displacement current and modified Ampere’s law by replacing J
by (Jc + Jd). i.e.
Curl H = J Original Ampere’s Law
Curl H = J + J Ampere’s law after Maxwell modification
8. What is the Poynting vector? Write its magnitude and
dimension.
ANSWER: The pointing vector represents the em energy flow per
unit area per unit time along the propagation of the wave. It is
given by the cross product of the field vectors i.e. = × !
The unit of S is Watt/m2 or Joule/sec- m2 and the dimension
SHORT ANSWER QUESTIONS
is"MT %& '.
9. What is the Poynting Theorem?
UNIT-1: RELATIVISTIC MECHANICS
ANSWER: John Henry Poynting, the famous British physicist
1. What is inertial and non-inertial frame of reference? Is earth
stated thatThe time rate of em-energy within a certain volume
an inertial frame?
plus the time rate of em-energy flowing out through the boundary
[ANSWER: A frame in which Newton’s laws hold good is
surface, is equal to the power transferred into the em field.i.e.
called inertial frame and if not hold good then is called non
∂
− ( ∇. S dV = ( E. JdV + 2( U45 6 dV
inertial.
∂t
.
.
.
Earth is an non-inertial frame due to its rotational and axial
10. What is the displacement current?
motion.
ANSWER: The displacement current is that current which comes
2. Two photons are approaching to each other. What is their
into play in the region in which the electric field and hence the
relative velocity?
OR
electric flux is changing with time. It is represented by i and is
Show that no signal can travel faster than light.
OR
;<
given by i = ℇA : >
Show that the velocity addition theorem is consistent with
;=
Einstein’s postulates of special theory of relativity.
11. What is the skin depth or penetration depth?
[ANSWER: From velocity addition theorem: If two particles
ANSWER: The skin depth or the penetration depth is the
came forward towards each other than the relative speed
distance over which a plane em wave is attenuated by 1/e of its
′
initial amplitude at the surface. It is denoted by ? and for good
= ` + /1 +
@
conductor it is given by ? = =
If the two particles are photons then u`= c and v = c,
G L BAJAJ ITM, GR NOIDA
ATOM – 2021-22
ENGINEERING PHYSICS
(KAS-101T)
B TECH: 1st Year, 1st SEM
A
3.
4.
5.
6.
BCD
After giving these values in the above formula, we obtain,
UNIT-3: QUANTUM MECHANICS
u = c. So speed of light is constant
What was the objective of Michelson- Morley experiment?
12. What is black body?
[ANSWER: Object: To prove the existence of ether, a
[ANS: A black body is a body which absorbs radiations of all
hypothetical medium present in the space.
wavelengths incident upon it. It neither reflects nor transmits
Conclusion: There is no ether in space.
Any of the incident radiations and therefore appears black
What are the postulates of Special theory of relativity?
whatever be the colour of incident radiation. Its coefficient of
[ANSWER: (i) All fundamental laws of physics remain same in
absorption and coefficient of emission both are 1.
all inertial frames. (ii) The speed of light is constant.
It is an ideal conception. There is no known surface which can
Show that time dilation is a real effect.
act as perfect black body. Lamp black and platinum black is
[ANSWER: The real effect of time dilation can be given by the
nearest approach to a black body as the coefficient of
decay μ-mesons at the height of about 10km from the surface of
absorptions for these two are 96% and 98% respectively.
the earth. They have a typical speed of 0.998c with life time of 13. Explain Bohr’s quantization rule on the basis of
2×10-8second. Therefore in this life time they can travel a
de’Broglie hypothesis.
distance of only 600m. Then how do μ-mesons travel a distance
[ANS: According to Bohr’s quantization
of 10km to reach the earth?
rule, only those orbits are allowed as the
The answer of this problem is by relativity. The μ-mesons has a
stationary orbits in which the angular
life time of 2×10-8second in its own frame. In observer frame on
momentum of electron is an integral
earth this time is lengthened by the relation
multiple of h/2π.
GH
-5
i.e.
E=
F = , Here n =
= / 1 − = 3.17×10 second,
I
1,2,3,4……
which enhance the traveled distance to be 9.5km.
But according to de’Broglie for the existence of stationary orbits
What are the mass less particles?
2KF = LM
[ANSWER: The particles which has zero rest mass (
= 0)
O
O
H
are known as the massless particles like Photon and Graviton et.
But
N=
=
Therefore
2KF = LM = L : >
PQ
R
S
GH
UNIT-2: ELECTROMAGNTIC FIELD THEORY
This equation may be written as E =
F= ,
I
7. Why Maxwell modify Ampere's law?
Here n = 1,2,3,4……
OR
14. What are the matter waves or de Broglie waves? OR Write
Why Maxwell proposed that Ampere's law require
some properties/ characteristics of matter waves.
modification?
[ANS: de Broglie matter waves have the following properties: 1ANSWER: Maxwell found that the Ampere’s circuital law is
Lighter particles have greater wavelength than heavier particles.
logically inconsistent with time varying fields. To make this law
2- The smaller the velocity of the particle, the greater is the
valid for both type of fields, Maxwell introduce a new concept of
wavelength associated with it.
What is wave function? Give its properties.
OR
What are properties of an acceptable wave function?/What
condition must it fulfilled?
[ANS: The quantity whose variations make up the matter waves
is called the wave function.
Properties:
• It should be normalized
• It should be single valued
• It should be continuous and finite]
16. Why is the physical significance of the wave function ( ψ )?
[ANS: The wave function ψ itself, has no direct physical
significance but the square of its modulus gives the probability
of finding the particle.
Probability = |UU ∗ |
15.
UNIT-4:INTERFERENCE & DIFFRACTION
17. Explain why two independent sources could not produce
coherent sources?
[ANSWER: Since the light from two independent sources is not
in phase due to atomic transitions.
18. Explain the necessity of broad source in interference due to
division of amplitude?
[ANSWER: The whole interference pattern of a film by a broad
source can be seen by placing the eye at a suitable position.
On the other hand the whole pattern due to point source can’t be
seen by placing eye at one place but one has to move the eye
from one point to other.
19. Why an excessively thin film appears black in reflected light.
[ANSWER: The path difference in reflected light is
M
Δ = 2X cos F +
2
\
For thin film t = 0, then Δ = ,Which is the condition of dark.
20. Explain the following effects in Newton’s ring experiment:
(1) Introduction of liquid: Diameters of dark and bright rings
are reduces and so whole pattern become more congested.
(2) Removal of glass plate with mirror: The reflected and
transmitted patterns are complementary to each other so
uniform illumination is observed.
21. Why Newton’s rings are circular?
[ANSWER: As the locus of equal film thickness is circle so the
pattern is in the form of rings.
22. What is the condition for missing spectra in grating?
[ANSWER: The condition for missing order is given by
L (^ + _)
=
^
Here (e+d) = Grating element, e is the slit width and d is the slit
separation.
23. Define resolving power of an optical instrument.
[ANSWER: The ability of any optical instrument to resolve any
closely spaced objects id called resolving power.
24. Write the Rayleigh criterion for the resolving power.
[ANSWER: According to the Rayleigh two objects are said to
be jus resolved if the principal maxima of diffraction pattern due
to one object falls exactly at the first minima adjacent to the
principal maxima of the other object and vice versa.
UNIT-V: FIBER OPTICS & LASER
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
What is the stimulated emission of radiation in a LASER?
[ANSWER: In certain material the upper stage has longer life
time of the order of 10-3 sec, known as metastable state and a
material having metastable state is called as active material. If
atom in excited state are compelled or forced to jump down to
ground state by photon of energy hv = E2 – E1, then it jumps to
ground state along with the emission of photon having energy hv
= E2 – E1,this process is called stimulated emission.
