides the pump energy to rise the chromium atom to higher energy level. The parallel ends rod forms an optical cavity so that the photon traveling along the axis of the optical cavity gets reflects back and fro the end surfaces. Working: The energy level of + ions on the crystal lattice. Consists of three level systems. Upper energy level is short lived state. Figure(11): Figure(12) When a flash light falls upon the ruby rod, the Å radiation photon are absorbed by + ions which are S t u d y n a m a .c o m
- 61. Unit-II (LASER) pumped to the exited state � . The transition from � to � is the optical pumping transition. Now the + ions in the exited state give a part of their energy to the crystal lattice and decay to the meta stable state � . Hence the transition from � to � is radiation less transition. Metastable state � is long lived state; hence the number of + ions goes on increasing, while due to pumping the number in the ground state � goes on decreasing. Population inversion is established between the � and � . The spontaneous photon emitted by + ion at � level is of the wave length of about Å. Drawbacks : 1) Efficiency of ruby Laser is very low. 2) The Laser Output is not continuous occurs in the form of pulse of microseconds duration. 3) The Laser requires the high pumping power. 4) The defects due to crystalline impurities are also presents in the laser. Figure(13): Ruby Laser output S t u d y n a m a .c o m
- 62. Unit-II (LASER) Gas Laser: Gas Lasers are most widely used Lasers. The ranges from low power Lasers like Helium-Neon Laser to high power Laser like laser. These lasers operate with rarified gases as the active medium and are excited by and electrical discharge. In gases the energy levels of the atoms involves the lasing process are narrow and as such require sources with sharp wavelength to excite atoms. Most common method to excite gas molecules is by passing an electric discharge through the gas electrons present in the discharge through the gas electrons presents in the discharge transfer energy to atoms of laser gas by collision. He-Ne Laser: Helium-Neon Laser was first gas Laser to be invented by Ali-Jawan in 1961. The pumping method employed in He-Ne Laser is electrical pumping method and is based on four level pumping scheme. Since He-Ne laser is a gas laser so He-Ne laser have sharp energy levels. Construction: It consists of a long discharge tube made up of fused quartz which is − �in the length and in the diameter. The tube is filled with �� and � gases under the pressure of Hg and . of Hg respectively. And are filled in the ratio ranging from : :. Neon is the active center and have energy levels suitable for laser transition. While He atoms help in exiting Neon atom. The electrodes are provided in the discharge tube to provided discharge in the gas which are connected to a high power supply. The optical cavity of laser consists plane and highly reflecting mirror at one end of the laser tube and a Plano- concave output mirror of an approximately % transmission at the other end. To minimize reflection Laser the discharge tube edges are cut at the angle. This arrangement causes the laser output to be linearly polarized. Working: A high voltage is applied across the gas mixture produces electrical breakdown of the gas into ions and electrons. Fast moving electrons are collide with Helium and Neon atoms and exit them to high energy level. �� atom are more easily excitable than Ne atoms as they are lighter. The life time of the energy levels � and � of He is more therefore these levels of He becomes densely populated. As the � energy levels � and � are close to the exited levels � and � of He. The probability of the atoms transferring their energy to Ne atom by inelastic collision is greater than the probability of coming ground state � by spontaneous emission. Since the pressure of the He is 10 times greater than the pressure of Neon, the levels � and � of Neon are densely populated than any other energy levels. Photons with the energy ℎ� stimulate the transition from � to � , � to � and � to � . During these S t u d y n a m a .c o m
- 63. Unit-II (LASER) transition radiation are emitted with the wavelength of . � , Å and . � respectivly. Figure(16): He-Ne Laser Figure(17): Energy level diagram of He-Ne Laser From the energy levels � spontaneous emission occurs in the energy level � . Since the energy level � is the lower Metastable state then the possibility of atom in the level � getting de-exited to the level � may occur, if it happened then number of atoms in ground state will go on diminishing and the laser ceases to function. This can be protected by reducing the diameter of the tube so that atoms in � follows direct transition to the level � through collision with the walls of tube. The He-Ne Laser operate in continuous wave mode. Application of Laser: 1) The laser beam is used to vaporize unwanted materials during the manufacturing of electronic circuits on semiconductors chips. 2) Laser is used to detect and destroy the enemy missiles during war. 3) Metallic rod can be melted and joined by means of laser beam. S t u d y n a m a .c o m
- 64. Unit-II (LASER) 4) Low price semiconductor lasers are used in CD players, laser printers. 5) High power lasers are used to leasing thermo nuclear reactions which would become the ultimate exhaust little power source for human civilization. 6) Laser is also being employed for separating the various isotopes of an element. 7) Laser beam are also been used to the internal confinement of plasma. S t u d y n a m a .c o m
- 65. Unit-3 (Fiber Optics) Unit-3 Fibre Optics Syllabus: Fibre Optics: Light guidance through optical fibre, types of fibre, numerical aperture, V-Number, Fibre dispersion (through ray theory in step index fibre), block diagram of fibre optic communication system S t u d y n a m a .c o m
- 66. Unit-3 (Fiber Optics) Fibre Optics: Fiber optics is the technology in which signals are converted from electrical into optical signals transmitted through a thin glass-fiber and re-converted into electrical signals. Definition: An optical fiber is a transparent medium as thin as human hair, made of glass or clear plastic designed to guide light waves along its length. Total Internal Reflection: When light waves goes into denser medium through rare medium then they goes away from the normal. If the angle of incidence exceeds the critical angle then the refracted ray comes back in to the same medium, this phenomenon is called the total internal reflection. Figure(1):Total internal reflection Principle of optical fiber cable: The propagation of light in the optical fiber from one end to another end is based on the principal of total internal reflection (TIR). When light enters through one end it suffers successive TIR from side walls and travels along the fiber length in a zigzag path. Construction: An optical fiber is cylindrical in shape and has three co-axial regions. The inner most region is the light guiding region known as core, whose diameter is of the order of . It is surrounded by a co-axial middle region known as cladding. The diameter of cladding is of the order of , the refractive index of cladding is always lower than that of the core. The purpose of the cladding is to make the light to be confined to the core. Light launched into the core and striking the core cladding interface at an angle greater than critical angle will be reflected back into the core. The outermost region is called sheath or jacket, which is made up of plastic or polymer. The sheath protects the cladding and core from abrasion and the harmful contamination of moisture and also increases the mechanical strength of the fiber. Optical fiber is used to transmit light signal over long distance. Optical fiber requires a light source for launching light into the fiber at its input and a photo detector to receive light at its output end S t u d y n a m a .c o m
- 67. Unit-3 (Fiber Optics) . As the diameter of the optical fiber is very small, LEDs and laser diodes are used as light source. At the receiver end semiconductor photodiodes are used for detection of light pulses and convert the optical signals into electrical form. Figure(2):Optical fiber cable S t u d y n a m a .c o m
- 68. Unit-3 (Fiber Optics) Light Propagation in the Fibber: Let us consider the light propagation in the optical fiber. The end at which the light enters the fiber is called the launching end. Let the refractive index of the core is and that of cladding is as < . Let the outside medium from which the light is launched have the refractive index . Let the light ray enters the fiber at an angle �� with the axis and strikes core-cladding interface at an angle �. If � > �the ray will suffer total internal reflection and remains within the fiber. Figure (3): Propagation of light through optical fiber cable. Fractional Refractive Index: It is the ratio of the difference of the refractive index of core and cladding to refractive index of core. It is denoted by ∆ and is expressed as ∆= − Where = refractive index of the core = refractive index of cladding It has no dimension and its order is . this parameter is always positive because > . In order to guide light effectively through the fiber ∆≪ typically of the order of 0.01 Acceptance Angle: Applying Snell’s law at the laun hing end si �� si �� = sin �� = sin �� …………………………………………………………………... (1) Now In Δ � + �� + = ⟹ �� = − � So putting in equation (1) S t u d y n a m a .c o m
- 69. Unit-3 (Fiber Optics) sin �� = sin − � sin �� = cos � …………………………………………………………………... (2) Now �� = ��� when � = � Applying Snell’s law sin � = sin � sin � = ∵ sin = But cos � = √ − sin� cos � = √ − () cos � = √ − cos � = √ − ………………………………………………………… (3) Therefore, putting the value in equation (2) we get sin ��� = × √ − sin ��� = √ − Let air be the medium at launching end so = Then sin ��� = √ − � �� = sin− √ − The angle � �� is called the acceptance angle of the fiber. Acceptance is the maximum angle that are light rays can have relative to the axis of the fiber and propagate down the fiber. In 3D the light rays contained within the cone having a fall angle � �� are accepted and transmitted along the fiber. Therefore the cone is called the acceptance cone. Figure(4): Acceptance cone= � �� S t u d y n a m a .c o m
- 70. Unit-3 (Fiber Optics) Numerical Aperture: Numerical aperture determines the light gathering ability of the fiber. This is defined as the � of the angle of acceptance angle. � = sin ��� But sin ��� = √ − , so � = √ − Relation between �� and ��� We know that numerical aperture is given as � = √ − � = √ + − � = √ + − × � = √ + − But + ≈ and − = ∆ so � = √ ∆ � = √ ∆ Normalized frequency V- number Optical fiber is characterized by a parameter caused V-number or normalized frequency. Normalized frequency is the relation between refractive indices and wavelength, and is given by � = �� √ − Where =radius of core =free space wavelength But we know that √ − = � = √ ∆ so � = �� � � = �� √ ∆ � = � √ ∆ Where = V- number helps in determining the number of modes that can propagates through a fiber from above relation number of modes that can propagates through a fiber increase with increase in � . Maximum number of modes in multi-mode step index fiber is given by � = � maximum number of mode in multi-mode graded index fiber is given by S t u d y n a m a .c o m
- 71. Unit-3 (Fiber Optics) � = � Also For single mode fiber, � < . For multi-mode fiber, � > . The corresponding wavelength is called cut-off wavelength. S t u d y n a m a .c o m
- 72. Unit-3 (Fiber Optics) Mode of Propagation: Figure(5): Mode of Propagation The total possible number allowed path in an optical fiber is known as modes. When light propagates at an angle close to the critical angle are high order modes and when modes propagates with angles longer than critical angle are low order mode. The zero order rays travels along the axis are known as axial ray. On the basis of modes of light propagation optical fiber are of two types: 1) Single mode fiber: - It supports only one mode of propagation. 2) Multi-mode fiber: - It supports number of modes for propagation. Refractive Index Profile: It is a plot of refractive index drawn on one of the axis (say-X) and the distance from axis of the core other axis (say-Y). On the basis of refractive index profile, there are two types of fibers- 1) Steps index fiber: In this refractive index of the core is constant throughout the core. 2) Graded Index Fiber: In this the refractive index of core varies smoothly over the diameter of the core. Types of the optical fiber: Based on the profile and modes of propagation optical fiber are of three types- 1) Single mode step index fiber: The diameter of typical SMSIF is about − which is of the order of wavelength of light used. SMSIF has a very thin fiber, the refractive index changes abruptly at the core-cladding interface for which it is called step index fiber. In this fiber light travels along the axis of the fiber. The � (i.e. numerical aperture) and ∆ (i.e. fractional refractive index) have very small values for single mode fiber and thus have very low acceptance. Therefore the light occupying in fiber becomes difficult. Costly laser diodes are used to launch the light into the fiber. A single mode fiber has very small value of ∆ and allows only one mode to propagates through them therefore intermodal dispersion does not exists in single mode fiber and thus have high data transfer rate. 2) Multi-mode step index fiber: This fiber is similar to single mode fiber only it has a large diameter of S t u d y n a m a .c o m
- 73. Unit-3 (Fiber Optics) the order of − . Large compared to the wavelength of light. In multi mode fiber the light follows a zigzag path. It allows more than one but finite number of modes to propagate through them. The NA is larger because of large core diameter the signal having path length along the axis of the fiber is shorter while the other zigzag path longer resulting in higher intermodal dispersion which means lower data rate and less efficient transmission. LEDs or laser source can be used for launching of light in this kind of fiber. This kind of fiber. This kind of fiber is used for short range communication. 3) Multi-mode Graded Index fiber: Multimode fiber have a core having refractive index at the center is very high and decreases as we move towards the cladding, such profile causes a periodic focusing of light propagation to the fiber. It allows more than one mode to propagate through them and the core diameter ranges from − the acceptance angle and � decreases with distance from the axis. The number of modes in this fiber is half that of multimode step index fiber. Therefore gives lower dispersion. Since the � of this fiber is less than multimode step index fiber, it makes coupling fiber to the source more difficult. Hence LEDs or laser light source can be used for launching the light in them; these are used in medium range communication. Refractive index profile: Index profile is the refractive index distribution across the core and cladding of fibre. Some fibre has a step index profile, in which the core has one uniformly distributed index. Other optical fibre has a graded index profile, in which refractive index varies gradually as a function of radial distance from the axis of the fibre. Multimode Step Index (MMSI OFC) Multimode Graded Index (MMGI OFC) Single mode Step Index (SMSI OFC) Fibre cross- section � Large Gradually decreases with distance from axis Very small ∆ Large Very small Acceptance angle � Large acceptance angle Gradually decreases with distance from axis Low acceptance angle Number of modes Allow finite number of modes � = �/ Number of mode are half of MMSI OFC i.e. � = �/ Only single mode is possible Range Short range communication Medium range communication Long range communication Data rate Lower data rate Lower data rate Higher data rate Efficiency Lower efficient Lower efficient Highly efficient S t u d y n a m a .c o m
- 74. Unit-3 (Fiber Optics) Light source LED LED Costly LASER diode Coupling Comparatively easy Very difficult difficult S t u d y n a m a .c o m
- 75. Unit-3 (Fiber Optics) Pulse Dispersion: High pulse launched into a fiber decrease in amplitudes as it travels along the fiber decrease in amplitude as it travels along the fiber due to laser. It also spreads during travel so its output pulse become wider than input pulse these are of three types: 1) Intermodal Dispersion: It is due to difference in propagation time in different modes. 2) Intramodal Dispersion: It results due to difference in wavelength, since fiber light consists of groups of waves. 3) Wave guided dispersion: It happens due to wave guiding properties of fiber. Fiber Losses: The losses in optical fiber may be due to following causes: 1. Rayleigh scattering losses: The glass in optical fiber is an amorphous solid that is formed by allowing the glass to cool from its molten state at high temperature, until it freezes. During the forming process, some defects are causes in fiber which allows scattering a small portion of light passing through the glass, creating losses. It affects each wavelength differently. 2. Absorption Losses: The ultraviolet absorption, infrared absorption and ion resonance absorption these three mechanisms contribute to absorption losses in glass fiber. The oxygen ions in pure silica have very tightly bounded and only the ultraviolet light photons have enough energy to be observe. Infrared absorption takes place because photons of light energy are absorbers by the atoms within the glass molecules and converted to the random vibration. 3. Micro bend Losses: Due to small irregularities in the cladding, causes light to be reflected at angle where there is no further reflection. 4. Macro bend Losses: It is a bend in the entire cable which causes certain modes not to be reflect and therefore causes losses to the cladding. Figure (6): Macro and micro band losses in optical fiber cable. S t u d y n a m a .c o m
