Subject: PHYSICS-I
Subject Code: PHY-101F
L T P
3 1 0
Internal (Sessional + Class performance) 50 Marks
Exam
100 Marks
Total
150 Marks
Duration of Exam
3 Hrs.
SYLLABUS
(Ist Semester)
1)Interference
2)Diffraction
3)Polarization
4)Laser
5)Fiber Optics
6)Dielectrics
7)Special Theory of Relativity
8)Superconductivity
Section A
Interference: Coherent sources, conditions for sustained
interference. Division of Wave-Front - Fresnel’s Biprism,
Division of Amplitude- Wedge-shaped film, Newton’s
Rings,
Michelson
Interferometer,
applications
(Resolution
of closely spaced
spectral lines,
determination of wavelengths).
Diffraction: Difference between interference and
diffraction
Fraunhofer
and
Fresnel
diffraction.
Fraunhofer diffraction through a single slit, Plane
transmission diffraction grating, absent spectra,
dispersive power, resolving power and Rayleigh criterion
of resolution.
Section B
POLARISATION
Polarised and unpolarised light, Uniaxial crystals double
refraction, Nicol prism, quarter and half wave plates,
Detection and Production of different types of polarized
light, Polarimetry; Optical and specific rotation, Biquartz
and Laurent’s half shade polarimeter.
LASER
Spontaneous and Stimulated emission, Laser action,
characteristics of laser beam-concept of coherence ,
spatial
and
temporal
coherence,
He-Ne
and
semiconductor lasers (simple ideas), applications
Section C
FIBRE OPTICS
Propagation of light in optical fibres, numerical aperture,
V-number, single and multimode fibres, attenuation,
dispersion, applications.
DIELECTRICS
Molecular theory, polarization, displacement vector,
electric susceptibility, dielectric coefficient, permittivity &
various relations between these, Gauss’s law in the
presence of a dielectric, Energy stored in a uniform
electric field, concept of local molecular fields and
Claussius Mossotti relation.
Section D
SPECIAL THEORY OF RELATIVITY
Michelson’s Morley Experiment, Postulates of Special
Theory of
Relativity, Lorentz transformations,
Consequences of LT (length contraction and time
dilation), addition of velocities, variation of mass with
velocity, mass energy equivalence.
SUPERCONDUCTIVITY
Introduction (Experimental survey), Meissner effect,
London equations, Hard and Soft superconductors,
Elements of BCS Theory.
Text Books:
Fundamentals of Engineering Physics
Modern Physics for Engineers
M. S. Khurana, MR Pub, Delhi
S P Taneja; R Chand Publication
Engineering Physics
Satya Prakash, Pragati Prakashan
Engineering Physics
R. K. Gaur & S. L. Gupta
Reference Books:
Concepts of Modern Physics
Beiser
Optics,
A. Ghatak,
Optics
Eugene Hecht
Fundamentals of Optics
Jenkins & White
Lectures on Physics
Feynman
Practical Books:
Fundamentals of Engineering Physics – M. S. Khurana, MR Pub, Delhi
Advanced Practical Physics – B.L. Worshnop and H. T. Flint
Practical Physics – S. L. Gupta & V. Kumar (Pragati Prakashan).
Advanced Practical Physics Vol.I & II – Chauhan & Singh.
Course Objective:
The objective of this course is to apply
the concepts of physics
(i) to different optical phenomena
(i) devices based on these phenomena
INTERFERENCE
Interference?
Why we need to study?
There are two possibilities:
1.
2.
Maximum intensity position
Minimum intensity position
Young’s Experiment:
crest
A
Trough
S
(Coherent Source)
B
(Slit)
(Screen)
Coherent sources ?
Conditions for sustained interference:
Two light sources must be coherent.
Two coherent sources must be narrow, otherwise a single source
will act as a multi sources.
The amplitude of two waves should be equal so that we can get
good contrast between bright and dark fringes.
The distance between two coherent sources must be small.
The distance between two coherent sources and screen should be
reasonable. The large distances of screen reduce to intensity.
Theory of Interference
P
A
O
d
S
(Coherent Source)
B
D
Slit
Screen
The path difference is an even multiple of

2
, the waves arrive
at a point in the same phase and produce maximum intensity.
If the path difference is an odd multiple of

2
, the two waves
arrive at a point in the opposite phase, they cancel out and the
resultant intensity should be minimum.
How to calculate the fringe width:
To determine the spacing between the bands/ fringes and
the intensity at point P.
P
x
S1
d 2
S
(Coherent Source)
M
O
d
d 2
N
S2
D
Slit
Screen
The path difference can be calculated from fig.
Path difference (Δ) = S2P-S1P
To calculate S2P, consider the ∆S2NP
S 2 P 2  S 2 N 2  NP 2
d

