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PHYSICS-I
Course Code - 15B11PH111
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1
Teaching Process
The whole course will be covered through
1. Lectures (40)
2. Tutorials (13)
Evaluation scheme
There will be three exams all over the semester.
20 marks for the Test-I (1 hr).
20 marks for the Test-II (1 hr).
35 marks for the End Semester Exam (2 hr).
25 marks for internal assessment and class tests(
includes surprise tests
,assignments and attendance)
2
BOOKS
Text Book:
•
(available in library)
1.OPTICS by Ajoy Ghatak
2. A Text Book of Optics by BrijLal and Subramanyam
3.Concepts of Modern Physics by A. Beiser
Reference Book:
•
OPTICS, Eugene Hecht, Pearson Education.
•
Fundamental of Optics, Jenkins & White.
3
15B11PH111
1. INTERFERENCE
DIFFRACTION
PHYSICAL OPTICS
POLARIZATION
2 RELATIVITY
3 Quantum Mechanics
4 RADIATION
MODERN PHYSICS
5 ATOMIC STRUCTURE
6 Thermodynamics
4
PHYSICAL OPTICS :
LIGHT WAVES
• A light wave is a harmonic electromagnetic
wave consisting of periodically varying electric
and magnetic fields oscillating at right angles to
each other
• And also to the direction of propagation of the
wave.
• Electric field is defined by vector E and
magnetic field by vector B.
• The vector E is generally referred to electric or
optic vector.
5
Quantum theory of Light
All types of radiations are packets of energy i.e. quanta
Eαν
E = hν
h = Planck’s constant = 6.626 x 10-34 Js
ν = Frequency of the electromagnetic waves
E = energy of each quantum (photon)
Dual Nature of Light
Both as wave and as particle : characteristics under
different conditions
6
COHERENT SOURCES
Sources emitting light waves of the same frequency,
nearly same amplitude and always in phase with each
other or having a constant phase relationship between
them are called Coherent sources
Note : No two independent sources are coherent
7
TWO WAYS TO GET COHERENT SOURCE
•Division of wave front.
Young’s double slit
Loyd Mirror
•Division of amplitude ( Intensity)
Thin film interference
Note: In practice it is not possible to get independent two sources
which are coherent. But two virtual sources formed from one single
source can act as Coherent sources.
8
i n t er f er e n c e
9
SUPERPOSITION OF TWO WAVES:
INTERFERENCE
10
Optical Path: It indicates the
number of light waves that fit into
that path.
Δ = N λ where λ is optical path
length and N is an integer.
11
INTERFERENCE
• If two or more light waves of the same
frequency overlap at a point, the resultant
effect depends on the phases of the two waves
as well as their amplitude and is governed by
the principle of superposition.
• The combined effect is obtained by adding the
amplitudes of the individual waves.
12
There are two types of interference
Constructive
Two waves in phase
Destructive
Two waves 180° out
of phase
13
CONDITION FOR CONSTRUCTIVE INTERFERENCE
Phase difference   0, 2 , 4 ,.......  2n
or path difference x  0,  , 2 ..........n
CONDITION FOR DESTRUCTIVE INTERFERENCE
Phase difference    , 3 ,........  (2n  1)
 3
(2n  1)
Path difference x  ,
.........
2 2
2
Note: In each case n = 0,1,2….
14
Interference
• Let us take two waves of the same amplitude.
1. At certain points two waves may be in phase. The
amplitude of the resultant wave will be equal to
sum of the two.
AR = A + A = 2A
Hence the intensity of the resultant wave is
I R α A R 2 = 22 A 2 = 22 I
So resultant Intensity is greater than the sum of the
intensities due to individual waves
IR > I + I = 2I
15
• Therefore, the interference produced at this
point is known as constructive interference. A
stationary bright band of light is observed at
point of constructive interference.
• At other points. The waves may meet in
opposite phase. Then
AR = A – A = 0
IR α o2 = 0 IR < 2I .
Destructive interference. A stationary dark band.
16
• So when two or more waves are
superimposed, the resultant effect is a
band of alternate bright and dark
regions.
• These bands are called
interference fringes.
17
• Let us consider two sources of light S1 and S2 of
same wavelength and are in phase at S1 and S2 .
Lights from these sources travel along different
paths and reach at point P.
P
r1
S1
r2
S2
18
• Let the geometrical path S1P = r1 and the path
S2P = r2 which are different in length. Also the
media of travel is different. So as a result optical
path lengths will be different.
• If the refractive index of the media of ray r1 is
µ1 then the optical path length will be
µ1r1.
Similarly for the beam r2 it is µ2r2.
19
The optical path difference between two
waves at the point P is (µ1r1- µ2r2).
Although, the waves started with same
phase, they may arrive at P with
different phases because they traveled
along different optical paths.
20
• If the optical path difference Δ = (µ1r1- µ2r2) is
equal to zero or an integral multiple of λ, then
the waves arrive in phase. i.e Δ = n λ where n is
integer = 0,1,2,3-----.
• At P overlapping produces constructive
interference or brightness.
• If the optical path difference Δ = (µ1r1- µ2r2) is
equal to an odd integral multiple of half