- 76. Unit-3 (Fiber Optics) 5. Temperature Changes: A temperature change from 0 to ℃ could add as much as to the cable losses. Stress (Strain and tension) could add another . 6. Attenuation Losses: Attenuation losses of an optical fiber is defined as the ratio of optical output power � from a fiber of length � to the output power �� . In symbol � is expressed attenuation in � / . � = � log [ �� � ] In case a fiber is an ideal when �� = � , therefore � = which means that there will no attenuation loss. In actual practice, a low loss fiber may have � = /. S t u d y n a m a .c o m
- 77. Unit-3 (Fiber Optics) Calculation of dispersion for step indeed fiber: Figure(7): Propagation of light through the optical fiber cable is the time taken by ray to travel + by velocity � then = + ………………………………………… (1) If be the refractive index of core and is speed of light in vacuum, then = � � From the figure in ∆ � � sec �� ⟹ � sec �� and � sec �� ⟹ � sec �� Putting the values in equation (1) we get = � sec �� + � sec �� � + �sec �� sec� ………….. 2 As the ray in the fiber propagates by a series of total internal reflection at the interface, the time taken by the S t u d y n a m a .c o m
- 78. Unit-3 (Fiber Optics) ray in traversing an axial length of the fiber will be � = .c s �� ………………………………… (3) Time taken by rays making zero angle with fiber axis will be minimum i.e. � � = . cos = …………………… (4) The maximum time is given by � �� = .c s �� ………………………………… (5) Now y Snell’s law si �� si �� = But �� = for �� = � (i.e. critical angle) sin � sin = sin � = ……………………………….......... (6) From the figure is clear that � + �� + = � = − �� So by equation (8) sin − �� = cos �� = ………………………………... (7) Putting the value of cos �� in equation (5) � = . � � � = . S t u d y n a m a .c o m
- 79. Unit-3 (Fiber Optics) Fibre Optics communication System: The optical fibers are widely used for communication purpose. The fiber optics communication system is almost similar to ordinary communication system. Simply the systems consist of transmitter, information channel and recover. Transmitter: The transmitter converts electrical signals (Information signal) into optical signals. Mainly transmitter consists of 1) Transducer: If input signal is other than the electrical signals, we use a transducer which consists a non- electrical message into electrical signal. 2) Modulator: The output of transducer is connected as the input of modulator, with the help of modulator electrical messages are converted into the desired form. There are two kinds of modulators; digital and analog. 3) Light source: The function of light source is to generate carrier waves on which the information signal is impressed and transmitted. The light sources used are light emitting diodes (LEDs) or LASER diodes. These are known as optical oscillators. 4) Input channel coupler: It transfers the signals to information channel i.e. optical fiber in a proper manner. 5) Information Channel: It is a link between transmitter and receiver. Figure(8): The optical fiber communication system Receiver: Receiver converts the signals into electrical signals; it consists of- 1) Output channel coupler: its main function is to direct the light emerging from optical fiber into the photodetector. 2) Photodetector: The photodetector converts the light wave into an electric current. The detector output includes the message, which is separated from the carrier in next step. S t u d y n a m a .c o m
- 80. Unit-3 (Fiber Optics) 3) Signal processor: the information from the carrier wave is separated by signal processor includes amplifiers and filters. The optical signal, if needed, amplified and undesired frequencies are filtered by the processor. 4) Signal restorer: while traveling through the optical fiber the signal progressively attenuated and distorted due to various laser and dispersion occurring in the fiber. Thus the signal should be amplified and restorers are used for this purpose. S t u d y n a m a .c o m
- 81. Unit IV- Quantum Mechanics Page 1 Unit-4 Quantum Mechanics Syllabus: Black body radiation, ultraviolet catastrophe, Crompton effect, plates theory of radiation, phase and group velocity, particle in a box, uncertainty principle, well-behaved wave equation, Schrodinger equation, application to particle in a box. S t u d y n a m a .c o m
- 82. Unit IV- Quantum Mechanics Page 2 Black body: A black body is one which absorbs all types of heat radiation incident on it when radiations are permitted to fall on black body they are neither reflected nor transmitted. A black body is known as black body due to the fact that whatever may the colour of the incident radiation the body appears black by absorbing all kind of radiations incident on it. A perfect black does not exists thus a body representing close proximity to perfect black body so it can be considered as a black body. A hollow sphere is taken with fine hole and a point projection in front of the hole and is coated with lamp black on its inner surface shows the close proximity to the black body, when the radiation enter through hole, they suffer multiple reflection and are totally absorbed. Figure(1): Black body Black Body radiation: A body which completely absorbs radiation of all radiations of all wavelength/frequencies incident on it and emits all of them when heated at higher temperature is called black body. The radiation emitted by such a body is called black body radiation. So the radiation emitted form a black body is a continuous spectrum i.e. it contains radiation of all the frequencies. Distributions of the radiant energy over different wavelength in the black body radiation at a given temperature are shown in the figure. Black body radiation is a common synonym for thermal radiation. Figure(2): Black body radiation Radiation: Radiation is a process which the surface of an object radiates its thermal energy in the form of the electromagnetic waves. Radiations are of two types Radiation Ionising radiation Non-ionising radiation S t u d y n a m a .c o m
- 83. Unit IV- Quantum Mechanics Page 3 Emissivity: The emissivity of a material is the irradiative power of its surface to emit heat by radiation, usually it is shown by or . It is the ratio of energy radiated by a material to the energy radiated by the black body. True black body has maximum emissivity � =� (highly polished silver has an emissivity for about . at least.) Plank’s Quantum Hypothesis: Plank assumes that the atoms of the wall of blackbody behave as an oscillator and each has a characteristic frequency of oscillation. He made the following assumption- 1) An oscillator can have any arbitrary value of energy but can have only discrete energies as per the following relation � =� ℎ� Where � =� , , , � …� ..and � and ℎ a e k o as f e ue a d Pla k’s o sta t. 2) The oscillator can absorb or emit energy only in the form of packets of energy ℎ� but not continuously. � =� ℎ� Average energy of Plank’s Oscillators: If be the total number of oscillations and as the total energy of these oscillators, then average energy will be given by the relation. ̅ = ……………………………………………………………………………. (1) Now consider ,� ,� …� …� …� …� .� .�as the number of oscillators having the energy values ,� ℎ�,� ℎ�� …� .� .� ℎ�espe ti el . The the Ma ell’s dist i utio fo ula = +� +� +� ⋯� …� …� …� …� .. = � +�− ℎ� �� +� − ℎ� �� +� ⋯� …� …� .� = ( � −�− ℎ� ��) .............................................................. (2) And the total energy = ×� + ×� ℎ�+ ×� ℎ�+� ⋯� …. = ×� +� ( − ℎ� �� � ×� ℎ�)� +� (− ℎ� �� � ×� ℎ�)� +� ⋯ = − ℎ� �� � ×� ℎ�[ � +� − ℎ� �� +� − ℎ� �� +� ⋯� …� .� .] = − ℎ� �� ℎ� ( � −�− ℎ� ��) ………………………………………….. (3) S t u d y n a m a .c o m
- 84. Unit IV- Quantum Mechanics Page 4 Putting the value of and from above equations in equation (1) we get- ̅ = ̅ = ℎ� − ℎ� �� ( � −�− ℎ� ��) ̅ = ℎ� ( ℎ� �� −� ) …………………………………………..………. (4) This is the e p essio fo the a e age e e g i Pla k’s os illato s. Plank’s radiation formula: The average density of radiation � in the frequency range � and �� +� �depending upon the average of an oscillator is given by- � � = �� �� ×�̅ …………………………………………..……… (5) � � = �� ℎ� ( ℎ� �� −� ) � � � = �ℎ � ( ℎ� �� −� ) � …………………………………………..……… (6) The a o e elatio is k o as the Pla k’s adiatio fo ula i te s of the f e ue . This la a also e expressed in terms of wavelength � of the radiation. Since �� = � for electromagnetic radiation, �� = − � �. Further we know that the frequency is reciprocal of wavelength or in other words an increase in frequency corresponds to a decrease in wavelength. therefore � � = − � � � � = − �ℎ � − � � ( ℎ ��� −� ) � � = �ℎ � ( ℎ ��� −� ) � …………………………………………..… (7) The a o e elatio is k o as the Pla k’s la i te s of a ele gth � Wien’s law and Rayleigh-Jeans law: With the help of Pla k’s adiatio Wie ’s la a d Ra leigh-Jens law can be derive. When the wavelength � and temperature � are very small, then ℎ� ��� ≫� . Therefore, can be neglected in the denominator of equation (7). � � = �ℎ � − ℎ ��� � S t u d y n a m a .c o m
- 85. Unit IV- Quantum Mechanics Page 5 By substituting �ℎ � =� and ℎ � =� , we get � � = � − �� � …………………………………………..………. (8) This is k o as Wie ’s la , hi h is alid at lo te pe atu e � and small wavelength �. For higher temperature � and large wavelength �, ℎ� ��� can be approximated to � + ℎ ��� . Then we have from equation (7) � � = �ℎ � � + ℎ � � −� � � � = � � � � ………………….…………………………………………..… (9) This is known as Rayleigh-Jeans law. Ultraviolet Catastrophe: One of the nagging questions at the time concerned the spectrum of radiation emitted by a so-called black body. A perfect black body is an object that absorbs all radiation that is incident on it. Perfect absorbers are also perfect emitters of radiation, in the sense that heating the black body to a particular temperature causes the black body to emit radiation with a spectrum that is characteristic of that temperature. Examples of black bodies include the Sun and other stars, light bulb filaments, and the element in a toaster. The colours of these objects correspond to the temperature of the object. Examples of the spectra emitted by objects at particular temperatures are shown in Figure 3 Figure 3: The spectra of electromagnetic radiation emitted by hot objects. Each spectrum corresponds to a particular temperature. The black curve(dotted line) represents the predicted spectrum of a 5000 K black body, according to the classical theory of black bodies S t u d y n a m a .c o m
- 86. Unit IV- Quantum Mechanics Page 6 At the end of the 19th century, the puzzle regarding blackbody radiation was that the theory regarding how hot objects radiate energy predicted that an infinite amount of energy is emitted at small wavelengths, which clearly makes no sense from the perspective of energy conservation. Because small wavelengths correspond to the ultraviolet end of the spectrum, this puzzle was known as the ultraviolet catastrophe. Figure 27.1 shows the issue, comparing the theoretical predictions to the actual spectrum for an object at a temperature of 5000 K. There is clearly a substantial disagreement between the curves S t u d y n a m a .c o m
- 87. Unit IV- Quantum Mechanics Page 7 Matter wave: According to Louis de-Broglie every moving matter particle is surrounded by a wave whose wavelength depends up on the mass of the particle and its velocity. These waves are known as matter wave or de- Broglie waves. Wavelength of the de-Broglie wave: Consider a photon whose energy is given by � =� ℎ�� = ℎ � [ � � =� ��……………………………………… (1) Where ℎ is Pla k’s o sta t. � ×� − , � is the frequency and � is the wavelength of photon. No Ei stei ’s ass e e g elatio = ……………………………………… (2) By equation (1) and (2) = ℎ � � = ℎ � = ℎ Where � =� In place of the photon a material particle of mass is moving with velocity then � = ℎ …………………………….…… (3) (i) Now we know that the kinetic energy of the material particle of mass moving with velocity � is given by- = = � = [ � � =� ] = √ So by equation (3) � = ℎ √ (ii) According to kinetic theory of gasses the average kinetic energy of the material particle is given by � = � where � =� . � ×�− / i.e. Boltzmann constant S t u d y n a m a .c o m
- 88. Unit IV- Quantum Mechanics Page 8 = � = � = � � � = = √ � So by equation (3) � = ℎ √ � ……………………………… (4) Group or Envelope of the wave: When a mass particle moves with some velocity than it emits the matter waves, those waves interacts each other and where there they interfere constructively they form an envelope around the particle which is known as wave group or simply envelope. Figure(2): Formation of the wave packet Group velocity: Group velocity of a wave is the velocity with which the overall shape of the a e’s amplitudes (modulation or envelope) of the wave propagates through space. It is denoted by �. Phase velocity: The phase velocity of a wave is the rate at which the phase or the wave propagates in the space. It is denoted by �. Expression for Group velocity and phase velocity: Let us suppose that the wave group arises from the combination of two waves that have some amplitude but differ by an amount ∆ in angular frequency and an mount ∆ in wave number. = � cos � −� ……………………………… = � cos[ � +� ∆� − � +� ∆] ……………………………… By the principle of superposition S t u d y n a m a .c o m
- 89. Unit IV- Quantum Mechanics Page 9 = +� …………….…………………………… (3) = [cos � −� +� cos{ � +� ∆� − � +� ∆}] Using the identity cos� � +� cos� � =� � cos� ( � +� )� cos� ( � −� ) And cos −� =� cos� � � +� � = � −� +� { � +� ∆ � −� � −� ∆ } � +� � = � −� � +� ∆ � −� ∆ � +� � = �+∆ �− � −∆� � +� � = +∆ �− �+∆� � −� � = � −� −� { � +� ∆ � −� � −� ∆ � −� � = � −� � −� � −� ∆ � +� � + � −� � = −∆ � +� ∆ � −� � =� − ∆ � −� ∆ � =� � [ � cos{ � +� ∆� − � +� ∆ } .� cos{ ∆ � −� ∆ }] Let � +� ∆ � =�and � +� ∆ � =� So we have = [cos� ( � −� ) .� cos� ( ∆ � −� ∆ )] ⟹ = [cos � −� .� cos� ∆ � − ∆� ]…………………… (4) This is the resultant wave equation of superposition of two waves having the amplitude � cos� ∆ � − ∆� and phase cos � −� where and are mean values of angular frequency and prapogation constant of the wave. Phase velocity: Since phase � −� = Differentiating with respect to we get � −� = ⟹ � = � But � = � phase velocity S t u d y n a m a .c o m
- 90. Unit IV- Quantum Mechanics Page 10 ⟹ � = � = � ...................................................... (5) Group Velocity: ⟹ ∆ � − ∆� = ⟹ ∆ = ∆� ⟹ ∆ ∆� = � So the group velocity � = � = � ........................................................ (6) S t u d y n a m a .c o m
- 91. Unit IV- Quantum Mechanics Page 11 Relation between Group velocity and phase velocity 1. For dispersive and non-dispersive medium: But by equation (5) i.e. � = � ⟹� � =�� Putting into equation (6) we get ⟹ � = ( �) ⟹ � = �.� � +� � ⟹ � = � � +� ( � � ) � � � � ⟹ � = � � +� ( � � ) � � �− ⟹ � = � � +� ( � ) � −�− .� � ⟹ � = � −� � � � Different cases: 1) If �� � =� i.e. if the phase velocity does not depends on the wavelength then � =� �, such a medium is called the non dispersive medium. 2) If �� � ≠� i.e. if it has positive values then � <� �, then such a medium is called the dispersive medium. 2. Relativistic particle: Let us consider a de-Broglie wave associated with a moving particle of rest mass and velocity , then the and will be given by ⟹ = �� ⟹ = � ℎ � �� = ℎ ⟹ = � ℎ . √ � − ……………………………….. (8) And ⟹ = � � ⟹ = � ℎ � � �� = ℎ � S t u d y n a m a .c o m
- 92. Unit IV- Quantum Mechanics Page 12 ⟹ = � � ℎ√ − � � ...................................................(9) � � = √ − � � Now phase velocity � = � So � = ( �� ℎ . √ − � � ) ( � � ℎ√ − � � ) � = � ℎ . √ − � � � ×� ℎ√ − � � � � � = � ………………………………………………………………… (10) Now group velocity � = � The expression can be written as � = ……………………………………………………………… (11) In order to find the value of � we have to solve the following terms- ⟹ = [ � ℎ . √ � − ] [By equation (8)] ⟹ = � ℎ √ � − − ⟹ = � ℎ (− ) � − − .� (− ) ⟹ = � ℎ � − − ……………………………………… (12) Again ⟹ = [ � ℎ√ � − ] ⟹ = � ℎ [ √ � − ] S t u d y n a m a .c o m
- 93. Unit IV- Quantum Mechanics Page 13 ⟹ = � ℎ . [ √ � − .� � −� .( � − ) − .� − ( � − ) ] ⟹ = � ℎ . [ √ � − � +� . √ � − ( � − ) ] ⟹ = � ℎ . [ � − + ( � −� )� .√ � − ] ⟹ = [ � ℎ � −� − ] …………………………… (13) Putting the value from (13) and (14) into (11) we get ⟹ � = [ { � ℎ � ( � −) − } { � ℎ ( � − ) − } ] Group velocity � = ………………………………………………………………………… (14) By equation (10) we have �.� = By equation (14) i.e. � =� So �.� � = 3. Non-Relativistic free Particle: According to the de-Broglie hypothesis � = ℎ �� Now the total energy = � ……………………………………………………………………… (1) And = ℎ� ……………………………………………………………………… (2) By equation (1) and (2) ℎ� = � S t u d y n a m a .c o m
- 94. Unit IV- Quantum Mechanics Page 14 ℎ� = �� ℎ And phase velocity � =� �� So we have � = �� ℎ × ℎ �� � = �� S t u d y n a m a .c o m