Why He-Ne is superior than ruby LASER?
[ANSWER: (1) He-Ne has low power consumption (2) The
efficiency of He-Ne is more than ruby laser.
What is population inversion, pumping and meta stable
state?
[ANSWER: Population Inversion: It is the situation in which
there are more atom in the excited state and less no of atoms in
ground state i.e. N1< N2.Pumping: It is the process to obtain
and maintain the population inversion. .Metastable State: It is
an exited state of an atom which has longer life time.
Distinguish between spontaneous and stimulated emission?
[ANSWER: The stimulated emission provide highly coherent
beam while the radiation from spontaneous emission process is
incoherent.
What is function of Cr+3 ions in ruby laser?
[ANSWER: Cr+3 provide the active centers for lasing action. It
has mata stable energy level and the electronic transitions
between different energy levels gives laser output.
What is the principle of optical fiber?
[ANSWER: The principle of optical fiber is based on the
phenomenon of total internal refraction (TIR).
The overall performance of step index -single mode fiber is
best. Why?
[ANSWER: For overall performance in terms of data rate and
attenuation, the single mode step index fiber has the best
characteristics because of:
In step index -single mode ncore-ncladding ≈ 0 and this
•
causes a large critical angle (almost 90 degree).
• The size of core is greatly reduced so all light paths
through core are practically of same length.
The net effect of these two changes is that all light rays traveling
through core travel the same distance, thereby reducing the
modal dispersion.
What is acceptance angle and acceptance cone?
[ANSWER: The acceptance angle is defined as the maximum
angle that a light ray can have relative to the axis of the fiber and
propagate down the fiber.
NA= Sin α5bc = dn@ − n = n@ √2∆
What is fiber loss or fiber attenuation?
[ANSWER: The loss of intensity of light signal is called the
@
k
attenuation. It is given by: Fiber Loss = −
log l
g(hi)
km
What is dispersion?
[ANSWER: The term dispersion describes the pulse broadening
effect and produced by the properties of core material and the
line width of light signal passing through the fiber.
The light pulse width produced by an input pulse of zero line
width.”
Dispersion is measured in the unit of time (nano second or pico
second.]
****************************************************
LONG ANSWER TYPE QUESTIONS
UNIT-1: RELATIVISTIC MECHANICS
1. What is Length contraction and Time Dilation in special
theory of relativity? Deduce expression for both .prove that
time dilation is real effect.
[ANSWER: EXPRESSION FOR Time Dilation:
A moving clock always appears to be going slow for the
stationary observer by the factor 1 −
n
Suppose a clock is situated in the system S’ and read the time
interval of an event t0,
where t0 = t’2 – t’1
………(1) Here is to the proper time.
When this event is noted by the observer from S the
t = t2 – t1
…………(2)
Using Inverse Lorentz
transformations:
.t =
pqo
r
=o %
(3)
p
r
@%
t=
nco
1−
n
−
t@s −
nco
1−
n
=
p
r
@%
Substituting value of equation 4 in equation 1 and 3, we may
obtain, ƒ s =
„% …
v2
1− 2
c
, † s = †, ‡ s = ‡ ˆL_
t s − t@s
1−
n
tt
=
OR
n
1−
t=
tt
1−
n
EXPRESSION FOR LENGTH CONTRACTION:
A moving rod always appears to be contracted for the
stationary observer by the factor 1 −
n
Let a rod AB is placed in S’
whose
coordinates
are (x@ , y@ , z@ ) and (x s , y s ,
z s ) w.r.t. S and S’
respectively.
The proper length i.e. the
length in S’ frame is
Lt = x s − x@s .(1)
However the length from S frame is L = x − x@
}c %n= ~
p
@%
r
= (x − x@ )/ 1 −
2.
=
…% „/
v2
1− 2
c
Equations are known as Lorentz’s transformation equations
obtained by Dutch physicist Henderik Antoon Lorentz in
1905.
Show how the relativistic invariance of the law of the
conservation of momentum leads to the concepts of variation
of mass with velocity and the equivalence of mass and
energy.
[ANSWER
Let S and S’ are
two
inertial
frames S and
S’, where S’ is
moving
with
velocity v in positive x direction. Let two identical particle A
and B of equal mass m moving towards each other in the same
line with equal velocity u’ and –u’ and collide inelasticity in
frame S’. Then velocity of particle as seen from frame S will be
n
−
=
}c• %n= ~
p
r
@%
€
p
r
@%
OR
……
(1)
……
(2)
……(2)
Using Lorentz transformations:
Eq (1) Lt = x s − x@s =
s
3. Derive formula for the variation of mass with velocity
with example.
OR
Using equation (3) in
equation (2)
ts −
x = α (x’+ vt’) ………… (2)
From equation (1) and (2) we can write
x = α [α (x-vt) +vt’]
•c
@
OR
t s = αt s − :1 − >
……..(3)
n
n
Now let a light pulse produces at time t = t’ = 0 at common
origin O and O’. After some time the position of pulse and the
frame S’ is shown in figure.
x = ct and
x’ = ct’
therefore equation (1) and (2) becomes,
ct = α t’(c+v) and t’ = α t(c-v)
@
Multiplying eqs c tt′ = α tt′(c + v)(c − v) or α =
(4)
L = Lt 1 −
n
Therefore a moving object is always shortened to a rest observer.
Deduce Lorentz transformation equation.
[ANSWER: Lorentz’ Transformation:
Let S and S’ are two inertial frames, where S’ is moving with
velocity
v in
positive
x
directio
n w. r.
t. S.
Let x’ = α (x – vt) ……….. (1)
Where α is proportionality constant depends on v only. Also y’ =
y and z’ = z
The inverse relation may be written as
Applying conservation of momentum before collision and after
the collision from frame S is
m1u1 + m2u2 = (m1 + m2) v ……………….(3)
Substituting the value of u1 and u2 from equation (1) and (2) in
equation (3) we get
Now from equation 1 we have
1−
u
c
2
1
2

u ′2  
v2 
1 − 2  1 − 2 
c 
c 
= 
′
u
v


1 + 2 
c 

1−
u
c
2
2
2
Similarly we can write

u ′2  
v2 
1 − 2  1 − 2 
c 
c 
= 
u ′v 

1 − 2 
c 

Thus equation 3 becomes
Where Q is the total electric charge given by the sum of free
charge and the bound charge. In terms of charge density ρ ,
…4
Now if particle B is at rest in frame S we have u2 = 0 and rest
mass of particle B m2 = mo
Then
Generalizing the above equation we get
4.
Deduce the relativistic velocity addition theorem.
OR
Obtain the expression for velocity transformation.
[ANSWER: Let S and S’ are two inertial frames, where S’ is
moving with velocity v in positive x direction w. r. t. S.
Let a particle P is moving with velocity u’ in S’ frame under any
external force. An observer read this velocity of the particle u
from S frame.
Here u = uc ı̂ + u‹ ŒJ + u• k• and us = usc ı̂ + us‹ ȷ̂ + us• k•
c
‹
•
uc = , u‹ = , u• =
, and usc =
=
=
=
From Lorentz transformation equation
, y’ = y, z’ = z
co
=
s
o , u‹ =
‹o
=
s
o , u• =
•o
=o
and
Differentiating these equations we get
, dy’ =
dy,
dz’
=
dz
Q = ›. ρ dV
From Eqs. 1 & 2, we get
›œ ∈ E. dS = Q OR
›œ D. dS = ›. ρ dV
Converting the surface integral into volume integral on the LHS,
we obtain
›. ∇.D dV = ›. ρ dV
Since the volume is arbitrary here so we can compare the integrand
of both side: ∇ . D = ρ
2nd Maxwell’s Equation: –—˜ ¡ = ¢
Statement: The net outward flux of magnetic induction vector
through any surface enclosing a volume is equal to zero.