wavelength λ/2 ----- (2n  1) then waves arrive
2
out of phase at P.
21
• The waves are inverted w.r.t each other and
produces destructive interference,
Δ = (2n+1)λ/2
or darkness.
• These regions of darkness and brightness are
also called regions of maxima and minima.
22
Theory of interference
Suppose that electric field components of two waves
arriving at point P vary with time as
where
is the phase difference between them.
According to Young’s principle of superposition, the
resultant electric field at the point P is
23
Phase shift
25
Above equation shows that the superposition of two
sinusoidal waves having same frequency but a phase
difference produces a sinusoidal wave with the same
frequency but with a different amplitude.
Let
where E is the new amplitude of resultant wave and
is new initial phase angle .
26
Square and add above two equations, we get
Thus square of the amplitude of the resultant wave is not a
simple sum of the squares of the amplitudes of the
superposing waves, but there is an additional term which is
known as interference term.
27
Intensity distribution
The intensity of a light wave is given by the square of its
amplitude
Using above equation in equation (*), we get
i.e. resultant intensity is not the just sum of the intensities
of two waves.
The term
is known as interference term.
ε0 =permittivity of free space
28
Now, when
, intensity of light is maximum
i.e.
When
Thus resultant intensity is more than the sum of intensities
of individual waves.
When
minimum
then
, intensity of light is
29
i.e.
When
then
Thus resultant intensity is less than the sum of intensities of
individual waves.
At points that lie between maxima and minima, when
we get
30
Above equation shows that intensity varies along the screen
in accordance with the law of cosine square. Figure shows
the variation of intensity as a function of phase angle.
31
Plot shows that intensity varies from zero at the fringe
minima to
at the fringe maxima.
32
INTERFERENCE
We have seen in the last article that
E2 = E12 + E22 +2E1 E2 cos φ
When φ = 2nπ, cos φ=1
Therefore,
E2 = E12 + E22 +2E1 E2 = ( E1 +E2 )2 = Imax
Similarly, When φ = (2n+1)π, cos φ=-1
E2 = E12 + E22 -2E1 E2 = ( E1 -E2 )2 =Imin
33
( E1 +E2 )2
Iav = I1+I2
( E1 -E2 )2
2
Iav 


0
Id
2
0
d


2
0
(E12  E 22  2E1E 2 cos )d

2
0
(E12  E 22 )2

 E12  E 22
2
 I1  I 2
d
Iav = 2I
If I1  I 2  I then I av  2I
34
• This result establishes that, in the interference
pattern, intensity of light is being simply
redistributed.
• i.e. energy is being transferred from regions of
destructive interference to the regions of
constructive interference.
• No energy is being created or destroyed in the
process.
• Thus the interference of light obeys the principle
of energy conservation.
35
Two coherent sources whose Intensity ratio is 100:1
produce interference fringes.
Deduce the ratio of
maximum and minimum intensity in fringe system.
Imax/Imin=121/81
Two coherent sources of Intensity ratio 
interfere. Prove that
Imax  Imin 2 