- 95. Unit IV- Quantum Mechanics Page 15 Uncertainty Principle: It is impossible to determine the exact position and momentum of a particle simultaneously. Let us consider a particle surrounded by a wave group of de-Broglie wave as shown in the figure Figure(3): particle surrounded by a wave packet Let us consider two such waves of angular frequency and and prapogation constant and traviling along the same direction are- = � sin � −� …………………………………………… (1) = � sin � −� …………………………………………… (2) According to the principal of superposition = +� = � sin � −� � +� � sin� −� = [sin � −� +� sin � −� ] = [ � sin� ( � −� � +� � −� )� .� cos� ( � −� � −� � +� )] � sin� � +� sin� � =� � sin� ( � +� )� .� cos� ( � −� ) = [sin� ( +� � − +� )� .� cos� ( −� � − −� )] S t u d y n a m a .c o m
- 96. Unit IV- Quantum Mechanics Page 16 Let � = +� (3) And � = +� ……………………………………......................... And ∆ � =� −� And ∆ � =� −� So we have = [sin � −� .� cos� ∆ � − ∆� ] ………………………………………………………… (4) The resultant wave is plotted in the figure (4). The position of the particle cannot be given with certainty it is somewhere between the one node and the next node. So the error in the measurement of the position of the particle is therefore equal to the distance between these two nodes. Figure(4): The envelope created by the superposition of two waves. The node is formed when cos� ∆ � − ∆� � =� � cos� � =� � ⟹� � =� +� � ⟹ ∆ � − ∆ = � +� � Thus and represents the positions of two successive nodes, then at any instant , we get- ∆ � − ∆� = � +� � ………………………………………………… (5) ∆ � − ∆� = � +� � ………………………………………………… (6) Now on subtracting (5) from (6) we get ∆ � − ∆� − ∆ � + ∆� = . � + � −� . � − � ∆� −� = � ∆ = � ………………………………………………… (7) But � = � � S t u d y n a m a .c o m
- 97. Unit IV- Quantum Mechanics Page 17 ∆� � � � = � Again � = ℎ � = ℎ � So � { ℎ � } = � ℎ .� .� = 1 .� = ℎ .� ℏ Where ℏ� = ℎ/ � Energy and time uncertainty principle: Let ∆ be the width of the wave packet moving along the x-axis, let � be the group velocity of the wave packet and is the particle velocity along x-axis. Now if the wave packet moves through ∆ in ∆ time. Since ∆ is the uncertainty in the x-coordinates of the particle and ∆ is the uncertainty in the time i.e. given by ∆ = ∆ � ∆ = �.� ∆ ……………………………………………………………… (1) If the rest mass of the particle is then the kinetic energy is given by = = .� = .� = .� = �� ....................................................................... (2) If ∆ and ∆ are the uncetainity in the momentum and energy respectively, then differentiating (2) we get ∆ = ��.∆�� ∆ = ��.∆�� ∆ = ∆� �� But =� So ∆ = �� ∆ .................................................................. (3) Now by(1) and (3) ∆ .� ∆ = �.� ∆ . �� ∆ S t u d y n a m a .c o m
- 98. Unit IV- Quantum Mechanics Page 18 But � =� so we have = ∆ .� ∆ .................................................................. (4) We know that ∆ ∆ ℏ ................................................................................ (5) So by (4) and (5) ∆ .� ∆ ℏ S t u d y n a m a .c o m
- 99. Unit IV- Quantum Mechanics Page 19 Application of uncertainty Principle: Determination of the position of a particle with the help of a microscope: Let us consider the case of the measurement of the position of an electron is determined. For this the electron is illuminated with light (photon). Now the smallest distance between the two points that can be resolved by microscope is given by ∆ = � � sin� � .................................................................. (1) From the above equation it is clear that for exactness of position determination improves with a decrease in the wavelength � of liaght. Let us imagine that we are using a � −� microscope of angular aperture �. Figure(5): microscope In order to observe the electron, it is necessary that at least one photon must strike the electron and scattered inside the microscope. The scattered photon can enter in the field of view +� to –� �as shown in the figure. Rough � =� ℎ� ��� =� So � = ℎ � …� …� …� …� …� …� …� .� .� And � =� …� …� …� …� …� …� …� So = ℎ� � � =� ℎ � Figure(6): Scattering of an photon by an electron S t u d y n a m a .c o m
- 100. Unit IV- Quantum Mechanics Page 20 The momentum of the scattered photon is ℎ � then the momentum along the x-axis is –� � sin� �and + � sin� �. But when the photon of wavelength �′ collied to the electron then this photon recoile the electron by giving some momentum to it. Now the uncertainty in the momentum transfer to the electron will be � =[ � +� ] − [ � −� ] = ℎ �′ + ℎ � sin� � � −� ℎ �′ − ℎ � sin� � = ℎ � sin� � ................................................................. (2) By equation (1) and (2) .� = � � si � � × ℎ� si � � � .� = ℎ .� ℏ Where ℏ� = ℎ � Diffraction of electron beam by a single slit: Suppose a narrow beam of electron passes through a narrow single slit and produces a diffraction on the screen as shown in figure. Figure(7): Diffraction pattern of electron beam by single slit But the theo of F au hofe ’s diffraction at a single slit � sin� �� =� ± �the first minima is given by � sin� �= � …………………………………….……………………………………… (1) In producing the diffraction pattern on the screen, all the electrons have passed through the slits but we S t u d y n a m a .c o m
- 101. Unit IV- Quantum Mechanics Page 21 can-not say definitely at what place of the slit. Hence the uncertainty in determining the position of electron is equal to the width of the slit then by equation (1) = � � si � � …………………………………………….…………………… (2) Initially the electrons are moving x-axis and hence they have no component of momentum along y-axis. After diffraction on the slit, they are deviated from their initial path to form the pattern. Now they have a component � sin� �.As y component of momentum may be anywhere between � sin� �and –� � sin� �. Hance the uncertainty in component is = � sin� �� −� − � sin� � = � sin� � = � ℎ � � sin� � [ � � = ℎ � ] .............................. (3) By equation (2) and (3) .� y = � � si � � × ℎ � sin� � .� y ℎ .� y ℏ S t u d y n a m a .c o m
- 102. Unit IV- Quantum Mechanics Page 22 Compton Scattering: When a beam of monochromatic radiation of sharply define frequency incident on materials of low atomic number, the rays suffers a change in frequency on scattering. This scattered beam contains two beams one having lower frequency or greater wavelength other having the same frequency or wavelength. The radiation of unchanged frequency in the scattered beam is known as unmodified radiation while the radiation of lower frequency or slightly higher wavelength is called a modified radiation. This phenomenon is known as Compton effect. Figure(8): Compton Scattering The energy and momentum S.N. Quantity Before collision After collision 1. Momentum of radiation ℎ� (Where � is the frequency of radiation) ℎ�′ 2. Energy of radiation E� =� hϑ(Where � is the frequency of radiation) E� =� hϑ′ 3. Momentum of electron 4. Energy of electron � =� (Where is the rest mass of the electron) � =� (Where is the moving mass of the electron) By the principle of the conservation of momentum along and perpendicular to the direction of the incidence, S t u d y n a m a .c o m
- 103. Unit IV- Quantum Mechanics Page 23 we get In x-direction � � = � � ℎ�� +� = ℎ�′ cos� �� +� � cos� �…………………………………. (1) In y-direction � +� = ℎ�′ sin� �� −� � sin� �…………………………………. (2) By equation (1) � cos� �= ℎ�� −� ℎ�′ cos� � …………………………………. (3) By equation (2) � sin� �= ℎ�′ sin� � …………………………………. (4) Squaring equation (3) and (4) then adding we get sin �� +� cos� = ℎ�� −� ℎ�′ cos� � + ℎ�′ sin� � = ℎ � +� ℎ�′ cos �� −� ℎ��′ cos� �� +� ℎ�′ sin � = ℎ � +� ℎ�′ cos �� +� sin� −� ℎ��′ cos� � = ℎ � +� ℎ�′ −� ℎ��′ cos� � ℎ � +� ℎ�′ −� ℎ��′ cos� � = …………………………………. (5) Now by conservation of energy � � = � � ℎ�� +� = ℎ ′ +� …………………………………. (6) ℎ�� +� −� ℎ′ = ℎ�� −� ℎ�′ +� = Squaring both side [ℎ�� −� ℎ�′ +� ] = As we know that � +� � +�=� +� +� +� � +� � +� so � −� � +�=� +� +� −� � −� � +� ℎ � +� ℎ�′ +� −� ℎ��′ −� ℎ�′ +� ℎ� = ℎ � +� ℎ�′ −� ℎ��′ +� +� ℎ�� −� �′ .� = …………………………………. (7) Subtracting equation (5) from (7) we get ℎ � +� ℎ�′ −� ℎ��′ +� +� ℎ�� −� �′ .� −� {ℎ� +� ℎ�′ − ℎ ��′ cos� �} = −� ℎ � +� ℎ�′ −� ℎ��′ +� +� ℎ�� −� �′ .� −� ℎ� −� ℎ�′ +� ℎ��′ cos� � = −� −ℎ .� ��′ � −� cos� �+� ℎ�� −� �′ +� = −� S t u d y n a m a .c o m
- 104. Unit IV- Quantum Mechanics Page 24 But � = √ − � � So = − � � = −� So −ℎ .� ��′ � −� cos� �+� ℎ�� −� �′ +� = −� −� − ℎ ��′ � −� cos� �+� ℎ�� −� �′ +� = − ℎ ��′ � −� cos� �+� ℎ�� −� �′ = ℎ �� −� �′ = ℎ ��′ � −� cos� � �� −� �′ = ℎ ��′ ℎ � −� cos� � (�−�′) ��′ = ℎ � −� cos� � � ��′ − �′ ��′ = ℎ � −� cos� � �′ − � = ℎ � −� cos� �………………………….. (8) Multiplying by both side �′ − � = ℎ � −� cos� � �′ −� � = ℎ � −� cos� � ………………………….. (9) � = ℎ � −� cos� � � = ℎ � sin � � = ℎ sin � ………………………… (10) Where � is the change in the wavelength Equation (10) shows that 1) If �� =� � ⟹� �� =�i.e. there is no scattering along the direction of incidence. 2) If �� = � ⟹� �� = ℎ = . × − × − × × 8 =� . Å, this wavelength is known as Compton wavelength and it is a constant quantity. 3) If �� =� �� ⟹� �� = ℎ =� . Åso the change in the wavelength waries in accordance to the scattering angle � and this is shown in figure. S t u d y n a m a .c o m
- 105. Unit IV- Quantum Mechanics Page 25 Figure(9): Graph between angle of incidence and wavelength Importance of Compton effect: 1) It provides the evidence of particle nature of the electromagnetic radiation. 2) This verifies the Pla k’s ua tu h pothesis. 3) This provides an indirect verification of the following relation � = √ − � � and � =� Direction of the recoil electron: We know that � cos� �= ℎ�� −� ℎ�′ cos� � …………………………………………… (3) � sin� �= ℎ�′ sin� � …………………………………………… (4) Dividing (4) by (3) we get tan� � = ℎ�′ sin� � ℎ�� −� ℎ�′ cos� � tan� � = �′ sin� � �� −� �′ cos� � Again by equation (8) i.e. �′ − � = ℎ � −� cos� � �′ = � + ℎ sin ( � ) Multiplying by � we get � �′ = � + ℎ� sin ( � ) �′ = � + ℎ� sin � � �′ = � � + ℎ� sin � Then by equation (11) and (12) we have S t u d y n a m a .c o m
- 106. Unit IV- Quantum Mechanics Page 26 tan� � = [ � � + ℎ� sin � ] .� sin� � �� −[ � � + ℎ� sin � ] .� cos� � Let ℎ� =� � then tan� � = [ �� sin� � � +� �� sin � ] �� −[ �� cos� � � +� �� sin � ] tan� � = [ �� sin� � � +� �� sin � ] � [ � − cos� � � +� �� sin � ] tan� � = [ sin� � � +� �� sin � ] [ � +� �� sin � � −� cos� � � +� �� sin � ] tan� � = sin� � � +� �� sin � � −� cos� � tan� � = sin� � � −� cos� � � +� �� sin � tan� � = .� sin� � � .� cos� � .� sin � � +� �� sin � S t u d y n a m a .c o m
- 107. Unit IV- Quantum Mechanics Page 27 tan� � = .� sin� � � .� cos� � .� sin � � � +� � tan� � = cot� � [ � +� �] tan� � = cot� � [ � + ℎ� ] S t u d y n a m a .c o m
- 108. Unit IV- Quantum Mechanics Page 28 Wave function and its properties: We know that height of the water surface varies periodically in water waves, the pressure of gas varies periodically in sound waves and the electric field and magnetic field varies periodically in light waves, similarly the quantities which varies periodically in case of matter waves is called the wave function. The quantity whose variations make up the matter waves. This is represented by . This has no direct physical significance and is not an observed quantity. However the value of wave function is related to the probability of finding the particle at a given place at a given time, wave function is a complex quantity i.e. � =� � +� � Conjugate of is ∗ =� � −� � And ∗ = | | =� +� | | at a time at a particular place is the probability of finding the particle there at that time and is known as probability density | | =� ∗ Let the wave function is specified in direction by the wave equation � =� −� �− � � Where � =� ��and � =� �� So = − ��� �− � �� = − �� ��− � � ……………………………… (1) As � =� ℎ� � = ℎ � .� �� � =� � ℏ � And �� = ℎ � �� = �. ℎ � � �� = � ℏ � � = � �ℏ Putting these values in (1) we get = − �� � �ℏ �− � �ℏ = − � ℏ ��−� ……………………………… (2) This is the wave equation for a free particle. Properties of wave function: 1) It must be finite everywhere: if is infinite at a particular point, then it would mean an infinitely S t u d y n a m a .c o m
- 109. Unit IV- Quantum Mechanics Page 29 large probability of finding the particle at that point, which is impossible. Hence must have a finite or zero values at any point. 2) It must be single valued: if has more than one value at any point, it means that there is more than one values of probability of finding the particle at that point, which is impossible. 3) It must be continuous: For Schrodinger equation must be finite everywhere. This is possible only where has no discontinuity at any boundary where potential changes. This implies that too must be continuous across a boundary. 4) must be normalised: must be normalised, which means that must to be zero as � ⟶� ±∞, � ⟶� ±∞, � ⟶� ±∞in order that ∫| | over all space be finite constant. If ∫ | | +∞ −∞ � =� � ⟹the particle does not exists but | | over all space must be finite i.e. the body exits somewhere it ∫ | | � =� ,� ∞,� −, complex are not possible. 5) Normalization: ∫ | | +∞ −∞ � =� As | | =� ∗ =� � � � 6) Probability between the limits and : This is given by = ∫ | | +∞ −∞ � � 7) Expected values: To correlate experiment and theory we define the expectation values of any parameter = ∫ . | | +∞ −∞ ∫ | | +∞ −∞ = ∫ ∗+∞ −∞ ∫ ∗+∞ −∞ If is a normalised wave function then ∫ | | +∞ −∞ � =� So =� ∫ | | +∞ −∞ � =� Orthonormal and Orthogonal wave function: For two wave function and if the condition ∫ ∗ � =� exists then they are said to be orthogonal wave function. Here ∗ is the complex conjugate of . The normalized wave function are defined by ∫� ∗ � =� S t u d y n a m a .c o m
- 110. Unit IV- Quantum Mechanics Page 30 The wave function satisfying both the conditions of normalisation and orthogonally said to be orthonormal. These two conditions simultaneously can be written as ∫� ∗ � =� = =� for � =� =� for � ≠� S t u d y n a m a .c o m
- 111. Unit IV- Quantum Mechanics Page 31 Operator: Operator ̂ is a mathematical rule which may applied to a function which changes the function in to an other function . So an operator is a rule by means of which from a given function, we can find another function for example: =� So an operator tells us that what operation to carry out on the quantity that follows it. Energy Operator: We know that the wave function is given as- = − � ℏ ��−� Differentiating partially with respect to we get � = − � ℏ − � ℏ ��−� � = − � ℏ = − ℏ � Hence energy operator ̂ = �ℏ Momentum Operator: Again by wave function i.e. = − � ℏ ��−� Differentiating equation with respect to we get = � ℏ − � ℏ ��−� = � ℏ ℏ � = −�ℏ ̂ −�ℏ Note: ℏ � =� Here is called an eigan function of the operator −�ℏ and are called the corresponding energy eigan values. S t u d y n a m a .c o m
- 112. Unit IV- Quantum Mechanics Page 32 Schrodinger’s wave equation: S h odi ge ’s a e e uatio s a e the fu da e tal e uatio s of ua tu e ha i s i the sa e sense as the Ne to ’s se o d e uatio of otio of lassi al e ha i s. It is the differential form of de-Broglie wave associated with a particle and describes the motion of particle. Figure(10): Schrodinger wave equations Schrodinger’s time dependent wave equation in 1-dimentional form: Let us assume that the for a particle moving freely in the positive x-direction is = − � ℏ ��−� ………………………. (1) Now the total energy = � +� � � …..………………… (2) And we know that the kinetic energy is related with the momentum as � = � So the equation (2) in terms of wave function , can be written as = � � � +� � …………………… (3) As we know that the energy and momentum operators are given by � =� �ℏ � and � = ℏ � Putting the values in equation (3) we get Schrodinger wave equation Time dependent Schrodinger wave equation (i) One Dimensional �ℏ = ℏ +� � (ii) Two Dimensional �ℏ = ℏ + +� � (iii) Three Dimensional �ℏ = ℏ + + +� � Time dependent Schrodinger wave equation (i) One Dimensional − ℏ � −� �� =� (ii) Two Dimensional + + − ℏ � −� �� =� (ii) Three Dimensional + + − ℏ � −� �� =� S t u d y n a m a .c o m
- 113. Unit IV- Quantum Mechanics Page 33 �ℏ = ( ℏ � ) � +� � �ℏ = − ℏ +� � ………..…………………….………… (4) S h odi ge ’s ti e i depe de t a e e uatio i 3-dimentional form: �ℏ � = ℏ + + � � +� � But the Laplacian operator is given as ∇ =� + + So the above equation can be written as- �ℏ � = ℏ ∇ � +� � ………………………….…………… (5) Schrodinger’s time independent wave equation in 1-dimention: Again from wave function- = − � ℏ ��−� = − � ℏ �� .� � ℏ � = � ℏ � .� − � ℏ �� = − � ℏ �� Where =� � ℏ � Now differentiating partially with respect to we get � = − �� ℏ − � ℏ �� …………………………………………… (7) Now differentiating partially with respect to we get = − � ℏ �� Again = − � ℏ …………………………………………… (8) Putting the value from (7) and (8) into equation (5) we get i.e. �ℏ � = ℏ ∇ � +� � �ℏ [− � ℏ − � ℏ ] = − ℏ [ − � ℏ ] +� � − � ℏ − � ℏ = − � ℏ �� [ ℏ +� � ] = ℏ +� � − � = ℏ � −� = ℏ ℏ � −� = − ℏ � −� = ………………………………………………………. (10) S t u d y n a m a .c o m
- 114. Unit IV- Quantum Mechanics Page 34 S h odi ge ’s ti e i depe de t a e e uatio i 3-dimention form: + + � − ℏ � −� � = ∇ − ℏ � −� = Where ∇ = + + S t u d y n a m a .c o m