Proof: Let us consider any arbitrary surface within em-field. The
no. of magnetic lines of force entering in this surface will exactly
equal to the no. of magnetic force lines leaving out. Therefore the
net
flux
will
be
zero.
Therefore
ϕ=0
OR
ݴ B. ds = 0
Where B is the magnetic flux density which is magnetic flux per
unit area. Converting the surface integral to volume integral
›. ∇.B dV = 0 Or
rd
∇.B = 0
¦¡
3 Maxwell’s Equation: Curl E = ¦§
Statement: The electromotive force around any closed loop
through any surface enclosing a volume is equal to the negative
rate of change of magnetic flux linked with that loop.
Proof: According to faraday’s induction law of electromagnetic,
the negative time rate of change of magnetic flux linked with a
circuit is equal to the emf induced in the circuit. i.e.
e=
−
©
=
=−
:›« ª. œ >
=
∮- E. dl = - ∂/∂t ›œ B. dS
Where E is the electric field strength and B is the magnetic flux
density. Converting the line integral to surface integral using
stokes theorem,
and
›œ (∇ ×E).dS = - ›œ − ∂B/ ∂t. dS OR
th
usc =
=
dx s
=
dt s
‘q %n
@%
p
.‘q
r
=
dx − vdt
dt −
n c
=
c
=
1−
OR
This equation is called velocity transformations.
UNIT-2: ELECTROMAGNTIC FIELD THEORY
−v
n
.
c
=
’′ =
“%
@%
”
“
•
1. Write down Maxwell’s equations in integral form and convert
them into differential forms. Give physical significance of each
equation.
ANSWER:
1st Maxwell’s Equation: –—˜ ™ = š
Statement: The net outward flux of electric displacement vector
through any surface enclosing a volume is equal to the net charge
density contained within that volume.
Proof: According to Gauss’s law in electrostatics:
›œ E. dS =
@
∈ž
Q
∇. E = − ∂B/ ∂t
¦™
4
Maxwell’s Equation: Curl H = J +
¦§
Statement: The magnetomotive force around any closed loop
through any surface enclosing a volume is equal to the conduction
current density plus the time rate of displacement vector linked
with that loop.
Proof: According to Ampere’s circuital law, the magneto motive
force around a closed path is equal to current enclosed by the path.
Mathematically, we obtain
›- H. dl = i
-------------
( 1)
Where H is the magnetic field strength and I is the current enclosed
by the path. If J is the current density, then
i = ›œ J. dS
------------From equations (1) & (2), we get
›- H. dl =
Converting
›œ J. dS
the
line
integral
( 2)
to
surface
integral
›œ ∇ ×H.dS = ›œ J. dS
Curl H = J
……..(3)
Taking div of this equation div (Curl H) = div J = 0
Therefore according to Ampere’s law div J = 0 always, however
;®
continuity equation states that div J = 0 only if
= 0, or ρ =
;=
constant. Hence Ampere’s law is valid only for static field. In
order to make Ampere’s law logically consistent with time varying
field, Maxwell replaced J by (Jc + Jd) in equation (3). Therefore
Ampere’s law after Maxwell modification becomes: Curl H = Jc
+ Jd ……(4)
Here Jd is known as displacement current density. Again taking
;¯
div of eq.(4), we obtain
Jd =
;=
Therefore, the modified Ampere’s law becomes
;¯
Curl H = Jc + Jd = Jc +
;=
2. Explain the concept of displacement current and show that it
led to the modification of Ampere’s law. (4th Maxwell Eq)
OR
Discuss the modification made by Maxwell in Ampere’s law
taking the displacement current into consideration. Explain
the displacement law and its implications. (4th Maxwell Eq)
ANSWER: The magnetomotive force around any closed loop
through any surface enclosing a volume is equal to the conduction
current density plus the time rate of displacement vector linked
with that loop.
Proof: According to Ampere’s circuital law, the magneto motive
force around a closed path is equal to current enclosed by the path.
Mathematically, we obtain
›- H. dl = i
-------------
( 1)
Where H is the magnetic field strength and I is the current enclosed
by the path. If J is the current density, then
i =
›œ J. dS
-------------
( 2)
From equations (1) & (2), we get ›- H. dl = ›œ J. dS
Converting
the
line
integral
to
surface
integral
3.
ANSWER: Maxwell’s equations are given by:
div E = 0, div H = 0,
;¿
Curl E = -X ,
;=
;¯
;<
Curl H =J+
= σE+∈
…….(1) a,b,c,d
;=
;=
Taking the curl of eq.(1c), we get,
;¿
; ‘Â- ¿
Ã
;<
FÁ( FÁ ) = FÁ :−X > = −X
= −X :σE+∈ >
OR
ÄFˆ_(_Å
OR ∇
− XÈ
à Ç
Ã…
)−∇
− XÆ
;=
= XÆ
ÃÇ
=0
Ã…
ÃÇ
Ã…
;=
+ ÈX
à Ç
Ã…
Ã…
;=
= −∇
…………….. (2)
à É
ÃÉ
Ll'ly we can obtain ∇ ! − XÈ
− XÆ = 0 ..(3)
Ã…
Ã…
Equations (2) and (3) are the wave equations in the conducting
medium.
In a region of free space: ∈=∈ , µ = μ , σ = 0,
Equations (2) and (3) becomes:
∇2 E = µo ∈o Ë /Ët2 ……..(4)
∇2 B = µo ∈o Ë Ì /Ët2 …….(5)
These are the wave equations for the E and H in free space.
These equations may be compare with the general wave equation
with a velocity , ∇ y – 1/ Ë y/Ët2 = 0
Comparing equation 4 & 5, we get,
@
……(6)
=
Putting
=
dBž ∈ž
%½²
∈o = 8.86× ½¢
@
dÐ.ÐÑ×½¢Ò½² × ÓI×@
Í/P and X = 4K × 10%Ï , we obtain
= 3 × 10Ö =
ÒÔ
Therefore, in free space the em-wave travels with light speed.
In a region of Non-conducting medium: ∈=∈ , µ = μ , ρ = 0
Equations (2) and (3) becomes:
∇2 E = µ ∈ Ë /Ët2 ……..(7)
∇2 B = µ ∈ Ë Ì /Ët2 …….(8)
These are the wave equations for the E and H in non-conducting
medium.
These equations may be compare with the general wave equation
with a velocity , ∇ y– 1/ Ë y/Ët2 = 0
@
Comparing equation 4 & 5, we get,
=
……(9)
›œ ∇ ×H.dS = ›œ J. dS
Curl H = J
……..(3)
Taking div of this equation div (Curl H) = div J = 0
Therefore according to Ampere’s law div J = 0 always, however
;®
continuity equation states that div J = 0 only if
= 0, or ρ =
√×B
;=
Putting
∈=∈
∈
and
X
=
X
X
Ø
Ù
Ø
Ù
constant. Hence Ampere’s law is valid only for static field. In
@
@
@
@
we obtain =
=
=
×
=
order to make Ampere’s law logically consistent with time varying
d∈l ∈Ú Bl BÚ
dBž ∈ž
√×B
√∈ÚBÚ
√∈Ú BÚ
field, Maxwell replaced J by (Jc + Jd) in equation (3). Therefore
Hence =
√∈ÚBÚ
Ampere’s law after Maxwell modification becomes:
@
Curl H = Jc + Jd
…………(4)
And the refractive index of the medium L = =
√∈Ú BÚ
Here Jd is known as displacement current density. Again taking
Therefore, in free space the em-wave travels with light speed.