Imax  Imin 1  
36
For incoherent light
I  4 I 0 cos  / 2
2
1
2
2
2
 4E   2E  ( E  E )
2
2
= sum of intensity
of constituent waves.
37
Interference by division of wave front
(Young’s double slit experiment)
38
Wavefronts emerged from S1 and S2 overlap and
produce dark and bright fringes.
39
Division of wave front can be achieved by
allowing a monochromatic light to fall on a
narrow slit S. Light (wave front) from S is
further divided by two closely spaced narrow
slits S1 and S2.
If S1 and S2 are equidistant from S and the
phases of waves emerging from S1 and S2 are
same, then S1 and S2 act as coherent
secondary sources.
These wave fronts overlap and produce dark
and bright fringes.
40
Optical path difference between waves
θ
41
Optical path difference between waves
Let P be any point on screen at a distance D from
the double slit. Let θ be the angle between MP
and MO. Let S1N be a normal on S2P.
Distance PS1 and PN are equal. Hence difference
in the path lengths of these two waves is S2N, i.e.
 Nature of interference at point P depends on
the number of waves contained in the length of
path difference S2N.
 If S2N contains integral number of wavelengths
constructive interference or maxima is the result
42
at P.
 If S2N contains an odd number of half
wavelengths destructive interference or minima is
the result at P.
Let point P be at a distance x from O. Then
and
Now
43
Or
Also from figure
44
Condition for bright fringe
 Waves from S1 and S2 produce bright fringes
when they interfere constructively.
 First bright fringe occurs at O, the axial point.
Second, third …………… bright fringes occurs
when integral number of waves interfere.
In general, condition of finding bright at P
45
Optical Path Difference =
xd
 m
D
where m is the order of fringe.
m D
x
d
m = 0 corresponds to zeroth order bright fringe,
i.e. path difference between two waves is zero. This
is the centre of the screen i.e point O.
m = 1 corresponds to first order fringe,
i.e. path difference between two waves is λ and so
on.
46
Condition for dark fringes
 Waves from S1 and S2 produce dark fringes
when they interfere destructively .
 Dark fringes occur when half integral number of
waves interferes i.e path difference is (2m+1) λ/2 .
First dark fringe occurs when
Second dark fringe occurs when
and mth dark fringe occurs when
47
Thus condition for finding the dark fringes is
xd