- 115. Unit IV- Quantum Mechanics Page 35 Application of Schrodinger’s wave equation: Energy level and wave function of a particle enclosed in one dimensional box of infinite height: Let us consider the case of a particle of mass moving along x-axis between two rigid walls A and B at � = and � =�. The potential energy � of the particle is given as �� = � <� � <� ∞ � � � � Within the box, the Schrodinger wave equation is given by − ℏ = [ � �� =�……………………………………. (1) Let = ℏ (2) −� = ………………………………………………. (3) This is a second order differential equation and its solution is given by = � sin� � +� � cos�……………………… (4) Where A and B are constants, the value of these constants can be calculated by the boundary conditions. By first boundary condition if � =� � ⟹� � =� Then by equation (4) we get = � sin� � +� � cos� ⟹ = For second boundary condition � =� at � =� then by equation (4) = � sin� � +� .� cos� ⟹ � sin� = But � ≠� so sin� � =� � ⟹� � =� � � =� , , , � …� …� …� As we know that = � ........................................................ (5) By equation (2) and (5) � ℏ = � � For representing the �ℎ energy level replacing BY we have � ℏ = � � = � � � × ℏ = � � � × ℎ � S t u d y n a m a .c o m
- 116. Unit IV- Quantum Mechanics Page 36 = ℎ � ......................................... (6) It is clear from expression (6) that inside an infinitely deep potential well, the particle can have only discrete set of energy i.e. the energy of the particle is quantised. The discreet energies are given by = ℎ � = ℎ � =� = ℎ � =� = ℎ � =� .. = .. .. .. .. = .. .. .. The constant A of equation (4) can be obtained by applying the normalization condition i.e. ∫ | | =� = = ∫ | � sin� | � = ∫ sin � � = ∫ −c s� � � � = ∫ � −� cos� � = [{ }� − { si � � � } � ] = [ � −� � −sin� � −� sin�] = [ � − sin � � −� sin�] = � � = � [ � − � −� ] = � sin� �� =� = = √ � …………………........................................... (7) Now the wave function will be given by = √ � sin � � � =� , , ,� …� …� .� S t u d y n a m a .c o m
- 117. Unit 5: Wave Optics Page 1 Unit-5 Optics Syllabus: Interference, division of amplitude & division of wave front, double slit experiment, thin film interference, Newton Ring Experiment. Diffraction: Difference between interference and diffraction, types of diffraction, single slit, double slit & n-slit diffraction, Resolving power of grating. S t u d y n a m a .c o m
- 118. Unit 5: Wave Optics Page 2 Interface: When two waves of approximately same amplitude and frequency going in the same direction in the same medium, generally coming from the same source, then the intensity of light at different places will be different. This phenomenon of light is known as interference. Interference can be obtained by two ways: Interference may be of two types: Figure(1): Interference Hierarchy tree Figure(2): types of Interference Constructive Interference: Locus of all the points where the crest of one wave falls on the crest or the through of the one wave falls on the through of the other, the resultant amplitude is the sum of the individual waves. So the constructive interference takes place at those points and the intensity at these points will be maximum. Figure(3): Constant Phase difference Figure(4): Waves in same phase Interfecence By the division of the wavefront 1. Youns double slit experiment 2. Fresenl's biprism By the division of the amplitude 1. Newton's ring 2. micleson's interferometer Interfecence Counstructive interference Distructive interferenc S t u d y n a m a .c o m
- 119. Unit 5: Wave Optics Page 3 Destructive Interference: Locus of the points where the crest of one wave falls on the through of the other wave the resultant intensity become the difference of the waves and at these places the intensity become minimum. At these points destructive interference will take place. Figure(5): Waves opposite phase Coherent sources: Two sources are said to be the coherent if they emit continuous light waves of the exactly same frequency/wavelength, nearly same amplitude and having sharply define phase difference that remains constant with the time. In practice it is impossible to have two independent coherent sources. For experimental purpose virtual sources formed by a single source and acts as coherent sources. Figure(6): You g’s Dou le Slit e peri e t Figure(7): Llo d’s Mirror Figure(8): Fresnel double mirror Figure(9): Fres el’s i-prism S t u d y n a m a .c o m
- 120. Unit 5: Wave Optics Page 4 Figure(10): Mi helso ’s I terfero eter Relation between phase difference and path difference: The difference between optical paths of two rays which are in constant phase difference with each other is known as the path difference. Suppose for a path difference the phase difference is � So � = � …………………..……….. Δ = ……………………………. by equation (1) and (2) � Δ = � � = � Δ Figure(11): Phase and path difference Principal of superposition: When two or more waves reaches at the same point of a medium then the displacement at that point becomes the vector sum of displacement produced by the individual waves. i.e. = + + … … …. S t u d y n a m a .c o m
- 121. Unit 5: Wave Optics Page 5 Mathematical treatment of interference: Let two waves of amplitude and and angular frequency � super imposes and re-unit at a point after traveling different path and ,let the phase difference of these two waves is � If and are two waves then = sin � ………… = sin � + � ………… (2) By the principle of superposition of waves , the resultant waves will be = + = sin � +sin � + � sin + = sin cos + cos sin = sin � +[sin � cos � + cos � sin �] = sin � +sin � cos � +cos � sin � = sin � [+ cos �] + cos � [sin �] ………… (3) Let cos �= + cos � ………… sin �= sin � ………… by the equation , and we get- = sin � . cos � + cos � . A sin � = [sin � cos � + cos � sin �] = sin � + � ………… Here and � are constant and can be given by equation and as + cos � + sin � = cos � + sin � + cos � + cos � +sin � = [cos � + sin�] + cos � + sin� + cos � = [cos � + sin�] + + cos � = cos� + sin� = = + + cos � Now the resultant intensity at any point is given as � ∝ for simplicity let � = So � = + + cos � Condition for maxima: For maximum intensity cos � = S t u d y n a m a .c o m
- 122. Unit 5: Wave Optics Page 6 then � = � This is the condition for constructive interference in terms of phase � Then by equation by � = + + � = + So the path difference Δ = � × � Δ = � × � Δ = Δ = I.e. the path difference is the even multiple of , this is the condition of constructive interference in terms of path difference ∆ Figure(12): � Condition for the minima: Again the intensity will be minimum when- cos � = − then � = + � This is the condition for destructive interference in terms of phase � Then by equation � = + − � = − And path difference Δ = � × � Δ = � + � Δ = + i.e. the odd multiple of the half wavelength, this is the condition of destructive interference in terms of path difference ∆ Now the average Intensity: S t u d y n a m a .c o m
- 123. Unit 5: Wave Optics Page 7 �� = ∫ � � ∫ � � �� = ∫ + + cos � � ∫ � � �� = [ � + � + sin �] � [�] � �� = + . � � �� = + �� = � + � The average intensity is the average of the maximum and minimum intensities. It can be given by- Now if = = then, �� = The average intensity is equal to the sum of the separate intensities. Whatever the intensity disappears at the minima is actually appears at the maxima. Thus there is no violation of the law of conservation of energy in the phenomena of interference. Condition for the sustained interference of light. 1. Two sources of light must be coherent. 2. Difference in the amplitudes of the two waves must be small. 3. Sources should be narrow or point source. 4. The separation between two sources should as small as possible. 5. If the interfering waves are polarised then the plane of polarisation must be same. 6. The sources should be monochromatic. 7. Interfering waves from two coherent sources of light should travel in the same direction.S t u d y n a m a .c o m
- 124. Unit 5: Wave Optics Page 8 Fringe width: Consider a narrow monochromatic source and two parallel narrow slits and very close together and equidistance from . Let be the distance between two slits and and be the distance of screen from coherent source. The path difference between the rays reaching from and to is zero so the point has maximum intensity. Considering a point at a distance from . The wave reaches at the point from and hence = − and = + Figure(13): Measurement of fringe width − = [ + ( +) ] − [+ ( −) ] + − = [ + + + . − { + + − .}] + − = + + + . − − − + . + − = − = + Now from the figure If the point is very close to point so − = Δand ≈ = Δ = + Δ = S t u d y n a m a .c o m
- 125. Unit 5: Wave Optics Page 9 Δ = 1. Bright Fringes: For bright fringes the path difference is the integer multiple of the i.e. = . = = This equation gives the distance of the bright fringes from the point . Hence for the ℎ bright fringe (replacing by ) = For next bright fringe + = + Therefor the distance between any two consecutive bright fringes + − = + − � = 2. Dark Fringes: For dark fringes the path difference is an odd multiple of So = + = + Hence the ℎ dark fringe (replacing by ) = + And for the + ℎ dark fringe + = [ + + ] + = + Therefore the distance between two consecutive dark fringes + − = + − + S t u d y n a m a .c o m
- 126. Unit 5: Wave Optics Page 10 + − = [ + − −] + − = . �′ = As the distance between two consecutive bright or dark fringes is same and is called fringe width and denoted by �. � = i. The fringe width is directly proportional to the wavelength of the light used i.e. � ∝ ii. The fringe width is directly proportional to the distance of the slits from the screen i.e. � ∝ iii. The fringe width is inversely proportional to the distance between the slits i.e. � ∝ Shape of the interference fringes: Actually these interfering fringes are hyperbolic in shape, but the eccentricity of fringes is quite large and hence these hyperbolic fringes appear more or less strength lines. Angular Fringe Width: The angular fringe width is defined as the angular separation between consecutive or dark fringes and is denoted by �. Figure (14): Angular fringe width = � � = � + − � � = + − � = + − � = β But � = so � = λ S t u d y n a m a .c o m
- 127. Unit 5: Wave Optics Page 11 Fresnel’s Biprism: The prism is a device to obtain two coherent sources to produce sustained interference. Fresnel used a biprism to show the phenomenon of interference. A biprism is usually a combination of two prisms placed base to base. In –fact this combination is obtained from an optically plane glass plate by proper grinding and polishing. The obtuse angle of the prism is about 9 and other angles are about ′ each. To show the phenomenon of interference a horizontal section of the apparatus is shown in the figure. Figure : Fres el’s Bipris Measurement of � Figure(16): Measurement of d by displacement of lens A bi-convex lens of short focal length is mounted between the bi-prism and the eyepiece by moving the lens along length of bench, two positions and are obtained such as for which the image of sources formed at the same place. For position = ………… For position = = [ = = ………… So on multiplying & we get S t u d y n a m a .c o m
- 128. Unit 5: Wave Optics Page 12 = × = = √ S t u d y n a m a .c o m
- 129. Unit 5: Wave Optics Page 13 Determination of the thickness of a thin sheet of transparent material: Figure(17): Shift in fringes on introducing the thin film Distance travelled by the light in air = −(by the velocity c) Distance travelled by the light in film= (by the velocity ) Time taken by the light to cover this distance = − + But = � ℎ = So we have = − + = − + = − + = + − Thus the path to �. . is eqivelent to an air path + − Now the path difference at Δ = ℎ � � − ℎ � � Δ = − [ + − ] Δ = − − − But − = � � So we have ∆ = − − … … S t u d y n a m a .c o m
- 130. Unit 5: Wave Optics Page 14 but for ℎ maxima, Δ = … … So by equation (1) and (2) − − = = [ + − ] … … Where is the position of the ℎ maxima Now in absence of the plate �. . = The ℎ maxima ′ =[ � ] (in the absence of ) ………….. 1. Displacement of the fringes: If denotes the displacement of the ℎ maxima by introducing the mica sheet, then − ′ = [ + − ] − = � + � − − � ` = − This equation is free from so the displacement of each maxima will be same. 2. Thickness of mica sheet: The displacement of any maxima by introducing a mica sheet of thickness is given by = × − 3. Refractive index of the material of prism: Once if we know the displacement of the fringes and thickness of the film we can calculate the refractive index of the material of the film as- − = × = × + S t u d y n a m a .c o m
- 131. Unit 5: Wave Optics Page 15 Stoke’s treatment of phase change: When a light wave is reflect from the surface of an optically dens medium, it suffers a phase change of � �. .a path difference of Let is an interface separating the denser medium (below i.e. glass) to rare medium (above i.e. air) it. A ray of light of amplitude " " incident on the interface is partially reflecte along the path and patially refracted into the denser medium along .Let is the coefficient of reflection and is the coefficient of transmission then the amplitude of reflected and transmitted wave will be ′ ′ and respectively. Then in case of no absorption of light + = Now if the reflected and refracted rays are reversed the resultant should have the same amplitude ′ ′ as that of the incident ray Figure(18): Reflection and refraction through a surface Figure(19): Ray diagram on reversing the direction of incidence When is reversed it is partly reflected along and partially refracted along as shown in figure. Similarly when the ray is revesed it is partly refracted along and partially reflected along . Now the content along should be zero and that along should be equal to a �. . + ′ = … … … . . … … … … = − ′ = −′ This equation indicates displacement in the opposite direction so according to Stoke’s la , he a light a e coming from a rare medium an additional phase � is introduced in it. S t u d y n a m a .c o m
- 132. Unit 5: Wave Optics Page 16 Interference in thin film: Consider a thin film of equal thickness and refractive index > . A monochromatic light ray incident at angle � is partially reflected and partially transmitted as shown in figure Figure(20): Reflation and transmission of light through a thin film Reflected system: In reflected system two waves & are in the position to interfare so the path difference between & Δ = ℎ � � − ℎ � �� Δ = + − Now from figure it is clear that = Δ = − But from Δ = sec Then = sec = sec And from Δ = sin � S t u d y n a m a .c o m
- 133. Unit 5: Wave Optics Page 17 = sin � But = + = + sin � …….…. (1) But from Δ and Δ = tan ⟹ = tan ⟹ = tan and = tan ⟹ = tan ⟹ = tan but = tan = tan …….…. (2) putting the value of and in from so we have = tan + tansin � = tan � . sin � = . sin cos . sin � Multiplying and dividing by sin we get = . sin cos . sin � sin . sin = . sin cos . = sin � sin = . sin cos Therefor ∆ = − Putting the value of and , we get ∆ = sec − . sin cos ∆ = sec − sin cos ∆ = cos − sin cos ∆ = − sin cos ∆ = cos cos ∆ = cos …….…. (3) S t u d y n a m a .c o m
- 134. Unit 5: Wave Optics Page 18 A. In Reflected system: The ray undergoes a reflection from the densor medium so a additional path difference of must be added, then ∆ = cos + 1. Condition for constructive interference: For constructive interference ∆= So cos += cos = − cos = − (where = , , , … … 2. Condition for the destructive interference: For destructive interference ∆= + So cos += + cos = + − cos = (where = , , , … … B. In transmitted system: In the transmitted system there will be no additional path difference so ∆ = cos 1. Condition for constructive interference: Condition constructive interference is ∆= then cos = Where = , , , … … 2. Condition for destructive interference: Condition for the destructive interference is ∆ = + Where = , , , … . . then cos = + So the reflected and transmitted interference patterns are complimentary. Colour in thick film: A thick film do not show the any colour in reflected system when illuminated with an extended source of light. S t u d y n a m a .c o m
- 135. Unit 5: Wave Optics Page 19 Wedge shape film: A wedge shape film is one whose surfaces are inclined at a certain small angle. Figure shows a thin wedge shape film of refractive index bounded by two plane surfaces and inclined at an angle �. Let a parallel beam of monochromatic light falls on the upper surface normally and the surface is viewed in the reflected and refracted system then alternate dark and bright fringes becomes visible. Figure(21): Reflection and refraction through a wedge shape film Let the light is incident nearly normally at a point on the film, the path difference between the rays reflected at the upper and lower surface is = where is the thisckness of the film at . Reflected system: Figure(22): Reflection through a wedge shape film The condition for the maximum intensity (bright fringes): In the reflected system according to the Stokes treatment an additional path difference of is introduced in the ray reflected from the upper surface. Hence the effective path difference between the two rays will be ∆= +and the condition for the bright fringes is ∆= So + = S t u d y n a m a .c o m
- 136. Unit 5: Wave Optics Page 20 = − = − The condition for the minimum intensity (Dark fringes): The condition for the destructive interference is Δ = + + = + + = + = + − = Transmitted System: Figure(23): Refraction through a wedge shape film In the transmitted system there will be no additional path difference so the effective path difference will be ∆= The condition for the maximum intensity (bright fringes): The condition for the maxima is given as ∆= so = S t u d y n a m a .c o m
- 137. Unit 5: Wave Optics Page 21 The condition for the minimum intensity (Dark fringes): The condition for the minima in interference is ∆= + So = + Fringe width: For ℎ dark fringe let this fringe observed at a distance from the edge, where the thickness of fringe is From figure (23-B) it is clear that = � then = � So = So � = …….. Similarly for + ℎ fringe Figure(23-B) + � = + …….. By equation (1) and (2) + � − � = + − + − � = + − But + − = � so �� = � = � S t u d y n a m a .c o m