;¯
div of eq.(4), we obtain
Jd =
;=
UNIT-3: QUANTUM MECHANICS
Therefore, the modified Ampere’s law becomes
;¯
5. Obtain time independent and time dependent Schrodinger’s
Curl H = Jc + Jd = Jc +
;=
wave equation. Explain ‘ψ’
Show that in free space the wave equations for the field vectors
OR
E and B of an em-field are given by:
Obtain
the
equation
of
motion
of the matter wave associated
°2 E = µo ∈o ¦² ³/¦t2
and °2 B = µo ∈o ¦² ¡ /¦t2
with the material particle of mass m and moving with the
Prove that the velocity of plane electromagnetic wave in
speed of v under the potential energy field V.
vacuum is given by: v = 1/d´¢ ∈¢ = µ
[ANSWER: Schrödinger’s time independent equation: Let us
OR
consider a particle of mass m moving with velocity v. Let the
@ Ã Û
Show that in good conducting medium the wave equations for
equation of matter waves in 1-D is: ∇ U =
….(1)
Ü…
the field vectors E and B of an em-field are given by: ° ² ¶ −
Solution of equation (1) is U = U ÝÅLÞ = U ÝÅL2Kß …(2)
¹² ¶
¹¶
¹² ¼
¹¼
·¸ ² − ·» = ¢
and
° ² ¼ − ·¸ ² − ·» = ¢ .
Differentiating eq. (2), we get
¹º
¹º
¹º
¹º
ÃÛ
à Û
OR
= U 2Kß àÝ2Kß and
= −U (2Kß) ÝÅL2Kß
Ã…
Ü…
Obtain the plane wave equations in non-conducting medium
Ã
Û
½
= −4K ß U =−4K
U = −4K á U
(lossless dielectric) and show that Q =
Ü…
\
√·¾
i.e the speed of em wave is less than speed of light in vacuum in
dielectric medium. Also find the expression for refractive
index.
= −4K
S
H
ã
U=−
S
ℏ
ã
U
}
â”
..(3)
~
@ Ã Û
6.
g
U = õ ÝÅL
S
SÇ
ℏ
SÇ
ƒ + Ì àÝ
ℏ
ƒ ………….
(2)
Where A & B are constant
Applying boundary condition U = 0 at x = 0 in equation (2) we
get 0 = 0 + B implies that B = 0
Thus equation (2) becomes
U = õ ÝÅL
SÇ
ℏ
ƒ ……
(3)
Applying second boundary condition i.e. U = 0 at x = L in
equation (3) we get
0 = õ ÝÅL
SÇ
ℏ
ƒ
Implies that either A = 0 which is not acceptable as total wave
function becomes zero
Or
ÝÅL
÷² ø² ℏ²
SÇ
ℏ
E=0
⇒
SÇ
ℏ
E = LK
³ =
²ùú²
known as energy eigen value for nth state
Thus the wave function ( or eigen function ) becomes
GI„
U = õ ÝÅL
g
To find the value of A we apply normalization condition
OR
i.e.
› UU ∗ _ƒ = 1
g
or,
› õ ÝÅL
GI„
_ƒ = 1
or,
g
g
û
GI„
û
∇ U=
=U
………..(4)
Ã…
ℏ
› }1 − cos g ~ _ƒ = 1 or E = 1
Now we know that the total energy E = Kinetic energy +
@
⇒ õ=
Therefor the normalized eigen function of the
+ åà or
= 2 ( − åà)
Potential energy =
g
Therefore, equation (4) may be modified as, ∇ U = −
GI„
S(Ç%æØ)
²P(¶%èé)
particle is U =
ÝÅL
g
g
U or ° ² ç +
ç = ¢ ……..(5)
²
ℏ
ℏ
Equation (5) is known as Schrödinger’s time independent
equation.
UNIT-4: INTERFERENCE & DIFFRACTION
Schrödinger’s time dependent equation:
7. Describe and examine the formation of Newton’s rings in
Let us start with Schrödinger’s time independent equation giving
reflected monochromatic light. Obtain the expression for the
by equation (5)
diameters of bright and dark rings.
S(Ç%æØ)
ℏ
[ANSWER: Newton’s rings are the special case of wedge shape
∇ U+
U = 0 or
∇ U + ( − åà) U = 0
ℏ
S
thin film interference. When a plano-convex lens is paced on a
ℏ
or ê− ∇ U + åàë U = U or H ç = ¶ ç
….(6)
plane glass plate in such that the convex surface is in contact
S
with the plate the a wedge shape film is produced between the
ℏ
Here ! = − ∇ U + åà
S
gap.
Equation (6) is known as Schrödinger’s time dependent
equation.
A particle is in motion along a line between x = 0 and x = a
with zero potential energy. At points for which x < 0 and x >
a, the potential energy is infinite. Solving Schrödinger’s
equation, obtain energy eigen values and normalized wave
function for the particle.
OR
A particle of mass m is moving in a 1-D potential of sides L
¢ , îOïðï ¢ ≤ ì > E
defined as:
è(ì) = í
∞ , îOïðï ¢ ≤ ì < E0
Solve the Schrödinger’s equation, obtain energy eigen values
Theory of Interference:
From wedge shaped interference we have that the path
and normalized wave function for the particle.
SOLUTION: 1-D Potential Well Problem:
difference between interfering reflected waves is
∆=
\
Let us consider a particle moving inside a box along positive X2X üàÝ (F + ý) +
axis between x = 0 and x = L
For normal incidence i = r = 0 and for small wedge angle
The potential V of the particle in 1-D box is
ý ≅ 0 Ýà üàÝ (F + ý) = 1
V(x)
=
0
for
0<x<L
\
Therefore
∆= 2X +
=
∞
for
x ≤ 0 and x ≥ L
As particle cannot move outside the box, so its wave function
Condition for Maxima:
ψ = 0 for x ≤ 0 and x ≥ L
We know that the maximum intensity is obtained at those
So wave equation for free particle within the box ( V = 0) is
points at which the path difference is an even multiple of
G\
à Û
SÇ
λ/2. i.e.Δ =
Here n = 1,2,3,4…..
+
U =0
……………….
(1)
Ä
ℏ
\
G\
The general solution of this differential equation is
Therefore ∆= 2X + =
⇒ 2X = (2L − 1)M/2
Putting the value from equation (3) in equation (1), we get
Condition for Minima:
We know that the maximum intensity is obtained at those
points at which the path difference is an odd multiple of λ/2.
( G @)\
i.e.Δ =
\
( G @)\
G\
Therefore ∆= 2X + =
2X =
Expression For Diameters:
Let us consider a plano-convex lens of radius of curvature R
placed on a plane glass plate at O. Let any nth order ring from
the X and Y at which the
thickness of film is tn.
From Geometry
Ì×Ì =
õÌ × Ì
FG × FG = (2 − G) × G
FG = 2 G − G ≅ 2 G
⟹ G
FG
=
2
Let us now obtain the diameters of Bright rings and darks rings:
Diameter of Bright Rings
8.
9.
( G%@)\
If d = 2e, n = 3m = 3,6,9…….. orders are missing
We know that the condition for bright rings is2X G =
10. What do you mean by the dispersive power of plane
(2L − 1)M
FG
G
transmission grating? Derive an expression for it.
2X
=X
=
⟹ G = (2L − 1) × 2M /X
2
4
2
OR
G = √2L − 1 × d2M /X
Two spectral lines have wavelengths λ and λ + dλ,
Clearly the diameter of bright rings are proportional to the
respectively. Show that if dλ < λ, their angular separation dθ
square root of odd natural numbers.
\
in grating spectra is given by: _ý = ( )
Diameter of Dark Rings
%\
G\
We know that the condition for dark rings is2X G =
[ANSWER: The dispersive power of the grating is defined as
FG
2LM
G
the rate of change angle of diffraction with the change in
=X
=
⟹ G = L × 4M /X
2X
2
4
2
wavelength used.