S2 P2  D2   x  
2

1
2
2
 
d 
x

 
 
2

 
S 2 P  D 1 
or

D2 




Expending by binomial theorem and neglecting higher power term of D
as D>>  x  d 
2

2
We have
2


d

x 


1
2
        
S 2 P  D 1  
2
 2

D




d

D is very very large in comparison of  x  
2

Therefore, higher power term of D can be neglected. Then we get
2
 
d 
 x   
2 
S 2 P  D 1  

2D 2 




or
2

d

 
x   

2 
S 2 P  D  

2D 




Similarly, we can calculate S1P, consider the ∆S1MP
2

d 

x  

2 
S1 P   D  

2D 




Then the path difference is (Δ) = S2P-S1P putting the value
of S2P and S1P then we have
Path difference 
For the
nth
xd
D
xn d
fringe the path difference =
D
(a) Bright Fringes:
We have already pointed out that for bright fringe, the path
difference should be equal to n
.
xn d
 n
D
D
xn  n
d
where n = 0, 1, 2, 3, 4, ----------------
x  0
Therefore, the distance between two consecutive fringes is also known as fringe width.
D
n  1
d
D
xn  n
d
xn 1 
x n 1  x n 
D
n  1  D n
d
d

D
D
D
n    n
d
d
d
x n 1  x n 
D

d

D

d
(b)
Dark Fringes: Similarly, for dark fringes, the path difference
should be equal to 2n  1

2
Point P to be dark;
xn d



 2n  1
D
2
D
x n  2n  1
2d
where n = 0, 1, 2, 3, 4, --------------
x n 1  x n 
2 D D 

2d
d
D
 
d
From the above equations, it is clear that fringe width
depends upon following factors:

It is directly proportional to the distance between two
coherent sources and screen   D 
It is directly proportional to the wavelength of light    
It is inversely. proportional to the spacing between two coherent sources
1





d


Type of interference:
Interference by division of wave front
Interference by division of amplitude
Fresnel’s Biprism:
Fresnel’s biprism is a device to produce two coherent
sources by division of wave front.
P
S1
G
Overlap
region
S
S2
O
H
a
b
D
Q
Construction:
A biprism consists of a combination of two acute angled prisms placed
base to base.
The obtuse angle of the biprism is 179º and other two acute angles are 30’.
30 
179°
30 
Applications of Fresnel’s Biprism:
Determination of thickness of thin sheet of transparent
material like glass or mica.
or
How to calculate the displacement of fringes when a mica
sheet is introduced in the path of interfering rays?
t
m
x
S1
Po
d
S2
P
O
D
Let the point P is the center of the nth bright fringe if the path difference is equal to nλ
The path difference between S2P and S1P is
  S 2 P  S1 P  t   mt   n
or
S 2 P  S1 P  m  1 t  n
We have already calculated that S 2 P  S1 P  
xn d
D
xn d
 m  1t  n
D
or
xn 
D
m  1t  n 
d
(1)
Where xnis the distance of the nth bright fringe from the central fringe in the absence of mica.
The position of the central bright fringe when the mica sheet is placed in
the path S1P is obtained by putting n=0 in equation (1) we get
x0 
Since
m
D
m  1 t 
d
(2)
>1 so that x0 is positive.
The fringe width is
   x n 1  x n

D
d
Using equation (1)
It means the fringe width is not affected by introduce of mica sheet.
D 

d 
Put these values in equation (2) we get,
t
x0 d
Dm  1
or
t
x0 
 m  1
Thus we can calculate the thickness of mica sheet.
(b) Determination of the distance between two virtual sources:
Displacement method is one of the methods to calculate the distance
between two virtual coherent sources:
m
n
n
m
According to the linear magnification produced by the lens:
size of the im ageI  dis tance between im ageand lens

size of the object O  dis tance between object and lens
d1
n

d
m
Further the lens moves towards the eyepiece and sets a potion of image of virtual sources S1 and S2 in
eyepiece again. This time the image separation of S1 and S2 should be appear different (d2) so that:
d2
m

d
n
From equation (1) and (2), we get
or
d2
d

d
d1
d 2  d1d 2
d  d1d 2
Interference due to Reflection:
(Reflected rays)
R1
R2
Source
i iN
M
A
r r
B
C
t
D
T1
T2
(Transmitted rays)
  Path ABC in film  path AN in air
  m  AB  BC  AN
(1)
BM
 cos r
AB