- 138. Unit 5: Wave Optics Page 22 Newton’s Ring: For atio of Newto ’s Ri g: When a Plano-convex lens of large radius of curvature is placed with its convex surface in contact with a plane glass plate, an air film of gradually increasing thickness is formed between the upper surface of the plan glass plate and the lower surface of the Plano-convex lens. If a monochromatic beam of light is allow to fall normally on the upper surface of the film then, alternative bright and dark concentric fringes with their centre dark are for ed. These fri ges or ri gs are k o as the Ne to ’s ri gs. Figure : For atio of the Ne to ’s ri gs Experimental arrangement: The experimental arrangement is shown in the figure. Light rays reflected upwards form the air film, superimpose each other and interference takes place, due to which the alternative bright and dark concentric rings are formed those can be seen by the telescope.S t u d y n a m a .c o m
- 139. Unit 5: Wave Optics Page 23 Figure(25): Ne to ’s ri g experimental arrangement The fringes are circular because the air film is symmetrical about the point of contact of the lens with the plane glass plate. Theory: The rings are formed both in reflected and refracted part. Reflected Part: As the films are obtained in the reflected part the effective path difference between the interfering rays is given by Δ = cos + ……. (1) Where is the refractive index of the film, is the thickness of the film, is the angle of incidence. The factor is account for the phase change of � on reflection from the lower surface of the film. For air = and for normal incidence = then Δ = + ……. (2) Central fringe: At the centre i.e. at the point of contact = So Δ = This is the condition for the minimum intensity, hence the central fringe will be dark. For Constructive interference (i.e. maxima): The condition for the constructive interference by thin film is given as Δ = S t u d y n a m a .c o m
- 140. Unit 5: Wave Optics Page 24 + = = − Then = − It is the condition for constructive interference Where = , , …. For destructive interference (i.e. minima): The condition for the destructive interference by the thin film is given as Δ = + + = + = It is the condition for minima Where = , , … .. Shape of the fringes: As in air film remains constant along the circle with its center at the point of contact, the fringes are in the form of the circles, since each film is the locus of the constant thickness of the air film. These fringes are known as the fringes of equal thickness. Figure : Dia eter of Ne to ’s ri g Figure : Shape of the Ne to ’s ri gs So the diameter of the bright ring is proportional to the square root of the odd number. Diameter of Bright ring: Let is the lens placed in the glass plate the point of contect is shown by . Let is the radius of the S t u d y n a m a .c o m
- 141. Unit 5: Wave Optics Page 25 cuvature of the curved surface of the lens. Let e the radius of the Ne to ’s ri g here the fil thi k ess is from the right angle Δ = − + = + − + As the air film is very thin so can be neglected = − + = = Substituting the value of in the equation for bright ring i.e. = + So = + Radius of ℎ brigth ring = √ + So the diameter of ℎ bright ring = So = √ + = √ + = √ √ + ∝ √ + Diameter of the Dark ring: Condition for dark ring is = And = So on comparing these two equation we get = S t u d y n a m a .c o m
- 142. Unit 5: Wave Optics Page 26 = = √ the diameter of the dark ring = = √ = √ = √ √ ∝ √ The diameter of dark ring is proportional to the square root of even number. So, as we go far from the centre the thickness of the ring reduces, this limits the number of rings in any pattern that means infinite number of ring can-not be seen. Newton’s Rings in transmitted part: In case of transmitted light, the effective path difference is cos Figure : Ne to ’s ri g refle tio s ste Figure : Ne to ’s ri g i Refra ted s ste Transmitted part: Constructive interference: = Destructives interference: = + S t u d y n a m a .c o m
- 143. Unit 5: Wave Optics Page 27 S.N. Name Figure Condition for Reflected part Condition for transmitted part1 ParallelThinfilm Constructive: (Maxima) cos =− cos= Destructive: (Minima) cos= cos= − 2 Wedgeshapefilm Constructive: (Maxima) =− = Destructive: (Minima) = =− 3 Neto’srig Constructive: (Maxima) =− = Destructive: (Minima) = = − S t u d y n a m a .c o m
- 144. Unit 5: Wave Optics Page 28 Applications of the Newton’s Ring: 1. Determination of wavelength of light Let and + respectively the diameters of the ℎ and + ℎ dark rings where is an integer. Then by equation = …………………………………………………… = � Similarly the diameter of + ℎ ring is given by + = + …………………………………………………… So by equation & + − = + − + − = + − + − = + − = = + − 2. Determination of refractive index of any liquid: For air film ( + − )�� = = � … … … … … . In liquid ( + − ) � � = � … … … … … … … . . By equation (1) and (2) ( + − )�� ( + − ) � � = � = ( + − )�� ( + − ) � � S t u d y n a m a .c o m
- 145. Unit 5: Wave Optics Page 29 Diffraction: 1. Bending of the light form the sharp edges of the obstacle is called the diffraction. 2. The intensity of light outside the geometrical shadow of an obstacle and presence of light within its geometrical shadow is called the diffraction of light. 3. The deviation of light from the rectilinear path is called the diffraction. S.N. Fres el’s diffra tio S.N. Frau hofer’s Diffra tio 1. Either the source of light or screen or both are at finite distance form obstacle or aperture. 1. Both the screen and source are effectively at infinite distance from the obstacle or aperture. 2. Wavefront may be of any type i.e. plane, spherical or cylindrical. 2. The incident wavefront is always a plane wavefront. 3. No need to use the lenses. 3. Lenses are required. 4. Diffraction pattern is the image of obstacle or aperture. 4. Diffraction pattern is the image of the source. 5. Intensity of light at any point is found by the half period zone method which is not accurate. 5. Intensity at any point is measured by the mathematical treatment which is more accurate method. Difference between diffraction and interference: S.N. Interference Diffraction 1. This phenomenon is the result of interaction taking place between two separate wave front originating from two coherent sources. 1. This phenomenon is the result of interaction of light between the secondary wavelengths originating from different points of the same wavefronts. 2. The regions of minimum intensity are usually almost perfectly dark. 2. The regions of minimum intensity are not completely dark. 3. Interference fringes may or may not be of same width. 3. Diffraction fringes are not of the same width. 4. All maxima are of same intensity. 4. The maxima are of varying intensities. S t u d y n a m a .c o m
- 146. Unit 5: Wave Optics Page 30 Fraunhofer’s diffraction at a single slit: Let parallel beam of monochromatic light of wavelength be incident normally upon a narrow slit =. A ordi g to Hu ge s’s theor a pla e a e fro t is i ide t or all o the slit . Each point of sends out sencodory wavelets in all directions. The rays proceeding in the same direction as the incident rays are focused on , while those diffracted through an angle � are focused at . Figure(27 : Frau hofer’s diffra tio To find the intensity at point , we drop a normal on the ray , the optical path from each point of the plane to point will be equal. Now the path difference between the wavelets reaching the point from point and is Δ = But from Δ = sin � = sin � = sin � Δ = sin � so ℎ � = � × Δ ℎ � = � × sin � … … …(1) Now if we consider n number of infinite point sources of secondary wavelengths on the plane wave front then this can be divided into equal parts, so phase difference between the waves obtained at the point from any two consecutive parts � = × � . sin � … … … S t u d y n a m a .c o m
- 147. Unit 5: Wave Optics Page 31 Now to find the intensity at point there are following two methods are available 1. Phase diagram Method. 2. Integral Method. Phase diagram method: In the figure, draw vectors , , … … … … ..such that the magnitude of each vector is and angle between the two consecutive vectors is � . The vector gives the resultant vector. Let the magnitude of the resultant vector is �. If is the centre of the polygon formed by the vector then by the simple geometry we can see that each vector substance � at the centre and the angle substaended by the resultant vector at the centre is �. Let and are the normal drawn of first vector and resultant vector from centre . Figure(28): Phase diagram from right angle triangle Δ = sin ( � ) = sin ( � ) But = = [ =] = sin ( � ) … … … Similarly from Δ = sin ( � ) = sin ( � ) But = �� so �� = sin � … … … S t u d y n a m a .c o m
- 148. Unit 5: Wave Optics Page 32 By (3) and (4) �� � = sin � sin � � = sin � sin � … … … Now putting the value of � from the equation no. (2) we get � = sin � sin � sin � sin � … … … Let � si � = Then � = sin sin Now is very small so sin ≈ Then � = sin So � = sin � = sin Where = Now the intensity � ∝ � � = � [Where is a constant � = [ sin ] � = [ � ] � = � [ sin ] … . … . . Where � = Conditions for maxima and minima: From the equation � = �[ si ] it is clear that the resultant intensity � at point on the screen depends on the angle of diffraction � or on . For maxima, the derivation of � with respect to must be zero. �. . S t u d y n a m a .c o m
- 149. Unit 5: Wave Optics Page 33 {� [ sin ] } = � . . ( sin ) ( cos − sin ) = …… Condition for Minima: For the minima, the first term in the above equation (8) should be zero i.e. sin = sin = = ± � Putting the value of we get- � sin � = ± � sin �= ± Where = , , , , … … … …. Condition for maxima: Now the second term of equation (8) will show the condition for maxima- i.e. cos − sin = cos = sin = sin cos = tan The condition for maxima is = tan To find the value of for which the above condition may hold, we draw two curves = … … …(9) = tan … … …(10) On the same graph as shown S t u d y n a m a .c o m
- 150. Unit 5: Wave Optics Page 34 Figure(29): Graph The value of at the points of intersection of these two curves satisfy the equation = tan At the central maxima: � = ⟹ = So � sin � = So, the intensity at the principle maxima � = � ( sin ) Applying the limits we get lim → ( sin ) = So � = � (maximum) So at the principle maxima the intensity will be maxima. Intensity for subsidiary maxima: For subsidiary maxima the value of sin must be maximum, for this the value of = � si � i.e. the value of sin �must be maximum �. . sin � = � = ± + � �. . � = ± � , ± � , ± � … … … so at the, S t u d y n a m a .c o m
- 151. Unit 5: Wave Optics Page 35 first subsidiary maxima � = � [ sin � � ] = � Second subsidiary maxima � = � [ sin � � ] = � Third subsidiary maxima � = � [ sin � � ] = � S t u d y n a m a .c o m
- 152. Unit 5: Wave Optics Page 36 Fraunhofer’s diffraction at a double slit: Let a parallel beam of monochromatic light of wavelength be incident normally upon two parallel slit and , each of width ′ ′ separated by opaque space of width d. Figure (30): diffraction at a double slit. Suppose each slit diffracts the beam in a direction making an angle � with the direction of the incident beam. From the theory of diffraction at a single slit the resultant amplitude is � = sin Where = � si � and is a constant These two slits can be considered as two coherent source placed at the centre of the slits. Then resultant intensity at point will be the result of interference between these two waves of same amplitude and phase difference � Now the resultant amplitude at point = √ + + cos � (interference) So = √ � + � + � � cos � = √ � + cos � = �√ . cos � S t u d y n a m a .c o m
- 153. Unit 5: Wave Optics Page 37 = � cos � But we know that � = si So = ( sin ) cos ( � ) … … …(1) Therefore the resultant intensity at point will be � ∝ � = Where is a constant Putting the value of from equation (1) � = . ( sin ) cos ( � ) … … … Let � = Hence resultant intensity � = � ( sin ) cos ( � ) … … … Condition for Minima: From the equation (3) it is clear that the intensity will be minimum when sin = ⇒ = ± � Where = , , … … … … …but ≠ So putting the value of we get ⇒ � sin � = � sin �= Where m= 1, 2, ,…… ut ≠ m = is the condition for the maxima Condition for maxima: From the equation (3) it is clear that the intensity will be maximum when term si will be maximum. sin = ⇒ = − � ⇒ � sin � = − � ⇒ sin �= − Where n=0,1,2, , ….. S t u d y n a m a .c o m
- 154. Unit 5: Wave Optics Page 38 Missing order maxima: The condition for the interference maxima is given as ∆= But Δ = sin � (from single slit) So the condition for the interference maxima will be + sin �= ……… And the condition for the diffraction minima is given as sin �= ………. Figure(31): Intensity graph of double slit diffraction For certain value of certain interference maxima become absent from the pattern. Let for some value of � the following two conditions be satisfied simultaneously Dividing the equation by + = Case I: If = Then + = ⇒ = ⇒ = If = , , … … … . . ⇒ = , , … … … … .. This means that , , … …etc order of interference maxima will be missed. S t u d y n a m a .c o m
- 155. Unit 5: Wave Optics Page 39 If = Case I: If = Then + = = ⇒ = If = , , , … … ….then = , , 9 … … .. This means that , , 9 … ….etc. will missed. S t u d y n a m a .c o m
- 156. Unit 5: Wave Optics Page 40 Fraunhofer’s Diffraction of � Parallel slit: Diffraction Grating: It is an arrangement consisting of several parallel and equidistant slits each of equal width. It is constructed by drawing the several equidistance parallel lines on an optically plane glass plate with a pointed diamond. The distance between two consecutive slits is + = which is called the grating element. Genrally the value of for the grating to be used with the visual light is of the order of − �. . lines drawn on length of the grating. Figure(32): Diffraction Grating Figure(33): Diffraction Grating Theory: In figure, is a grating of parallel and equidistance slits , , … … … …the width of each slit is ′ ′ and width of opaque space between the two consecutive slits is ′ ′. �. .the grating element = +. Let a plane wavefront of wavelength is incidents normally on the grating. Then diffracted by it is focused on a screen by means of a convergent lens on screen. Intensity distribution: It is clear from the figure that diffracted waves do not reach a point on the screen in the same phase since their optical paths are not equal. The path difference between the two consecutive wave is Δ = sin � Therefor the phase diff. � = � × ∆ Wave diffracted at an angle � from each slit is � = si where = � e si � [by single slits diffraction] Figure(34): Phase Diagram S t u d y n a m a .c o m
- 157. Unit 5: Wave Optics Page 41 Now we can find the resultant amplitude due to the superposition of such waves by phase diagram method. In figure we draw vector , , … ..such that magnitude of each vector is � and the angle between the consecutive vector is �. The vector which joints the initial points of first vector and final point of last vector is and this vector sustained an angle �. and are the normal plotted from the centre of polygon on first and resultant vectors. From the figure in Δ = sin � ⇒ = sin � But = = �� � = sin � … … … . Similarly in Δ ⇒ = sin ( � ) ⇒ = sin ( � ) But = = = sin ( � ) …………. (2) Dividing equation (2) by (1) � �� = sin �� sin � � = [ sin �� sin � ] = � [ sin �� sin � ] On substituting the value of � we get = ( sin ) [ sin �� sin � ] Multiplying and dividing by we get S t u d y n a m a .c o m
- 158. Unit 5: Wave Optics Page 42 = ( sin ) [ sin �� sin � ] So the resultant intensity will be given as � ∝ � = Where is the proportionality constant Putting the value of from (3) we get � = ( sin ) [ sin �� sin � ] � = � ( sin ) [ sin �� sin � ] … … …(3) Where � = In this expression the term si represents the intensity due to diffraction due to a single slit, while second term [ si �� 2 � si � 2 ] represents the intensity due to interference of wave obtained from slits. Condition for Principle Maxima: For principal maxima the path difference will be zero so the phase diff then sin � = ⇒ � = ± �where = , , … ..then sin �� is also zero and in the limit when sin � → the value of term [ si �� 2 si � 2 ] will be . Hence from equation the resultant intensity will be maximum. i.e. lim� 2 → [ sin �� sin � ] = So we have � = � ( sin ) …….. Which is the intensity at principle maxima i.e. similar to the intensity by a single slit. Condition for Minima: From the equation (3) it is clear that the intensity will be minimum when sin �� = but sin � ≠ i.e. sin ( � ) = ⇒ � = ± � S t u d y n a m a .c o m
- 159. Unit 5: Wave Optics Page 43 Where = , , , … … But � = � si � So � sin � = � … … …(5) sin �= This is the condition for the minimum intensity for N-slit diffraction. Condition for Secondary maxima: Condition for maxima is � � = So from equation (3) � [� ( sin ) { sin �� sin � } ] = � ( sin ) � { sin �� sin � } = � ( sin ) [{ sin �� sin � } sin � . � cos �� − sin �� .cos � {sin � } ] = sin ( � ) . cos ( � ) − sin ( � ) . cos ( � ) = sin ( � ) . cos ( � ) = sin �� . cos � ……… 6) ⇒ tan ( � ) = tan ( � ) … … … again by equation tan ( � ) cos ( � ) = sin ( � ) ⇒ sin ( � ) = tan ( � ) cos ( � ) sin ( � ) = tan � sec �� sin ( � ) = tan � √ + tan �� S t u d y n a m a .c o m