G = √L × d4M /X
Clearly the diameter of dark rings are proportional to the square
If the wavelength changes from λ to ( λ + dλ) and the
root of natural numbers.
corresponding change in the angle of diffraction is from θ to θ +
Define resolving power. Derive an expression for the
resolving power of a plane transmission grating.
dθ, then the dispersive power is defined as dθ/dλ.
[ANSWER: Resolving Power: The ability of any optical
Expression for Dispersive Power: The grating equation
instrument to distinguish the images of two very close objects is
(e + d) Sin θ = nλ
called the resolving power.
Differentiating this equation w.r.t. λ, we obtain
It is defined as ratio of the wavelength of any spectral lines to
G
G
G
@
=( )
=
=
= ( )
the smallest wavelength difference between very close lines for
Ø
\
(
)d@% G
%\
\
(
) @%
which the spectral can just be resolved. Mathematically RP =
(
)
\
Expression for resolving power of grating:
11. Discuss the thin film interference in transmitted and
Let us consider there are two closely spaced object represented
reflected light and show that the two interference patterns
by λ and λ +
are complementary to each other.
dλ
are
[ANSWER: Let us consider a thin parallel film PQRS of
placed
in
thickness t and of medium index µ. Let a light ray from S made
front
of
fall on this
grating,
film.
After
whose
multiple
diffraction
reflection and
patterns are
refraction, the
obtained on
rays 1 and 2 are
the screen.
obtained
as
Therefore
reflected and
let nth
transmitted
principal maxima due to λ + dλ fall in θ direction, then (e +d)
component. As
Sin θ = n (λ+ dλ) …….(1)
the rays 1 and 2
Also the first minima adjacent to the nth order principal maxima
satisfy
the
N(e + d) Sin θ = m λ = (nN+1) λ ……(2)
conditions for
Equating equation 1 and 2, nN (λ+ dλ) = (nN+1) λ
superposition
\
so interference
OR
= L = ^ÝàÁ ÅLÄ à ^F
\
pattern is obtained on the both sides.
What do you understand by missing spectra? Obtain a
(A)
Interference
in reflected light:
condition for the missing spectra in N- slit diffraction
The
path
difference
between ray 1 and 2 in reflected light is
pattern. What particular spectra would be absent if the
(õÌ
∆=
+ Ìü) gi − õ û = µ(õÌ + Ìü)û − õ û
width of the transparencies and opacities of the grating are
equal.
= µ(õÌ + Ìü) − õ
……….(1)
[ANSWER: In grating spectra for normal for certain value of θ,
Calculation of AB and BC:
if the maxima of the interference due to N-slits falls exactly at
In right triangle ∆, Cos r = BE/AB ⇒ AB = BE /Cos r = t/ Cos r
the minima due to single slit diffraction pattern, then the maxima
Also AB = BC ⇒ BC = t /Cos r ⇒
AB+ BC = 2 t/ Cos r
will disappear from the resultant pattern. This phenomenon is
Calculation
of
AM:
called Missing Spectra OR Absent Spectra.
Using the right triangle AMC, Sin i = AM/AC⇒AM = AC Sin i
The condition for the principal maxima is given by
(e + d) Sin θ1 = n λ
= (AE + EC) Sin i = (AE + EC) µ Sin r
The condition for the minima due to single slit diffraction is
AM = (t Tan r + t Tan r ) µ Sin r
given by:
e Sin θ2 = m λ
= 2µt Tan r Sin r = 2µtSin2r/Cos r
At a certain point on screen, if the two conditions are
=
G Â
Therefore Δ = 2µ
− 2µt
r = 2X Cos F
simultaneously satisfied then
θ1 = θ2 i.e (e + d) Sin θ = n λ
t¥ Â
t¥
(
)
G
Using Stokes approach for path difference in reflected light, the
and
e Sin θ = m λ
OR
=
S
This is the condition for the spectrum of order n to be
missing/absent.
For example
If d = e,
n = 2m = 2,4,6,……………. orders are missing
final result is Δ = 2X Cos F +
Condition for Maxima:
\
G\
We know for maximum intensity: Δ =
Δ = 2X Cos F +
Therefore
⇒ 2X üàÝ F = (2L − 1)M/2
Condition for Minima:
\
=
G\
Here n = 1,2,3…….
( G @)\
We know that for minimum intensityΔ =
Δ = 2X Cos F +
\
=
Therefore
( G @)\
⇒ 2X üàÝ F = 2LM/2
Here n = 1,2,3…….
(B) Interference in transmitted light:
the path difference in transmitted light is Δ = 2X üàÝ F
Now we are able to right the condition for maxima and minima:
Condition for Maxima:
Therefore Δ = 2X Cos F =
G\
Therefore Δ = 2X Cos F =
( G @)\
Condition for Minima:
Here n = 1,2,3…….
Here n = 1,2,3…….
From these results it is obvious that the two patterns thus
obtained are complementary to each other.
UNIT-4: OPTICAL FIBER & LASERS
12. Derive relation between Einstein’s coefficients and discuss
the results.
[ANSWER: If N1be the number of atoms per unit volume in the
collection of energyE1 then the absorption rate of transition shall
be proportional to both N1 and the number of photons available
per unit volume Uv at correct frequency v . The absorption rate
is given by @ = @ Ì@ ’ … … . . (1)
And the total emission rate is given by
P21 = N !A @ + B @ u" #………2
Constants B12, A21 and B21are called Einstein’s transition
coefficients.
When atomic system is in thermal equilibrium with the radiation
field, then rate of upward transition is equal to rate of downward
transition, that is, @ Ì@ ’ = N !A @ + B @ u" #
û
Solving for uv :
’ = $• •
:
$
> • %
•
The population density N in a given state depends on the
energy E of the state and temperature T given by Boltzmann’s
'
equation: = ^ %Ç/%& so • = ^ H(/h&
Hence
’ =
û •
á)/*+
• %
'
•
=
û •
•
. Ò,/-+
@
( • /
• )%@
From Planck’s radiation law we know that radiation field
density: number of photons per unit volume in the frequency
range v and v + dv is given by ’( =
On Comparing
û •
•
=
ÖIH(.
.
ÖIH(.
.
@
. á)
-+ %@
ˆL_ Ì@ = Ì @
From these results
1. Stimulated emission and absorption are competing processes.
2. The rate of spontaneous emission is far larger than stimulated
emission.
3. The ratio of A21 and B21 is proportional to cube of
frequency of the transition.
13. Describe construction and working of He-Ne/Ruby Laser.
[ANSWER:
A ruby
laser is
a solid-state
laser that
uses
a
synthetic ruby crystal
as its gain medium. The
first working laser was
a ruby laser made
by Theodore
H.
"Ted"
Maiman at Hughes
Research
Laboratories on May 16, 1960. .
Construction of Ruby Laser;
The ruby laser consists of a ruby rod, which is made of
chromium doped ruby material. At the opposite ends of this rod
there are two silver polished mirrors. Whose one is fully
polished and other is partially polished.
Operation of Ruby Laser
The optical pump raises the electrons from ground state E1 to E3.
At E3 these excited levels are highly unstable and so the
electrons
decays
rapidly to the level
of E2. This transition
occurs with energy
difference (E1– E2)
given up as heat
(radiation
less
transmission). The
level E2 is very
important for stimulated emission process and is known as Meta
stable state. Electrons in this level have an average life time of
about 5m.s before they fall to ground state. After this the
population inversion can be established between E2 and E1. The
population inversion is obtained by optical pumping of the ruby
rod with a flash lamp. When the flash lamp intensity becomes
large enough to create population inversion, then stimulated
emission from the Meta stable level to the ground level occurs
which result in the laser output.