BM = t
AB 
and also
BC 
t
cos r
t
cos r
  AB  BC
Now, for AN
AM
 tan r
BM
AM  BM tan r
or
AM  t tan r
 2t tan r
AC = AM + CM
(because AM = CM)
Now,
AN
 sin i
AC
or
AN  AC sin i
or
AN  2t tan r sin i
 2t
sin r
sin r
sin i 
cos r
sin r
 2 mt
sin r
 sin r
cos r
or
sin i  m sin r
sin 2 r
AN  2mt
cosr
So that,
t 
sin 2 r
 t
  m

  2mt
cosr
 cosr cosr 
2mt
sin 2 r

 2mt
cosr
cosr



2mt
1  sin 2 r
cos r

2 mt
cos 2 r
cos r
  2mt cosr
So that the actual path difference:
  2 mt cos r 

2
Production of colors in thin films:
When a thin film of oil on water, or a soap bubble, exposed to white light (such as
sunlight) is observed under the reflected light. The brilliant colors are seen due to
following reasons:   2mt cosr
The path difference depend µ or the wavelength. It means the path
difference will be different for different colors, so that with the white light
the film shows various colors from violet to red.
The path difference also varies with the thickness of film so that it
passes through various colors for the same angle of incidence when seen
in white light.
The path difference changes with the angle r and angle r with change
with angle i. So that the films assumes various colors when viewed from
different directions with white light.
Non uniform thickness (Wedge shaped film):
X
X’
E
M’
i
i
D
r
i
µ
t
B r
O
θ
N r+θ
M
C
I
I’
The path difference (Δ) = path [BC+CD] in film – path BE in air
Δ = µ (BC+CD)-BE
or
Δ = µ(BC+CI)-BE
because CD=CI
= µBI-BE
= µ (BN+NI)-BE
In right angle triangle BED,
sin i 
In right angle triangle BND,
sin r 
BE
BD
BN
BD
…………………(1)
…………………(2)
…………………(3)
From equations (2) and (3) we have
sin i
BE

sin r BN
or
m
BE
BN
or
Putting the value of BE in equation (1), we have
  m BN  NI   mBN
or
  mNI
In right angle triangle DNI,
NI
 cos( r   )
DI
We know
DI  DH  HI  t  t  2t , so that
NI  2t cos(r   )
Therefore,
  mNI  2mt cos(r   )
BE  mBN
Applying Stoke’s law,
Therefore, the net path difference is
  2mt cos( r   ) 


2
is very small, so can be neglected then the path difference is
  2 mt cos r 

2
Now, the condition for brightness is
2mt cos r 

2
 n
n = 0, 1, 2, 3, ……………
or
2mt cos r  2n  1
t
2n  1
4m cos r

2
Putting n = 0, 1, 2, 3,……………
t

4 m cos r
,
3
5
,
, …………………..etc.
4 m cos r 4 m cos r
The film will appear dark if
2mt cos r 

2
 2n  1

2
or
2mt cosr  n  1
n = 0, 1, 2, 3, …………
or
2mt cos r  n
t
n = 1, 2, 3, …………
n
2 m cos r
It is observed that in the direction of increasing thickness of the film there
will be alternate bright and dark fringes parallel to the edge of the film.
Fringe width:
t
θ
XN
Let us consider XN is the position of Nth bright fringe, where the
thickness of the film is t then
tan 
t
XN
Since θ is very small, so that

t
XN
or
Putting the value of t then we have (for bright fringe)
t   .X N
2n  1
4m cos r
  .X N
For the small angle of incidence cos r  1
because angle r will also be small then
XN 
2n  1
and
4m
X N 1 
Then the fringe width
   X N 1  X N


2m


2m
2n  3
4m
Similarly, we can calculate the fringe width for dark fringes.
We have already calculated that
t   .X N
Now we calculate the fringe width for dark fringe put the value of t
form equation (2), we have,
n
  .X N
2m cos r
Similarly for the small angle of incidence cos r  1
n
  .X N
2m
and
XN 
or
X N 1 
n
2m
n  1
2m
Then the fringe width is
   X N 1  X N
or


2m


2m
Newton’s Rings:
Source
Actually the path difference between the interfering rays is
   2mt cos r     
2
Therefore, the effective path difference
   2mt cos r  
2
For normal incidence cosr =1, then the path difference
For maxima
  2 mt 
or