- 160. Unit 5: Wave Optics Page 44 sin ( � ) = tan � √ + tan � By (7) On squaring both sides we get sin ( � ) = si � 2 c s � 2 + tan � sin �� sin � = [ + tan � ] . cos � sin �� sin � = [cos � + tan � . cos � ] sin �� sin � = cos � + si 2 � 2 c s2 � 2 cos � sin �� sin � = [cos � + . sin � ] sin �� sin � = − sin� + . sin � [ sin �� sin � ] = + − sin � [ sin �� sin � ] = + − sin � ……… (7) So by equation (3) and (7) � = � ( sin ) × + − sin � This is the expression for the intensity at the subsidiary maxima in N-slit diffraction. S t u d y n a m a .c o m
- 161. Unit 5: Wave Optics Page 45 Resolving Power of Optical Instrument: To distinguish two close object is called geometrical resolution and the ability of an optical instrument to distinguish the image of very close object is called the resolving power of that optical instrument. The ability of instrument to produce the separate diffraction pattern is known as resolving power. Raleigh’s criterio of resolutio : According to this criterion two sources are resolved by an optical instrument when the central maxima in the diffraction pattern is fall over the first minima in the diffraction pattern of the second maxima and vice versa. In order to illustrate the criterion let us consider the resolution of two wavelengths and . Figure shows the intensity curve of the diffraction pattern of two wavelengths. The diffraction in wavelength is such that their principal maxima are separately visible. There is a distinct point of zero intensity in between the two. Hence the two wavelengths are resolved. Figure(35): Two separate maxima In the case when there is small dip between the maxima of and such that the central maxima of wavelength coincide with the first minima of and wise versa as shown in the figure (36). The resultant intensity curve has a dip in the middle of the two central maxima. Thus two wavelengths can be distinguished from one another Figure(36): Condition of just resolution If the difference between the two wavelength and is so small that the maxima corresponding to wavelength come still closer as shown in the figure (37) the resultant intensity curve in this case is quit smooth without any dip, thus wavelengths cannot be resolved. Figure(37): Maxima that cannot be resolve Resolving power of Grating: The resolving power of a diffraction grating is defined as the capacity to form separate diffraction maxima of S t u d y n a m a .c o m
- 162. Unit 5: Wave Optics Page 46 two wavelengths without which they are very close to each other. This is measured by . Let is a plane diffraction gratting having grating element = +and total numbers of number of slits. Let a beam of light having two wavelengths and + is normally insidented on the gratting. is the ℎ primary maxima of spectral line of wavelength at an angle of diffraction � and is the ℎ primary maxima of wavelength + at diffraction angle � + � The principal maxima of in � direction will be + sin �= …………………………………………..…….. And the equation of minima + sin �= Where is an integer except , , … … ., because for these values of the condition for maxima is satisfies and we obtain diffraction maxima. Figure(38): Formation of diffraction pattern by a grating Now first maxima adjacent to ℎ principle maxima + sin � + � = + ………… (2) And first minima + sin � + � = + ………… (3) Now multiplying the equation (2) by we have + sin � + � = + ………… (4) By & + = + + = + = S t u d y n a m a .c o m
- 163. Unit 5: Wave Optics Page 47 S t u d y n a m a .c o m
- 164. Unit –V: Semiconductors Page 1 Unit-6 Semiconductors Syllabus: Crystalline and Amorphous solids, Band theory of solids, mobility and carrier concentrations, properties of P-N junction, Energy bands, hall effect, VI characteristics of photodiode, Zener diode and photovoltaic cell S t u d y n a m a .c o m
- 165. Unit –V: Semiconductors Page 2 Crystalline and amorphous solids: Solids can be broadly classified in to following three types- 1) Crystalline solids 2) Amorphous solids or non-crystalline solids 3) Polly crystalline solids Crystalline solids If the atoms or the molecules in a solid are arrange in some regular fashion then it is known as crystalline solids. Hence in a crystalline solid the atoms are arranged in an orderly three dimensional array that is repeated throughout the structure. This is shown in the figure (1-a). The metallic crystal are , , etc. the non-metallic crystals are , �, etc. Figure (1) Amorphous solids or non-crystalline solids: Amorphous means without form. When the atoms or molecules in a solid are arrange in an irregular fashion then it is known as amorphous solids which is shown in the figure (1-b). The examples are , � � , etc. Polly crystalline solids There are some solids which are composed of many small regions of single crystal material and are called polycrystalline solid. Hence the atoms in polycrystalline solids are so arranged that within certain sections some short of pattern of the atoms exists but the various sections are randomly arranged with respect to each other as shown in the figure. S t u d y n a m a .c o m
- 166. Unit –V: Semiconductors Page 3 Difference between amorphous and crystalline solids: Amorphous Solids Crystalline Solids 1. Solid those do ’t have defi ite geo etrical shape. 1. Crystalline solids have the characteristic geometrical shape. 2. Amorphous solids do not have particular melting point. They melt over a wide range of temperature. 2. They have sharp melting point. 3. Physical properties of amorphous solids are same in different direction. i.e. those solids are isotropic. 3. Physical properties of crystalline solids are different in different directions. This phenomenon is known as anisotropy. 4. Amorphous solids are unsymmetrical. 4. When crystalline solids are rotated about an axis there appearance does not changes. This shows that they are symmetrical. 5. Amorphous solids do not beak at fixed cleavage planes. 5. Crystalline solids cleavage along particular direction at fixed cleavage planes. How does the band forms in the solids: We know that the atoms are arranges in a periodic manner in a solid and they formed the crystal. In an atom the electron are revolves in different orbits according to their energy. If we take each individual atom and find the energy of electron then this energy becomes identical for each corresponding atom for every electron. But as in solids the atoms are not free but they interacts one-another so the energy become slightly more or less for some of the electrons and if we plot the energies we get an energy band in solids. There may be a number of energy bands in a solid but two of them are of our interest 1) Valance band: The energy band plotted by energy of the electrons those are revolving in the outermost orbit is called the valance band. 2) Conduction band: The electrons those are revolving in the outermost orbit are loosely bounded and can be separated by giving some energy to those electrons. Now those electrons are free to move inside the crystal and they are not concern to any individual atom. The energy band plotted by the energy levels of the free electrons is called the conduction band. Since these electrons are free to move inside the crystal and are responsible for conduction of electricity is known as conduction band. There is a gap in between the upper most energy level of valance band and lowest energy level of conduction band is known as forbidden energy gap. Because these energy levels cannot be occupy by S t u d y n a m a .c o m
- 167. Unit –V: Semiconductors Page 4 any electron. On the basis of this band theory we can classify the conductors, insulators and semiconductor. Types of materials on the basis of the electrical conduction: Materials can be classified into three different categories on the basis of their electrical conductivity. Figure(7): Classification of the materials Conductors: Conductors are those materials which have completely or partially filled conduction band and the forbidden energy gap between the conduction band and valance band is zero. So the electrons those are in valance band also available in conduction band to flow the current.Figure(8): Conductors Materials concudutor Semiconductors Intrinsic Semiconductor Extrinsic Semiconductors n-type Semiconductors p- type Semiconsuctors insulators S t u d y n a m a .c o m
- 168. Unit –V: Semiconductors Page 5 Insulator: The materials which does not allows to flow the current from them, are called the insulator. In case of insulator there is a large energy gap between the conduction band and valance band of about ~ � so it is impossible to lift the electron from valance band by giving some energy to the conduction band. Therefore materials are insulators. Figure(9): Insulator Semiconductor: Semiconductors are those materials which has there electrical conductivity somewhere between conductors and insulators. It means these materials behave as an insulator at low temperature while at the elevated temperature they shows some electrical conductivity. The semiconductor has totally empty conduction band at absolute zero, but at elevated temperature some of the electrons jumps from the valance band to Figure(10): conduction band as the forbidden energy gap between the semiconductors is of moderate size of about . ~ . � and this much amount of energy can be provided easily to the valance band electrons so the current can flow in this type of material. Semiconductors are of two types: 1) Intrinsic semiconductors 2) Extrinsic semiconductors Intrinsic semiconductors: Intrinsic semiconductors are pure semiconductor as and �. These materials have four electrons in their outermost orbit. To complete the octal an / � atom form the covalent bonds with four other neighbouring / � atoms as shown in the figure. S t u d y n a m a .c o m
- 169. Unit –V: Semiconductors Page 6 Due to this, no electron is available in conduction band at low temperature and therefor it behave as an insulator, but at elevated temperature, due to some thermal agitation some of the covalent bonds in the semiconductor material breaks, due to which an electron hole pair creates. The electron is now available in conduction band even at the room temperature and hole is available in valance band. The hole is a vacancy created in the valance band is filled by the neighbouring electron and thus electron and hole starts flowing in valance band and due to both electron and hole the electric conduction in material is now possible. Figure(11): Intrinsic semiconductor Extrinsic Semiconductors: In intrinsic semiconductors only 6 electrons per cubic meter contributes to the conduction of electric current hence these are of no particular use. If a small amount ~ �� of pentavelent or trivelent impurity is introduce into a pure / � crystal, then the conductivity of the crystal increases appreciably and the crystal becomes an extrinsic semiconductor. Again, extrinsic semiconductors are of two types 1) − extrinsic semiconductor 2) � − extrinsic semiconductor � − ����Extrinsic semiconductor If a pentavalent impurity as , �, is added to / � then four electrons of outermost orbit of these atoms creates covalent bonds while the extra electron which is free in the crystal enhanced the electrical conductivity of the materials. In this type of the crystal the current flows due to a negatively charge particle i.e. electron so the materials are known as − semiconductor materials. The impurity atom introduce discrete energy level for the electron just donates the extra electron in the crystal therefor these are called donor impurity levels. S t u d y n a m a .c o m
- 170. Unit –V: Semiconductors Page 7 Figure(12): − Semiconductors crystal Figure(13): Donor level in − extrinsic semiconductor � − ����Extrinsic semiconductors: When a trivalent , , , atom replaces an / � atom in a crystal �� , only three valance electrons are available to form covelent bonds with neighboring / � atoms. This result into an empty space or a vacant position called hole. When a voltage is applied to the crystal then an electron bound to a neighbouring / � atoms occupy the hole position there by creating a new hole. This process continues and holes moves in a crystal lattice. This type of semiconductor is called the � − semiconductor. The trivalent impurity atoms introduces vacant discrete energy levels just above the top of the valance band. These are called acceptor impurity level, which are only . � above the valance band in case of and . � in case of �. S t u d y n a m a .c o m
- 171. Unit –V: Semiconductors Page 8 Figure(14): p − typesemiconductor Figure (15): Acceptor level in extrinsic semiconductors. S t u d y n a m a .c o m
- 172. Unit –V: Semiconductors Page 9 Charge mobility: When an electric field, ⃗ is applied to a conductor or semiconductor then the electrons (in opposite to ⃗ ) and holes (in the same direction to ⃗ ) starts flowing with drift velocity, . This drift velocity is proportional to the applied field ⃗ , i.e. ∝ ⃗ = � ⃗ Here � is proportionality constant which is known as mobility of charge carriers. So � = | | |⃗ | So mobility relates the drift velocity to electric field. Mobility gauges how easily current carrier can move through a piece of conductor or semiconductor. Charge concentration: Intrinsic semiconductors: 1) Electrons in conduction band behave as free particle with effective mass 2) Number of conduction electrons per cubic meter whose energies lies between ⃗ and ⃗ + ⃗⃗⃗⃗⃗ is given as - = … . Where is the density of states at bottom of the condution band and it is given as per quantum mechanics as = � ℎ − … . Here is the energy at the bottom of the conduction band. And is the Fermi-Dirac probability function, which is given as- = + ( �−� �� ) … . . Where is the Fermi level, is the absolute temperature and is Boltzmann constant. Now since electron may have energies between to ∞ in conduction band so total number of electron will be given by integrating (1) = ∫ ∞ �� S t u d y n a m a .c o m
- 173. Unit –V: Semiconductors Page 10 = ∫ � ℎ − + ( �−� �� ) ∞ �� = � ℎ ∫ − + ( �−� �� ) ∞ �� As ≫ and − ≫ so + �−� �� ≈ �−� �� So = � ℎ ∫ − ( �−� �� ) ∞ �� = � ℎ ∫ − ( � −� �� ) ∞ �� = � ℎ ∫ − ( � −��+��−� �� ) ∞ �� = � ℎ ∫ − − �−�� �� ( � −�� �� ) ∞ �� = � ℎ � −�� �� ∫ − − �−�� �� ∞ �� Let �−�� �� = so that = . Limits:- As → ⟹ = And → ∞ ⟹ = ∞ so = � ℎ ( � −�� �� ) ∫ −� . ∞ = � ℎ ( � −�� �� ) ∫ −� ∞ But ∫ / −�∞ = √� = � ℎ � ( � −�� �� ) [ √� ] = ( � ℎ ) ( � −�� �� ) … . . This is the density or concentration of electron in conduction band in intrinsic semiconductor. Hole concentration in valance band: Since holes are created by removal of an electron so Fermi function will be − So let us calculate the Fermi-Dirac distribution for holes as S t u d y n a m a .c o m
- 174. Unit –V: Semiconductors Page 11 − = − + ( �−� �� ) − = − [ + ( �−� �� ) ] − − = − [ − ( �−� �� ) ] − = − + ( �−� �� ) − = ( �−� �� ) … . . And for the top of the valance band the density of the states will be given as = � ℎ ℎ � − Here ℎ is the effective mass of the hole near the top of the balance band. So the hole concentration will be given as ℎ = � ℎ ℎ � − ( �−� �� ) On integrating we get ℎ = ∫ � ℎ ℎ � − ( �−� �� ) �� −∞ ℎ = � ℎ ℎ ∫ � − ( �−� �� ) �� −∞ ℎ = � ℎ ℎ ∫ � − ( �−��+��−� �� ) �� −∞ ℎ = � ℎ ℎ ∫ � − �−�� �� ( ��−� �� ) �� −∞ ℎ = � ℎ ℎ ( ��−� �� ) ∫ � − − ��−� �� �� −∞ Let ��−� �� = ⟹ − = And limits As ⟶ −∞ ⟹ ⟶ ∞ And ⟶� ⟹ ⟶ ℎ = � ℎ ℎ ( ��−� �� ) ∫ −� − ∞ ℎ = − � ℎ ℎ ( ��−� �� ) ∫ −� ∞ S t u d y n a m a .c o m
- 175. Unit –V: Semiconductors Page 12 ℎ = − � ℎ ℎ ��−� �� ∫ / −� ∞ ℎ = � ℎ ℎ ��−� �� ∫ / −� ∞ ℎ = � ℎ ℎ ��−� �� √� ℎ = ( � ℎ ℎ ) ��−� /�� … … This relation gives the density or concentration of holes in the valance band of an intrinsic semiconductor. Intrinsic concentration of charge On combining the equation number (4) and (6) we get the following expression for the product of electron-hole concentration ℎ = ( � ℎ ) ( � −�� �� ) ( � ℎ ℎ ) ( ��−� �� ) ℎ = ( � ℎ ) ℎ ( � −��+��−� �� ) ℎ = ( � ℎ ) ℎ ��−�� �� ℎ = − � �� Where = ��� ℎ ℎ and � − = − (A) Fermi levels in intrinsic semiconductors: In an intrinsic semiconductor electron and holes are always generated in pair so = ℎ i.e. ( � ℎ ) ( � −�� �� ) = ( � ℎ ℎ ) ( ��−� �� ) −( ��−� �� ) . ( � −�� �� ) = ℎ ( � −��−��+� �� ) = ℎ Taking log of both sides ( − − � + ) = ln ℎ − − � = ln ℎ S t u d y n a m a .c o m
- 176. Unit –V: Semiconductors Page 13 = + � + ln ℎ Now if the effective mass of the electrons and holes are same then = + � This shows that the Fermi level lies exactly in the middle of the forbidden energy gap as depicted in figure. The Fermi level can also be defined as the energy level at which there is a . probability of finding an electron. It depends on the distribution of energy level and number of electron available. (B) Fermi level in extrinsic semiconductor: (i) n-type extrinsic semiconductor: At usual temperature all the donor level will be fully activated and the donor atoms will be ionised. It means the density of electrons will be increase. It means the density of electrons in the conduction band will be approximately equal to the density of donor atoms, i.e. = (density of donor atoms) Then = = [ � ℎ ] . ( � −�� �� ) Let [ � ℎ ] = = Then = . ( � −�� �� ) = ( � −�� �� ) = −( � −�� �� ) Taking log on both sides ln [ ] = − ( − ) − = − ln[ ] = − ln[ ] It shows that the Fermi level lies below the bottom of the conduction band, as shown in the figure. In intrinsic semiconductor, Fermi level lies in the middle of the forbidden energy indicating equal concentrations of free electrons and holes. When a donor type impurity is added to the crystal, then if we assume that all the donor atoms are ionised, the donor electrons will occupy the states near the bottom of the conduction band. Hence it will be more difficult for the electrons from the valance band to cross the energy gap by thermal agitation. Consequently, the number of holes of the valance band is decreased. Since Fermi level is a measure of the probability of occupancy of the allowed energy states, for n-type S t u d y n a m a .c o m