He-Ne LASER
The He-Ne laser
was the first gas
laser
to
be
invented by Ali
Javan, William
Bennet Jr. and
Donal Herriot at
Bell Laboratory,
USA in 1962.
Construction: The active/gain medium of the laser is a mixture
of helium and neon gases, approximately in the ratio of 5 : 1.,
contained at low pressure (typically – 300 Pa) in a glass
envelope. The energy of pump source of the laser is provided by
an electrical discharge of around 1000 V through an anode and
cathode at each end of the glass tube.
Working: The laser process in a He-Ne laser starts with
collision of
electrons
from
the
electrical
discharge
with helium
atoms in the
gas. Collision
of the excited
helium atoms
with
the
ground state
neon atoms
results in transfer of energy to the neon atoms, exciting neon
electrons into the 5s level.
The number of neon atoms entering the excited states builds up
as further collision between helium and neon atoms occur
causing a population inversion between the neon 5s, 3p and
other electronic levels. Spontaneous and stimulated emission
between the 5s (2P1/2) and 3p (2P1/2) states results in emission of
10
632.991 nm wavelength light, the typical operating wavelength
Attenuation (Loss) = −
log Po/Pi
of a He-Ne laser.
L(km)
14. What are the types of optical fiber?
For an ideal fiber Pi = Po, so loss = 10 log 1 = 0 dB
OR
The major contributor of loss of light in fiber during
What are light propagation modes of OF?
transmission are[ANSWER: OF can be classified into three groups according to (A) Material Loss: It is due to absorption of light by the fiber
the way of light propagation mode in core and index profile.
material. It includes absorption due to light interacting with the
These are
molecular structure of material as well as loss due to material
(1) Multimode-step index fiber: it is known as step index fiber. A
impurities.
multimode step index fiber is shown in figure, the core diameter
The loss due to atomic structure itself is relatively small.
of which is around 50µm. Some physical parameters like relative
However losses due to impurities can be reduced by better
refractive index, index difference, core radius etc determines the
manufacturi
maximum number of guided modes possible in a multimode
ng
fiber. A single mode fiber has a core diameter of the order of 2
techniques.
to 10µm and the propagation of light wave is shown in figure. It (B) Scattering
has the distinct advantage of low intermodal dispersion over
Loss: Light
multimode step index fiber.
is scattered
(2) A single mode step index fiber has a very thin core of uniform
by
the
refractive index of a higher value which is surrounded by a
molecules of
cladding of lower refractive index. The refractive index changes
the material
abruptly at the core cladding boundary as shown in the figure: ]
due to structural imperfection and impurities. The scattered light
does not propagate through the fiber and therefore lost out.
When light is scattered by an obstruction, it produces power
loss. In OF, obstruction refers to the density variation which
produces change in refractive index. Scattering can also be due
to macro bends i.e. fiber deformation. This type of fiber loss can
be made negligible by better manufacturing techniques. In
@
general
Scattering loss ∝ ã
2
Where λ is the
wavelength
of
light.
(C) Wave guide and
Bending
Loss:
This type of light
signal loss are due
to
imperfections
and
deformations (micro and macro
bends) in fiber structure which
cause radiation of light away
from the fiber.
The figure shows the light loss
caused by a change in diameter.
Clearly
in
the
(3) Multimode- graded index fiber: A graded index fiber is a
absence
of
multimode fiber with a core consisting of concentric layers of
deformation, the light
different refractive indices. Therefore, the refractive index of the
would have been
core varies with distance from the fiber axis. It has a high value
confined in the fiber.
at the center and falls off with increasing radial distance from the
A very small change
axis as shown in the figure above.
in core size (i.e.
For overall performance in terms of data rate and attenuation, the
micro bends) also cause radiation loss as well as back scattering.
single mode step index fiber has the best characteristics.
All of the above losses are dependent on wavelength of light
1- Explain different types of signal loss in optical fiber.
used. Combining all of loss phenomenon, we find only a small
[ANSWER: The reduction in amplitude, intensity and power of
loss in the following regions for glass fiber:
the light signal during its propagation through thee optical fiber (a) 0.80-0.90 micro meters range is known as “First loss window”.
(b) 1.26-1.34 micro meters range is known as “Second loss
is known as the fiber loss or signal loss or attenuation.
window”.
(c)
01.50-1.63
micro meters range is known as “Third loss
If Pi be the power input and Po be the output power, then the
window”.
signal loss or attenuation is given by:
Obviously, by care full selection of operating wavelength, the
Attenuation (Loss) = -10 log Po/Pi
(Decibels)
above losses can be minimized.
The loss increases with the length of fiber so it is in general
15. Derive the expression for Numerical aperture, Acceptance
expressed as dB/km as
angle, acceptance cone for an optical fiber.
uc =
[ANSWER:
The acceptance
angle is defined
as the maximum
angle that a light
ray can have
relative to the
axis of the fiber
and
propagate
down the fiber. It
u• =
‘q o n
@
8
=q op =2.4 x 10 m/s
r
‘A o ?@%n /
= op
@ q
r
•/
@
u‹ =
‘> o?@%n /
@
=q op
r
•/
@
= 2.4 m/s
= 7.2 m/s
Therefore, u= ux i+ uy j+ uz OR u= 2.4 x 108 i + 2.4 j + 7.2 k
4. The mass of a moving body is 11 times its rest mass. Find its
kinetic energy and momentum.
5ž
[ANSWER: It is given that m= 11 m , Hence
= 11m
p
r
@%
⟹ v = 2.99 × 10Ö m/sec
Now that kinetic energy K = (m − m )c = (11m − m )c =
In the figure the light AB is entering in fiber core from air at
10mt c = 8.2 × 10%@& joule
angle α. At point B, applying Snell’s law,
And momentum p=mv= 11m v = 11 × 9.1 × 10%&@ × 2.99 ×
nt × Sin α = n@ × Sin(90 − θ)
10Ö = 2.99 × 10% @ Kg
In order to fulfill the total internal reflection, we must have, 5. Calculate amount of work done to increase speed of an
θ ≥ θ , where θ is the critical angle.
electron from 0.6C to 0.8C. Rest energy of electron =
Let the maximum value of α which makes θ ≥ θ is denoted by
0.5MeV.
α5bc .
Nt × Sin α5bc = n@ × Sin(90 − θ )
[ANSWER: The kinetic energy K1 at v1 = 0.6c:
1
= n@ × cosθ = n@ × d1 − Sin θ
− 1)m c = 2 × 10%@Ó joule
K1 = (m − m )c = (
6
•
=n@ × 1 −
= dn@ − n
1−
6•
is denoted by α max as shown in the figure.
Here θ = Sin%@ } ~ = critical angle.
6
6•
The quantity Sin α5bc is known as the numerical aperture (NA).
Therefore,
NA= Sin α5bc = dn@ − n
****************************************************
NUMERICAL TYPE QUESTIONS
And the kinetic energy K2 at v2 = 0.8c:
1
K2 = (
− 1)m c = 5.35 × 10%@Ó joule
1−
Therefore the amount of work done = Change in kinetic energy
= K2- K1 ≅ 3.35 × 10%@Ó joule = 2.1 × 10E eV
At what speed should a clock be moved so that it may appear
to lose 1 min. in each hour.
=F
[ANSWER: From time dilation formulae, t =
p
6.
NUMERICALS ON RELATIVITY
1. Calculate the length and orientation of a rod of length 2
meter in a frame of reference which is moving with 0.6 c
@%G H
r
velocity in a direction making an angle of 30 0 with the rod.