2
2 mt  2n  1
For minima
2 mt 
or
or
2mt  2n  1

2

2

 n


2

2
 n  1
2mt  n
2
where n = 0, 1, 2, 3, 4, -------
2
 2n  1
   2mt  
where n = 0, 1, 2, 3, 4, -----
where n = 1, 2, 3, 4, -------
How to calculate the radius or diameter of the nth fringe:
Let  n be the radius of nth bright ring and t is the thickness of air film.
Let R be the radius of plano-convex lens. By the geometry of the circle
D
R 2   n2  R  t 
2
R    R  t  2Rt
2
or
But
2
n
2
R
2
O
 n2  2 Rt  t 2
R
R-t
t  2 R
A
B
so that the higher power terms are neglected.
Therefore, we have
 n2  2 Rt
or
t
 n2
2R
ρn
C
For constructive interference: We have,
2 mt  2n  1
or
2m
2R
 2n  1
n2  2n  1
or
or
 n2

2

2
R
2m
 R2n  1 

2m


n  
 R2n  1 

2m


Diameter Dn  2  n  2 
Dn2 
4 R2n  1 2 R2n  1

2m
m
Now, for the air film the refractive index m  1
Therefore,
Dn2  2R2n  1
n  p th
Now the diameter of
bright ring,
n  p   2R2n  p   1
D2
Therefore,
D2
n  p 
 Dn2  2R2n  p   1  2R2n  1
 4nR  4 pR  2R  4nR  2R
or
D2
n  p 

 Dn2  4 pR
D2
n  p 
 Dn2
4 pR
Another important application of Newton’s ring:
With the help of Newton’s ring experiment we can determine the
refractive index of a liquid. For this purpose, first of all the experiment
is performed with air film and diameter of nth and (n+p)th rings.
Liquid
We have already calculated that for air,  D2n  p   Dn2   4 pR

 air
-----(1)
Now the liquid is introduced between plano-convex lens and glass plate and the
same experiment is repeated. Let m be the refractive index of the liquid used,
then
4 pR
D 2
 Dn2 

-----(2)
 n  p 
 liquid
m
From equation (1) and (2), we have
D 2
 Dn2 

 n  p 
 liquid
D 2
 Dn2 


 n  p
 air
m
D 2
 Dn2 


n

p

 liquid
D 2
 Dn2 
 n  p 
 air
m
Michelson Interferometer:
M1
M 2'
M2
Monochromatic
Source (S)
B
Telescope (T)
C
Let the distance between M1 and M2’ is d, which is also known the thickness
of the air film and θ is the angle which reflected beam makes with the
normal. As the ray after reflection from M2 suffers reflection at the silvered
surface of the glass plate B and the additional path difference of
2
is introduced between the two interfering rays.So that the path
difference   is
  2 mt cos r 
  2d cos  

2

2
Now condition for bright and dark rings
For dark rings
For bright rings
Here
or t = d, r is θ and µ = 1 (for air) so that
Forms of the fringes:
(i)
Circular fringes: When M1 and
M 2' are exactly parallel to each other.
The fringes produce concentric circles.
(ii)
'
Curved and straight fringes: When M1 and M 2
are inclined so that air film is wedge shaped.
M 2'
M1
M 2'
M1
M 2'
Fig: Various types of fringes.
M1
APPLICATIONS OF MICHELSON INERFEROMETER:
(i)
Determination of wavelength of monochromatic light
When the mirror move through a distance  , one fringe shift and there
2
is a corresponding change of  in the path difference.
If N fringes shift across the cross wire, when the mirror moved
through a known distance x, then
xN


2
2x
N
By knowing x and N, the wavelength

can be determined.
(ii) Determination of refractive index or thickness of transparent sheet:
When a thin transparent sheet of thickness ‘t’ and refractive index ' m ' is introduced
in between glass plate and mirror M2 and adjustment are made so that the light pass through
it normally. In this position the path difference will be
m  1 t
As the beam travels the plate twice, an extra path difference m  1 t is added.
So that the total path difference = 2m  1 t
Therefore, the fringes must shift when the movable mirror is moved till the fringes
are back to their initial position. If the displacement of the movable mirror is x, then
2x  2m  1 t
or
t
x
m  1
(iii)
Determination of difference in wavelengths:
we can write,
2 x  n1
2 x  n  12
and
or
or
n
2x
1
n 1 
2x
2
Subtracting equation (1) from (2), we get
n 1 n 
2x
2

2x
1
or

1  2  1 2
2x
Since 1 and 2 are almost equal, thus, we have 12  2avg .
So that
1  2 
2avg .
2x
------(1)
------(2)