- 177. Unit –V: Semiconductors Page 14 semiconductors must move closer to the conduction band, as shown in the figure. (ii) P-Type extrinsic semiconductor: When an acceptor type impurity is added it also modifies the energy level diagram of semiconductor and makes the conduction easier. The presence of impurity creates new energy levels which are in the gap in the neighbourhood of the top of valence band of energies as shown in figure. Ambient temperature results in ionisation of most acceptor atoms and thus an apparent movement of holes takes place from the acceptor level to the valance band. The energies for holes are highest near the valance band decreases vertically upward in the energy level diagram. Alternatively, one may say that electrons are accepted by the acceptors and these electrons are supplied form the valance band, thus leaving a preponderance of holes in the valance band. If we assume that there are only acceptor atoms present and that these are all ionised, we have ℎ = = [ � ℎ ℎ ] . ( ��−� �� ) = � ( ��−� �� ) Where � = [ ��ℎ�� ℎ ] / = � = ( ��−� �� ) � = −( ��−� �� ) Taking log on both side, ln [ � ] = − � − / = � + ln[ � ] It shows that the Fermi level lies above the top of the valance band. S t u d y n a m a .c o m
- 178. Unit –V: Semiconductors Page 15 Hall Effect: According to Hall when a current carrying metal or semiconductor is placed in a transverse magnetic field, a potential difference is developed across it; the direction of the developed potential difference is perpendicular to the direction of both applied magnetic field and applied current. In a P type semiconductor slab the current is given by � = ….. Where = Concentration of holes = The charge on the hole = Area of cross-section = Drift velocity of the charge carrier Therefore the current density along the external applied electric field will be given by = � = ……. When a transverse magnetic field is applied, the hole experience a Lorentz force � which deflect them towards face (in our case). Because of this at face the holes starts gathering at surface and it acquires a positive polarity. An equivalent negative charge is developed at surface . Due to this potential difference developed between the faces and an electric filed � is produced. This field is called Hall field. This electric field produces a force � on the hole in opposite to Lorentz force � . 1. Hall Voltage �� : When a sufficient number of holes accumulates at the surface , the force � balance the Lorentz force i.e. � = � …. This equilibrium condition usually reached in − Now the Lorentz force on holes due to magnetic field is given by � = sin 9 � = …. Substituting the value of from equation (2) we have � = �⃗⃗⃗ � = � …. And the electric force on the hole due to Hall voltage � = � But � = ��/ � = �� …. S t u d y n a m a .c o m
- 179. Unit –V: Semiconductors Page 16 Where b is the width of the semiconductor slab. Putting the value of � and � in equation (3) we have �� = � But = �/ So �� = � �� = � …. If is the thickness of the semiconductor slab then = ∴ �� = � . �� = � …. Figure(17): Hall Effect 2. Hall coefficient �� : Hall coefficient � is defined as the Hall field per unit magnetic induction per unit current density. � = �� � = ��/ � Putting the value of �� from equation (7) � = � × × � � = � × � � = …. Again putting the value of from equation (9) into (8) we get- �� = � ×� S t u d y n a m a .c o m
- 180. Unit –V: Semiconductors Page 17 � = �� � .….(10) � = � � …. (11) 3. Hall Angle �� In the semiconductor the resultant electric field is the vector some of the applied field � and the developed Hall field � as showm in figure. If �� is the angle between the resultant electric field and the direction along which the current is flowing as shown in the figure then- Figure(18) tan �� = � � …. (12) But we know that � = �� = …. (13) And � = � � where � is conductivity …. (14) So by (12), (13) and (14) tan �� = ×� � tan �� = � But = � by equation (9) then tan �� = � � Now � � = � tan �� = � �� = tan− � …. (15) S t u d y n a m a .c o m
- 181. Unit –V: Semiconductors Page 18 P-N junction Diode: When P-type and N-type semiconductors are join together by some special techniques. A p-n junction is formed. P-N junction allows to flow of current in one direction only and this property is called rectifying action. Figure (19): Symbol of P-N junction diode. Figure(20): Diode There are two operating regions and three possible biasing condition of a P-N junction. i) Zero biasing: When no external potential is applied to the p-n junction the diode is said to be unbiased. The potential barrier discourages the diffusion of any majority carrier across the junction. However the potential barrier helps minority charier to drift across the junction. Then an equilibrium will be established. ii) Forward bias: The P-N junction is said to be forward bias when its p-side is connected to the positive terminal and the N-side to the negative terminal of the battery. If applied voltage become greater than the value to potential barrier, the potential barrier will overcame and current starts flowing. When an applied voltage is increased gradually more and more charge carrier of lower energy gain sufficient energy and current starts increasing. Figure(21): Forward bias circuit Figure(22): Forward bias characteristic iii) Reverse bias: When positive voltage is applied to the n-type and negative voltage is applied to the p-type semiconductor. The diode is said to be reverse biased. The depletion layer grows wide in this case only a small amount of current flows due to the minority charge carrier. The circuit of reverse bias P-N junction diode and reverse bias characteristics are shown in the figure. S t u d y n a m a .c o m
- 182. Unit –V: Semiconductors Page 19 Figure(23): Reverse bias PN junction circuit diagram Figure (24): Reverse bias characteristics curve. Zener Diode: Zener diode is a special purpose heavily doped PN-junction diode, designed to operate in the breakdown region. The symbol of Zener diode is shown bellow Figure(25): Symbol Zener Diode Construction: Zener diodes are like ordinary PN junction diode except that they are fabricated by varying the doping so that sharp and specific breakdown is obtained. Zener diode consists of two and � substrates diffused together and has metallic layer deposited on both sides to connect anode and cathode terminals. V-I Characteristics: The graph plotted between voltage taking on x-axis and current on the Y-axis is called the � −charactristics. Forward bias characteristics: The forward bias V-I Characteristics of Zener diode is shown below. It is almost identical to forward bias characteristics of PN junction diode. Figure(26): Forward bias Zener Diode Figure(27): Forward bias characteristics of Zener diode Reverse Bias Characteristics: The reverse bias characteristic of Zener diode is generally different from that of the PN-junction diode. As we S t u d y n a m a .c o m
- 183. Unit –V: Semiconductors Page 20 increase the reverse voltage, initially small current starts flowing due to thermally generated minority charge carriers. At a certain value of reverse voltage the reverse current will increase suddenly. This voltage is called Zener break down voltage. Once the break down occurs the voltage across Zener diode remains constant. Figure(28): Zener diode Reverse bias Figure(29): Zener diode reverse bias characteristics The sudden increase in the current may occurs due to the following reasons- i) Avalanche effect: This type of breakdown takes place when both side of junction are lightly doped in this case the electric field is not so strong to produce Zener break down. Here the minority carrier accelerates by the field, collides with the atoms of semiconductors due to the collision with valance electrons, covalent bonds are broken and electron hole pair are produced. This is called avalanche break down. At this point the device damages permanently and cannot be used again by removing the reverse voltage. ii) Zener effect: When both side of junction are very heavily doped and small reverse bias voltage is applied, a very strong electric field is set. This field is enough to break the covalent bonds. This is called the Zener effect or Zener break down. Due to which an abrupt increase in the reverse current occurs, and the device stats acting as a conductor. After the removal of reverse voltage the device will be available to use and at Zener voltage the device do not damages. Applications of Zener diode: i) As a voltage regulator ii) Switching operations iii) Clipping and clamping circuits. Zener diode as a voltage regulator: A simple voltage regulator use a Zener diode in reverse bias in parallel with the load � as shown in the figure- S t u d y n a m a .c o m
- 184. Unit –V: Semiconductors Page 21 Figure(30): Zener Diode as a voltage Regulator When the voltage in the circuit increases the voltage across Zener diode remains constant which appears across the load. The Zener diode draws more current and voltage across the diode remains constan. Photo Diode: The Photo diode is a PN junction semiconductor diode which is always operates in the reverse bias condition Construction: the construction of a photodiode and its circuit and symbol are shown in the figure. The light is always focused through glass lens on the junction of photodiode. As the photodiode is reverse biased the depletion region is quite wide. The photons incident on the depletion region will impart their energy to the ions present there and generate electrons hole pair. The number of electrons-hole pair will be depend on the intensity of the light. With increase in the light intensity number of electrons –holes pairs are produced and the photo current increase. Figure(33): Symbol Photo diode Figure(34):Construction of photo diode Photo diode Characteristics: V-I characteristics are shown below and the variation of photocurrent with light intensity is shown below in the figure S t u d y n a m a .c o m
- 185. Unit –V: Semiconductors Page 22 Figure(35): Photo diode V-I Characteristics Figure(36): Photo diode intensity/current characteristics Dark Current: It is the current flowing through a photodiode in the absence of light. Dark current flows due to thermally generated minority charge carrier and hence increase with increase in temperature. The reverse current depends on the intensity of light incident on the junction. It is almost independent of the reverse voltage. S t u d y n a m a .c o m
- 186. Unit –V: Semiconductors Page 23 Solar Cell: A solar cell is a photovoltaic device designed to convert sunlight (solar energy) in to the electrical energy. Construction: The solar cell is made from semiconductor materials like silicon. The p-type layer is made very thin so that the light radiation may penetrate to fall on junction. The doping level of p-type semiconductor is very high. As the photon reaches at the junction, here it is absorbed and an electron from valance band jump to conduction band this creates an electron hole pair. The electron produced in the p-side and the hole produced at the n-side works as minority carriers. These minority carriers cross the junction due to the depletion reign electric field cross the junction, even in the absence of applied voltage. This phenomenon is clearly depicted in figure. Figure(37): Generation of photo electrons Figure(38): Circuit of solar cell Thus a photo current flows in the circuit Figure(39): Solar cell circuit Figure(40): Symbol of solar cell Advantages: 1) It is a pollution free energy conversion system. 2) Cheap for solar power aircrafts. 3) Useful in remote areas where no other source of energy can be frequently transferred. 4) It is clean source of energy. S t u d y n a m a .c o m
- 187. Unit –V: Semiconductors Page 24 Disadvantages: 1) It does not convert all the solar energy in to the electrical energy. 2) Its efficiency depends on the temperature. 3) Requires large area for power applications. 4) The output is DC which cannot be transported through large distance without significant loss. Applications: 1) In space satellite. 2) In low resistance relay for ON and OFF applications. Characteristics of solar cell: 1) Voltage v/s intensity of incident light: - The voltage increases linearly with increase in the intensity of light. Figure(41):Voltage/current intensity 2) Current v/s intensity of incident light: The current v/s intensity at a given load resistance are shown in figure below. The current increases linearly first and after a certain point the current stops increasing Figure(42): Current/intensity graph 3) Voltage v/s current or V-I characteristics: Voltage current characteristics for fixed load resistance � is shown in the figure(43). S t u d y n a m a .c o m
- 188. Unit –V: Semiconductors Page 25 Figure(43): Voltage v/s current or V-I characteristics The nature of the V-I characteristic of a solar cell is similar to that of a photo diode. Typical V-I characteristics of a solar cell is shown in the figure. On the vertical axis � the applied voltage � is zero everywhere and therefore the point of intersection represents the short circuit condition. The point of intersection of the characteristic curve with � axis represents the short circuit current , , … ..and � , � … ….are open circuit current. 4) V-I characteristic as a function of load resistance: Figure(44):VI graph 5) Voltage v/s power i.e. V/W characteristics curve: the voltage v/s power characteristic curve as a function of load resistance � at fix light intensity is shown in figure below- Figure(45):Voltage/power graph S t u d y n a m a .c o m
- 189. Unit-7 Nuclear Physics Page 1 Unit-7 Nuclear Physics Syllabus: Nuclear composition, mass defect, binding energy, nuclear force, liquid drop model, elementary idea about nuclear fission and fusion S t u d y n a m a .c o m
- 190. Unit-7 Nuclear Physics Page 2 Nuclear constituents: By particle scattering, Rutherford conclude that the atom of any element consist of central core called nucleus and electrons moving around it. The entire mass of the atom and positive charge is concentrated inside the nucleus. The nucleus is supposed to consists of two particles the protons and neutrons. The protons are positively charged particles while the neutrons are neutral particles. The mass of the protons and neutrons are almost same and the charge on the proton is equal and opposite to that of electron. S.N. Name of the particle Mass of the particle Charge on the particle 1 electron . × − − . × − 2 proton . × − . × − 3 neutron . × − Both the neutrons and protons inside the nucleus together are called the nucleons. The number of protons �. . ℎ is called the � and the sum of the number of protons and neutrons is called the . The stability of the nucleus is depends on the relative number of neutrons and protons. The conventional symbol of the nuclear species follows the following pattern- �� Where =mass number =atomic number (i.e. the number of the protons/electrons) � =chemical symbol General Properties of the nucleus: (a) The nuclear mass: Mass of the nucleus is the sum of the mass of the protons and neutrons combined in a nucleus. This is usually expressed in terms of � � = . ×− . It is ℎ part of the nucleus. it is assume that the total mass of the nucleus should be the sum of the mass of the neutrons and protons i.e. ℎ = ( �) + � ℎ = ( �) + − � S t u d y n a m a .c o m
- 191. Unit-7 Nuclear Physics Page 3 Where � = � = ℎ = � = � (b) Nucleus charge: The charge on the nucleus is due to the protons contained in it. This can be given as ℎ = ×. × − (c) Nucleus radius: It has been observed that volume of the nucleus of directly proportional to the number of the nucleons , ∝ � ∝ ∝ � ∝ � / / = � / / where is the proportionality constant = / where = � / = . × − (d) Nuclear density- The density of the nucleus can be calculated as follows. � = ….. but ℎ = � ℎ = � . × − ( ) ℎ = . × − ℎ = × ℎ S t u d y n a m a .c o m
- 192. Unit-7 Nuclear Physics Page 4 ℎ = × . ×− now putting these values in the equation number (1) � = �× . × − 7 �× . × − � = . × / Spin and magnetic moment: Like electron neutrons, protons are also fermions with spin quantum number = therefore spin angular momentum ⃗ of magnitude of- = ℎ � √ + = ℎ � √ ( + ) = ℎ � √ In nuclear physics magnetic moments are expressed in nuclear magnetron �� �� = ℎ � . � Where � is the mass of the proton Classification of the nuclei: The atoms of different elements are classified as follows. (i) Isotopes: Isotopes are the nucleus with same number of atomic number , but different mass numbers . The nuclei , , , are all isotopes of oxygen. The isotopes of the element have identical chemical properties but different physical properties. (ii) Isobars: nuclei with the same mass number but different atomic number the nuclei , are the examples of the isobars. These elements have the different chemical and physical properties. (iii) Isotones: Isotones are the nuclei with equal number of neutron �. . = −. Examples S t u d y n a m a .c o m