Here to=1 hour = 60 min and t = 60 min + 1 = 61min
[ANSWER: The length and orientation of a rod along the
=
7
direction of the moving frame of reference = LOcos 300.
v = c 1 − F = 3.0 x 108 x 1 − } ~ = 5.4× 105ms-1
=
7@
The apparent length in a direction perpendicular to the direction
NUMERICAL ON ELECTROMAGNETIC THEORY
of motion(x-axis)
1. If earth receives 2cal/min/cm2 solar energy, what are the
( .7 )
n
√&
0
amplitudes of electric and magnetic field vectors?
Lx =Locos30 }1 − ~ =
1−
= 1.73x 0.8 = 1.38m
OR
Apparent length in a direction perpendicular to the direction of
If earth receives 1400 W/m2 solar energy, what are the
motion (y-axis)- No contraction
amplitudes of electric and magnetic field vectors?
@
LY = L0sin 300 = 2 x = 1 m
OR
Therefore the length of the rod in a moving frame in a direction
If earth receives 1.4K W/m2 solar energy, what are the
making an angle of 300 with rod
amplitudes of electric and magnetic field vectors?
SOLUTION: The energy flux per unit area per second is
LY = dL8 + L9 = d(1.38) + (1) = 1.704 m
S = 2cal min-1 cm-2 = 2×4.2×104/60joule m-2 sec-1
If the rod makes an angle θ with X-axis in the moving frame,
S = EH = 2×4.2×104/60 = 1400
------- (1)
then
@
-‹
But
E/H = dµà/È = 376.72
----- (2)
Tan θ = =
= 0.72 OR ý = tan-1(0.72) = 35.80
-q
@.&Ö
On solving E = 726.2 Vm-1 and H = 1.928 A/ m
2. How fast would o rocket have to go relative to an observer
Therefore the amplitudes of electric and magnetic fields of
for its length contracted to 99% of its length at rest?
radiation are
€
<<
[ANSWER: Here L = 99 LO /100, That is ,
=
€;
@
Eo = E√2 = 726.2√2 = 1026.8 V/m and
According to the Lorentz length contraction formula
L=
Ho = H√2 = 1.928√2 = 2.726 A/m
n
<<
.
2.
A 1000 watt sodium lamp radiating its power. Calculate the
Lt 1 −
i.e.
= }1 − ~
@
electric field and magnetic field strength at a distance of 2 from
<<
n
6
-1
the lamp.
OR } ~ = 1 - OR v = 0.141 c = 42.3 x 10 ms
k
@
@
SOLUTION: Poynting vector S = ÓIÙ = ÓI = 19.9044
3. A particle has velocity, u’ = 3i+4j+12k m/sec in a coordinate
But, we know that S = EH = 19.9044 ……..(1)
system moving with velocity 0.8c relative to laboratory along
+ve direction of x-axis. Find u in laboratory frame.
Also E/H = dµà/È
= 376.6--------- (2)
[ANSWER: Given u’ = 3i+4j+12k, ux’= 3, uy’ = 4, uz’= 12 and v=
On solving E = 86.59 V/m and H = 0.23 A/ m
0.8c
Therefore the amplitudes of electric and magnetic fields of
Therefore according to law of addition of velocity, the x, y and z
radiation are
component of u frame are:
Eo = E√2 = 86.59√2 = 122.456 V/m
And Ho = H√2 = 0.23√2 = 0.325 A/m
3. Show that for frequencies < 109Hz, a sample of silicon will act
as good conductor. For silicon, one may assume ε/εo = 12 and σ
= 2 mhos/cm. Also calculate the penetration depth for this
sample at frequency 106Hz.
6.
SOLUTION:
To prove Good Conductor:
We know that for good conductor
Hence
Æ
ÞÈ
=
Æ
2KJ×È0 ÈF
=
Æ
ÞÈ
2×102
≫> 1
2×3.14×109 ×8.85×10−12 ×12
≅ 300
Hence the silicon sample is a good conductor.
Calculation of Skin depth:
The skin depth or penetration depth for good conductor is given
(Let X = X ) ? =
@
A
=
BCD
M
5893
=
= 2075 Å
2X
2 × 1.42
(ii) Condition for bright film: 2X üàÝ F = (2L − 1)M/2
(2L − 1)M
5893
=
=
= 1037.5 Å
4X
4 × 1.42
Two plane glass surfaces in contact along one edge are
separated at the opposite edge by a thin wire. If 20
interference fringes are observed between these edges in
sodium light of λ=5890Ǻ of normal incidence, find the
diameter of the wire.
=
[ANSWER: = = 20 × 5890 ×
= 5.89 × 10%Ó
Light of wavelength 6000 Å falls normally on a thin wedge
shaped film of refractive index 1.4 forming fringes that are 2
mm apart. Find the angle of wedge in seconds.
\
[ANSWER: The fringe width is given by: ý =
OR
@ ÒT
G\
7.
=
U=
d2/2 × 3.14 × 106 × 4 × 3.14 × 10−7 × 2 × 102
BA
\
B
Given: λ2 = 6000 x 10-8 cm, μ= 1.4, β= 2 mm = 0.20 cm
= 0.0356 ^ ^F
θ= 10.71 x 10-5 rad = 10.71 x 10 -5 x 180/K
NUMERICALS ON INTERFERENCE
th
th
θ= 0.0061o OR θ= 0.0061 x 60 x 60 sec = 21.96 sec.
1. In Newton’s ring experiment diameter of 4 and 12 dark
rings are 0.400 cm and 0.700 cm .Deduce the diameter of 20th NUMERICALS ON DIFFRCTION
1. Light of wavelength 5500Ǻ falls normally on a slit of width
dark ring.
[ANSWER: G M − G = 4NM
……(1)
22 × 10 −5 cm . Calculate the angular position of the first two
n=4,
n+p=12,
minima on either side of the central maximum.
=
0.4
ˆL_
=
0.7
−
=
4
×
8M
[ANSWER: Given that e = 22 × 10%E
= 22 × 10%Ï
Ó
@
Ó
@
and
− Ó = 4 × 16M
and λ = 5500Å In single slit for minima ï VWPX = PN
sin θ = nλ / e
n solving, D20 = 0.0906cm
2.
Newton’s rings are observed normally in reflected light of
th
wavelength λ=6000Ǻ. The diameter of the 10 dark ring is
0.50 cm. Find the radius of curvature of the lens and the
thickness of the film.
[ANSWER:
=
ÓG\
= (0.50) /4 × 10 ×
D15 = 0.590cm, D 5 = 0.336cm, p = 10, R = 100cm
4.
2n + 1
5.
2.
EE
×@ ÒT
×@ ÒY
Second
= 0.25
order
ý@ =
n=2
A single slit is illuminated by light composed of two
wavelengths N½ andN² . One observes that due to Fraunhofer
diffraction, the first minima obtained forN½ coincides with the
second diffraction minima of N² . What is the relation
between N½ andN² .
[ANSWER:
For
single
slit
the
minima
occurs
e sin θ = ± nλ
e sin θ1 = λ 1 ………(1) and e sin θ 2 = 2λ 2 ….(2)
θ1 = θ 2 = θ So e sin θ1 = λ 1 = 2λ 2 So λ2 = λ1 / 2
3.
−5
λ= 5.88 × 10 cm = 5880Ǻ
White light falls normally on a film of soapy water whose
thickness is 1.5×10 -5 cm and refractive index 1.33. Which
wavelength in the visible region will be reflected strongly?
[ANSWER: When light falls normally r = 0o on a film, the
( G @)\
condition of maxima is: Δ = 2X Cos F =
4.
4µt
λ =
= 7.98 x 10 - 5 / (2n + 1)
For n=0, λ= 7.98 x 10 -5 cm
For n=1, λ= 2.66 x 10 -5 cm For n=2, λ=1.598 x 10 -5 cm
For n=0, λ= 1.14 x 10 -5 cm
Hence λ= 7.98 x 10 -5 is most strongly reflected.
Light of wavelength 5893A0 is reflected at nearly normal
incidence from a soap film of refractive index 1.42. What is
the least thickness of the film that will appear (1)- dark and
(2)bright
G\
[ANSWER: (i) For Minima: 2µ cos F =
When light falls normally r=0o, cos r = 1, n=1 for least thickness,
\
ÅLý@ = =
θ 2 = sin −1 ( 2λ / e) = 30°
6 × 10 =106cm and
= = 3 × 10%Ó
Ö
Newton’s rings are observed by keeping a spherical surface
of 100cm radius on a plane glass plate. If the diameter of the
15 th bright ring is 0.590 cm and the diameter of the 5 th ring
is 0.336 cm, what is the wavelength of light used?
ANSWER:
N = (O²P R − O²P )/QRR
%E
3.
G = 4LM
For first order n=1
14.3
Find the minimum number of lines in a plane diffraction
grating required to just resolve the sodium doublet (5890 Å
and 5896 Å) in the (i) first order, (ii) second order.
[ANSWER: R= λ/dλ=Nm, N=1/n(λ/dλ)
For first order: n=1, Mean wavelength λ=(λ1+λ2) / 2= 5893 Å;
dλ=6 Å,
N=982.
For second order: n=2, N=491.
What must minimum number of lines per cm in a half inch
width grating to resolve the wavelength 5890 Å and 5896 Å.
[ANSWER: R= λ/dλ=Nm, N=1/n(λ/dλ)
For first order: n=1, Mean wavelength λ=(λ1+λ2) / 2= 5893 Å;
dλ=6 Å,
N=982
Since the grating is half inch wide, therefore the number o lines
per inch= 982 ×2
Minimum number of lines per cm=982 × 2/ 2.54 =773
NUMERICALS ON LASER
1. In a ruby laser , total number of Cr+3 ions is 2.8 ×1019. If the
laser emits radiation of wavelength 7000Ao, calculate the
energy of laser pulse.
[ANSWER: Given n= 2.8 ×1019. And λ = 7000Ao So Energy
of laser pulse = n(hν) = n(hc/λ) = 7.94J
2.
Calculate the population ratio of two states in He-Ne laser
that produces light of wavelength 6000Ao at 300K
[ANSWER: Given λ= 6000Aoand T = 300K. Population ratio of
two states is
Z•
Z
= e
["^
\] =
e
[ ^
2\] =
G
0.00410
(5) V-Number/Normalized frequency/ Cut-off Parameter
(V):å =
I_
\
õ=
×&.@Ó× E×@ Ò`
× 0.1324 = 15.9898
@.&×@ Ò`
(@E.<Ö<Ö)
æ
= 127.83 ≅ 127
(6) No of Modes: S = =
(7) Maximum Radius for Single Mode: We know that for single
mode operation V < 2.405 OR
I_
×&.@Ó×_
õ=
× 0.1324 < 2.405 OR a = 3.76×10-6m
\
@.&×@ Ò`
NUMERICALS ON de’Broglie:
1. Calculate the de-Broglie wavelength associated with a
proton moving with a velocity equal to (1/20)th velocity of
light.
Ñ.Ѳ׽¢ÒaQ ײ¢
O
2.
Solution:-N =
=
= ². ÑQ × ½¢%½Q P
½.Ñb×½¢Ò²b ×a×½¢Ð
PQ
Find the expression for de'Broglie wavelength of a particle in
the terms of kinetic energy relativistically. What will be the
value of λ if Show that the de-Broglie wavelength for a
material particle of rest mass m 0 and charge q accelerated
from rest through a potential difference of V volts
Oµ
relativistically is given by N =
²Pé cè?½ cè/²Pé µ² @
Solution:- We know that the total energy of a particle is the sum
of kinetic energy and rest mass energy. Therefore
E = Kinetic energy + Rest mass energy = k + moc2
………(1)
Also from relativistic mechanics, the total energy of the particle
is given by
=N
+ Ø Ó ………(2)
Comparing equation (1) and (2),
=N
+ Ø Ó=
(d + Ø )
Or
H
: >
\
+
Ø
Ó
= (d +
Ø
)
e?e ²Pé µ² @
O
If e ≪ P¢ µ² , then we obtain N =
√²Pg
In the case of a charged particle, we put K = qe, and therefore
the above equation transform to
Oµ
Oµ
N=
hR
N=
eÖ = 5.54 × 10&Ó
NUMERICALS ON OPTICAL FIBER
1. An optical fiber has NA of 0.2 and a cladding refractive
index of 1.59. Determine the acceptance angle for the fiber in
water which has refractive index of 1.33.
[ANSWER: Given that NA= 0.2, n2 = 1.59, no =1.33,
Acceptance angle- αmax = ?
NA= dL@ − L then, NA= dL@ − L =0.2 , n1 = √ (NA)2+(n2)2,
n1 = √ (0.2)2+ (1.59)2 = 1.6025. When fiber is in water n0= 1.33
then NA= dL@ − L / n0
NNA= √(1.6025)2-(1.59)2/1.33 = 0.15. Now αmax = sin-1NA ,
Sin-1(0.15) = 8.60.
2. An optical fiber has core and cladding indices 1.466 and
1.460 respectively and is operating at 1.3µm wavelength.
Calculate:
(1) NA (2) Acceptance angle (3) critical angle (4) Fractional
refractive index difference (5) V-number/Normalized
frequency/Cut-off parameter (6) No of modes (7) maximum
acceptable radius for single mode operation. Given the
diameter of core is 50µm.
[ANSWER:
(1) NA= dL@ − L = [ (1.466)2- (1.460)2 ]1/2 = 0.1324,
(2) Acceptance Angle: αmax= Sin-1NA = Sin-1(0.1324) = 7.6140
(3) The Critical Angle: The critical angle is given by Sinθc =
n2/n1 , OR θc = Sin-1(n2/n1) , Sin-1(1.460/1.466)= 84.8140
G %G
(4) Fractional Refractive Index Difference: ∆= •
=
Oµ
N=
After solving, we may obtain
cè?cè ²Pé µ² @
²Pé cè?½ cè/²Pé µ² @
NUMERICALS ON SCHRODINGER’s EQUATIONS:
1. A particle confined to move along X – axis has the wave
function ç = iì between ì = ¢ and ì = ½ and ç = ¢
elsewhere. Find the probability that particle can be found
betweenì = ¢. Qj iPk ì = ¢. aj.
Solution:-P = › çç∗ kì = ›¢.aj i² ì² kì
i² a ¢.Qj i²
"ì '¢.aj =
"!¢. Qj#a − !¢. aj#a ' = ¢. ¢½Ñi²
=
a
a
Find the minimum energy (ground state energy) and the first
excited state energy of an electron moving in one dimension
in an infinitely high potential box of width 1Å.
¢.Qj
2.
Solution:- We know that P =
ÐPl²
For minimum energy taking n = 1,
P² O²
¶½ =
= Ñ. ¢a × ½¢%½Ð ménoï
ÐPl²
For Ist excited state n = 2,
= P² ¶½ = ². Q½² × ½¢%½b m
A particle is in motion along a line between x = 0 and x = a
with zero potential energy. At points for which x < 0 and x >
a, the potential energy is infinite. The wave function is
Pqì
ç = pVWP
l
Normalized the wave function for the particle.
Solution: we apply normalization condition :
g
g
GI„
i.e.› UU ∗ _ƒ = 1or, › õ ÝÅL
_ƒ = 1
or,
P ² O²
3.
û
g
› }1 − cos
GI„
g
û
~ _ƒ = 1 or
g
E = 1 ⇒ õ=
g
Therefor the normalized eigen function of the particle is
U=
LKƒ
2
ÝÅL
E
E
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