- 193. Unit-7 Nuclear Physics Page 5 are , & = . (iv) Mirror nuclei: Mirror nuclei have the same number of but the number of protons and neutrons are interchanged. (v) Examples are: = =, � = = Nuclear Liquid Drop Model: This model was proposed by Neil-Bohr in 1937. This model shows the analogy of the nucleus with the li uid d op that’s why this odel is k ow as li uid d op odel. A o di g to this odel followi gs a e the analogies between the liquid drop and the nucleus. 1. Both are spherical in shape. 2. Both liquid drop and nuclei filled with an incompressible substance. 3. Short range nuclear forces are analogous to the intermolecular forces in liquid. 4. The density of nuclear matter is very large which do not depends on the number of nuclei just as the density of the liquid drop not depends on the number of molecules. 5. Both nuclear and intermolecular forces are saturated forces. 6. Inside the nucleus the nucleons moves as an atom moves in a liquid drop. 7. Nucleons are bound with the nuclear forces just as the atom in a liquid bound together by the inter-molecular forces. This is why the nucleus is considered as a small drop of liquid and this model is called the liquid drop model. some other similarities in liquid drop and nucleus: S.N. Basic Liquid drop Nucleus 1. Shape Due to surface tension Due to nuclear force. 2. Density depends on volume Do not depends on the radius Do not depend on number of the nucleon. 3. Emission of particle Due to the mutual collision the kinetic energy of some molecules increases and leaves the liquid surface. −Particles emission takes place due to the collision of nucleons and extent the kinetic energy. 4. Analogy in the energy Latent heat of vaporisation. Binding energy per nucleons. 5. Absorption of the particle Condensation of the drop. Absorption of the particle striking the nucleus. On the basis of the liquid drop model, the nuclear fission can be explained. This model can also explain S t u d y n a m a .c o m
- 194. Unit-7 Nuclear Physics Page 6 radioactive decay , , � �, successfully. Moreover this model can also explain the nuclear quadruple moment. Un-success of the liquid drop model: This model could not explain the reason for most stability of the lighter nuclei such as , , etc. which have equal number of protons and neutrons. Similarly this model could not be successful to explain the magic numbers. Mass Defect: The actual mass of the nucleus is less than the total mass of protons and nucleons present inside it, because some mass is lost in the form the binding energy. This defect is called the mass defect and denoted by Δ . Δ = [ . � + − . �] − Where = Mass number of nuclei. = Number of protons. − = Number of neutrons � = Mass of protons. � = Mass of neutrons. = Measured mass. Curve between the mass defect and nuclear number: Figure shows the variation of mass defect per nucleons with atomic numbers. The mass defect per mass number in the nuclear is known as packing fraction and denoted by = Δ� � S t u d y n a m a .c o m
- 195. Unit-7 Nuclear Physics Page 7 Figure(2): Graph between packing fraction and atomic number From the graph, we can say 1. Mass defect is positive for elements having mass number less than 20.3 2. Mass defect/ packing fraction is negative for elements having mass number between . 3. The mass defect is again positive for elements having mass number above 200. The significance of packing fraction is that it is directly related to the availability and its stability of nuclear. Binding Energy: The energy required to remove any nucleon (neutron or proton) from the nucleus is called binding energy of nucleus. The binding energy is the energy equivalent to the mass defect. We know that the mass defect is given by Δ = [ . � + − . �] − Where all the symbols have their usual meaning. Now the binding energy = Δ = { . � + − . � − } S t u d y n a m a .c o m
- 196. Unit-7 Nuclear Physics Page 8 Figure(3): Curve plotted between Binding energy per nucleon and mass number The average binding energy per nucleon is found to be about for all elements. Some important features from binding energy curves- 1. Binding energy of very light nuclei like is very small. The curve rise sharply with increase in mass number and reaches a maximum value of . for . This makes iron most stable. 2. Binding energy decreases for element having > . It is found to be decreases to . for uranium. Semi-empirical mass formula: The actual mass is slightly less than the calculated mass of the nucleus, because some mass is lost in the form of the binding energy This mass defect is given by Δ =[ . � + − . �] − Where the symbols have their usual meanings. Now the binding energy = Δ = { . � + − . � − } S t u d y n a m a .c o m
- 197. Unit-7 Nuclear Physics Page 9 = [ . � + − . �] − On the basis of the liquid drop model the binding energy is considered to be the sum of independent different energy terms and is expressed as = + + + + … … … … . 1. Volume Energy: For most of the nuclei, the binding energy per nucleon is constant So ∝ ∝ = � … … … … 2. Surface Energy: ∝ − ∝ − = − ( ) = − … … … … Negative sign shows that the binding energy decreases due to surface energy. 3. Coulo b’s e ergy: The electrical repulsion between each pair of protons in the nucleus also contributes in decreasing its binding energy. The coulomb energy is proportional to the number of proton pair [ − ! ] In the nucleus, ]and inversely proportional to the nuclear radius ∝ so the coulomb energy � = − [ − ] the coulomb energy is negative, because if arises from an effect that opposes nuclear stability. 4. Asymmetric energy: We know that the light nuclei having same number of neutron and proton are more stable but as mass number increases, the number of neutron increases more as compare to the protons, due to which the stability of nucleus decreases. Thus it is concluded that if the number of neutrons increases as compare to protons the binding energy will decrease. = − = − − = − S t u d y n a m a .c o m
- 198. Unit-7 Nuclear Physics Page 10 So = − . − ……………. 5. Energy due to even odd effect: The nucleus having even number of neutron and proton are more stable as compared to the nucleus having odd number of neutron and proton while the nucleon having even number of proton and odd number of neutrons or even number of protons and odd number of neutrons are found to be either less stable or more stable. So = ± � − / …………… 6 Here (+) sign is for the even number of neutron and protons (-) sign is for odd number of neutron and protons And � = for even number of neutron and odd number of protons and vice-versa. On combining all the above terms we get = � + − � − [ � − − � − − + ] Isomerism: The existance of the atomic nuclei those have same atomoic number and the same mass numbers but different energy states. Magic numbers: The nuclei which have number of protons or number of nutrons equal to , , , , , are relatively stable. These numbers are called the magic numbers. The existence of magic numbers is established by the following facts. i) There is abundance of nuclei in nature which have number of nucleons equal to magic no. ii) From the binding energy curve it is clear that the nuclei = − =and = − =are especially stable. iii) The nuclei of even atomic number or atomic number more than 28 have isotopic abundance more than % are only = − =, = − =and = − = iv) There are not more than 5 isotones of all the nuclei except for = and = . v) = has ten stable isotopes and = has six stable isotopes. vi) After = ,the ∝ − energy shows the discontinious behabiour. S t u d y n a m a .c o m
- 199. Unit-7 Nuclear Physics Page 11 vii) − energy is very large when number of protons or neutrons in a radioactive nuclear is equal to the magic number. viii) The neutron absorption cross section for = and = nuclei is very small as compared to the other neighbouring nuclei. When the average binding energy per nucleon is plotted against mass number. The curve is not smooth, but several kinks are observed. These kinks correspond to sudden increase in binding energy. Thus stability is related to higher binding energy. Nuclear Forces: We know that the nucleus consists of protons. Due to the positive charges on protons, there will be repulsive electrostatic force between two protons and the resulting repulsive force between protons will tend to push the nucleus apart. Therefore, for the nucleus to have a permanent existence there must be some strong attractive forces demand. Moreover, these forces cannot be gravitational forces because they are much smaller than the force required. Moreover these forces cannot be electrical forces in nature because the strong repulsive forces between protons will lead to disruption of nucleus. Actually, these forces are short range attractive forces known as Nuclear Forces. The nuclear forces have the following properties: 1) These forces are attractive forces between (proton-proton, neutron-neutron or neutron-proton). 2) These forces are strongest forces in nature. 3) They are spin dependence. 4) These fo es a e sho t a g fo es, A g aph of Coulo ’s law of epulsio a d sho t a ge fo es of attraction is shown in figure. S t u d y n a m a .c o m
- 200. Unit-7 Nuclear Physics Page 12 Figure(4): Comperesion of Colum and nuclear forces. 5) These forces are independent of charge. i.e. the nuclear forces are same for protons –proton, proton-neutron, neutron-neutron. 6) In case of the nuclear forces, each nucleon attracts those nucleons which are immediate neighbouring just like the molecules in a liquid or the gas interacts with the neighbouring molecules. Yukawa mesons theory: Yukawa in 1935 proposed a theory to explain the binding forces between neutrons and protons known as meson theory of nuclear forces. Yukawa postulated the existence of a new particle called � meason having a rest mass greater then that of the mass of electron but less than that of a nucleon. Though this particle was discover much later, yet Yukawa showed that the interaction produced by mesons between nucleon were of the correct order of magnitude. According to this theory, all nucleons (protons and neutrons) consist of identical cores surrounded by a cloud by one or more � mesons. The mesons may be either neutral or may carry positive or negative charge. The charge on the mesons is equal to the charge on the electronic charge and according to their charge they designated as � , �+ or �− . The idea of this theory is that it is the mesons cloud which differentiate between a neutron and a proton. The mesons are supposed to exchange rapidly between S t u d y n a m a .c o m
- 201. Unit-7 Nuclear Physics Page 13 nucleons thereby changing their identity equally fast and are responsible for keeping them bound together. Thus nuclear force between a proton and a neutron is the result of the exchange of charged mesons (�+ and �− ) between them. When a �+ meson jumps from a proton to a neutron, the proton is converted into a neutron and vice versa. − �+ → or + �+ → Similarly when a �− meson jumps form a neutron to a proton, it is converted to a proton, and vice versa. − �− → or + �− → In the same way, the forces between two protons and those between two neutrons arise due to the exchange of neutral mesons between them. In this way nucleus is an ever-changing structure. It should be remembered that numbers of protons and neutrons remains the same in the nucleus. Nuclear Fission: The phenomenon of braking of a heavy nucleus in to two or more light nuclei of almost equal masses together with the release of a huge amount of energy is known as nuclear fission. The process of nuclear fission was first discovered by the German scientists, Otto Hahn and Strassman, in the year 1939. In this process, when uranium nucleus ( ) was bombarded with slow neutrons, this nucleus was found to split up in to two radioactive nuclei which were identified as isotopes of Barium ( ) and Krypton ( ). It is given by the following nuclear reaction- + → [ ] → + + + . Figure (5): Liquid drop model for nuclear fission It is not that Barium and Krypton are only isotopes to be obtained by the fission of . Actually, this is a S t u d y n a m a .c o m
- 202. Unit-7 Nuclear Physics Page 14 very complicated phenomenon and more than 100 isotopes of over 20 different elements have been obtained in it. All the elements fall in the middle 75 to 160 mass number region of the periodic table Theory (Liquid Drop Model): The mechanism of the nuclear fission was first explained by Bohr and Wheeler on the basis of the liquid drop model of the nucleus. According to this model, the nucleus is assumed to be similar to a liquid drop, which remains in equilibrium by a balance between the short-range, attractive forces between the nucleons and the repulsive electrostatic forces between the protons. This inter nucleon force gives rise to surface tension forces to maintain a spherical shape of the nucleus. Thus, there is a similarity in the forces acting on the nucleus and liquid drop. When nucleus drop captures slow or neutron, oscillations setup within the drop. These oscillations tend to distort the spherical shape so that the drop becomes ellipsoid in shape as shown in the figure. The surface tension forces try to make the drop return to its original spherical shape while the excitation energy tends to distort the shape still further. If the excitation energy and hence oscillations are sufficiently large, the drop attains the dumbbell shape. The Columbic repulsive forces then push the nucleus into similar drops. Then each drop tries to attain the shape for which the potential energy minimum, for example spherical shape. Nuclear Reactor: It is a device that produces a self-sustained and controlled chain reaction in a fissionable material. One type of nuclear reactor is shown in the figure. A modern rector has following important parts. Figure (6): Nuclear reacter 1) Fuel: The fuels play the key role in the operation of the reactor. The fissionable material is known as the fuel. Generally, and can be used as fuel. S t u d y n a m a .c o m
- 203. Unit-7 Nuclear Physics Page 15 2) Moderator: It is used to slow down neutron to thermal energies by elastic collisions between its nuclei and the fission neutrons. Heavy water graphite or barium oxides are commonly used for this purpose. Heavy water is the most suitable moderator. 3) Control Rods: To control the fission rate in the reactor, we use cadmium and boron rods. Cadmium and boron are good absorber of slow neutrons. These rods are fixed in the reactor- walls. When they are pushed into the reactor, the fission rate decrease and when they are pulled out the fission rate gets increased. 4) Shield: the various types of rays, like , , those are radioactive are emitted from the reactor. These rays may be injurious to the health of people working near the reactor. For protections the reactor is therefore surrounded by a concrete wall of about 2 meter thick and containing high protection element like iron. 5) Coolant: The reactor generates heat energy due to the fission reaction which is removed by means of cooling agent. For this purpose, air water, carbon-dioxide etc. are generally used as coolant. Coolant is circulated is circulated though the interior of the reactor by a pumping system. 6) Safety Device: If the reactor begins to go fast, a special set of control rods, known as shut-off rods insert automatically. They absorb all the neutrons so that chain reaction stops immediately. Working of nuclear reactor: to start the reactor, no external source is required. Even a single neutron is capable of starting fission, although few neutrons are always present there. The reactor is started by pulling out the control rods. Then the neutron strikes the nucleaus and fission it along with the emission of two or three fast neutrons. These neutrons are slow down by moderator (graphite), after which they induce further fission of . The reaction onece starts is controlled with the help of control rods by moving them inside and outside. Applications of Nuclear Reactor: The nuclear reactor are used mainly for the following purpose. 1) Generations of energy 2) Production of . 3) Production of neutron beam 4) Production of radioisotopes Nuclear Fusion: Nuclear fusion is nothing but the formation of heavier nuclide by the fussing of two light nuclei. In this process, the mass of the product nuclide is generally less than the sum of masses of the nuclides which are fused. The efo e, as pe Ei stei ’s ass energy relation= , an enormous amount of energy realeased which is called nuclear energy. The first artificial fusion reaction was the hydrogen bomb which was tested in November 1952. Fusin reactions are thermonuclear reactions which occurs at extremely S t u d y n a m a .c o m
- 204. Unit-7 Nuclear Physics Page 16 high temperatures. For example, in order to fuse deuterium ( ) and tritium ( ), the force of repulsion (called Coulomb potential barrier ) of these two positively charge particles must be overcome. The following fusion reaction is possible for the fusion of the two heavy hydrogen nucleoids + → + + . (energy) The nucleus of tritium can again fuse with heavy hydrogen nucleus + → + + . (energy) Thus the combine form is + + → + + + . (energy) From the above equation, it is clear that three deuterium nuclei fused together to form a helium nucleus and liberate 21.6 MeV energy which is obtained in the form of kinetic energy of proton ( ) and neutron ( ). S t u d y n a m a .c o m
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