 ( 2m  1)
D
2
x  (2m  1)
D
2d
where m is the order of fringes.
m=0 means the 1st order dark fringe.
48
Pattern of bright and dark fringes:
49
Fringe width (β)
 The distance between two successive bright or
dark fringes is known as fringe width and is same
everywhere on the screen.
For bright fringes ; mth order fringe occurs when
xd
 m
D
and (m+1)th order fringe occurs when
50
Fringe width (β) is given by
Similarly is for dark fringe.
51
Note:
Fringe width for bright fringe = fringe width for
dark fringe.
β is independent of the order of fringes.
1.
i.e.
fringes produced by blue light are closer
compared to those by red light.
2.
i.e. farther the screen wider the fringe separation.
3.
i.e. closer the slits wider the fringe separation.
52
Fresnel’s Biprism:
The results of Young’s double slit experiment
clearly explain interference and the wave
nature of light.
When the experiment was first done,
objections were raised about the sanctity of
the results since
there could have been diffraction effects
from the edge of the slits.
53
To counter this, Fresnel proposed an
interference experiment using Biprism.
Fresnel Biprism:
Here two virtual coherent sources of light
are created by refraction through a biprism.
Fresnel’s biprism consists of two acute
angle prisms placed base to base with each
other.
54
55
In reality the biprism is constructed from a
single plate of glass having obtuse angle of
1790 and acute angles of 30’ on both sides.
The prism is so adjusted in relation to the
source slit that
the two halves of the incident wave front
suffer separate and simultaneous refraction
through the prism; hence this single prism
is termed as biprism.
56
57
Light passing through the lower section is
refracted up, while light going into the top section
is refracted down.
This creates two virtual sources S1 and S2, with
an apparent separation ‘d’.
The waves from S1 and S2 interfere and give
interference fringes.
58
Fringe width in interference is given by
59
Determination of Fringe width (β)
The experiment is performed on a heavy metallic
optical bench about 2 meters in length and
supported on four leveling screws at the base.
The bench is provided with a scale on one side,
graduated in mm . The bench carries four uprights
for supporting the adjustable slit, the biprism, a
high power micrometer Ramsden’s eyepiece and a
convergent lens.
These uprights are capable of movement along
and also perpendicular to the length of the bench
and may be adjusted to any desired height.
60
When fringes are observed, set cross wire of eye
piece on any bright fringe. Note the position of eye
piece x (say).
Move the micrometer screw of eye piece and count
the number of bright fringes that pass across wire,
N (say). The position of cross wire is again noted,
let x’.
Then
61
Determination of separation of two virtual
sources (d)
Method 1:
Distance between two virtual sources can be
determined using lens displacement method.
A convex lens is introduced between the bi-prism
and the eyepiece and the latter is fixed at a
distance from the slit, greater than four times the
focal length of the lens.
A magnified distinct real image of two virtual
sources on the crosswire is observed. The
distance between the two images is measured. Let
it be d1.
62
Fig. (a)
Fig. (b)
63
The lens is then moved to the conjugate position
so that distinct diminished images are formed in
the plane of the crosswire. Separation d2 is
measured.
Since the magnification m1 in the first position is
just inverse of the magnification m2, we have
d1 , m  d 2
m1 
2
d
d
64
Deviation Method
Z
If δ is the small angle of deviation produced in the
path of ray by a thin prism
………….. (1)
where α is the refracting angle and µ is the
refractive index of prism.
d
2
z
 d  2 z

………….. (2)
65
or
d  2(  1)z
β = λD/d = λD/ 2(µ - 1)αz
If base angles are different, then
α2
µ
(d1+d2)
d  d1  d2  ( 1)(1  2 )z
α1
z
D
D
D


(d1  d 2 ) (  1) z(1   2 )
66
Fringes with white light
When white light is used the center fringe at O is white since all waves will
constructively interfere here while the fringes on the both side of O are colored
because the fringe width () depends on wavelength of light.
Position of nth bright fringe for different  are:
For green light,
For red light,
(x n )g 
n g D
2(g  1)z
n r D
(x n ) r 
2( r  1)a
67
Lateral displacement of fringes
Thickness of thin transparent sheet/film can be
determined by biprism experiment.
The thickness is measured by determining the
amount of displacement of fringes.
Suppose S1 and S2 are the two virtual coherent
monochromatic sources. Point O is equidistant
from S1 and S2, where central fringe is obtained.
i.e.
S1O =
S2O
68
69
when a transparent glass sheet of thickness ‘t’ and
refractive index ‘µ’ is introduced in the path of one
of the beam, say from S1.
Then
and central bright fringe is displaced from O to P.
The light from S1 travels partly in air and partly
through the glass sheet.
Distance travelled in air = S1P – t
Distance travelled through glass sheet = t
70
Optical path
 S1P  S1 P  t   t
Optical path
Optical path difference at P
Because in the presence of sheet, optical path
lengths S1P and S2P are equal and central fringe is
obtained at P.
71
Thus
Substitute the values of optical paths
But the path difference between two interfering
beams is
So from above equations
72
Or
where x is the lateral shift of the central fringe due
to the introduction of thin glass sheet.
path difference = nλ
73
Interference by division of amplitude
When a light beam is incident on a thin film, a part
of it is reflected and major part is transmitted.
There is further reflection and transmission in the
bottom part of film.
74
Interference due to multiple reflections from
the surface of transparent thin film was
observed by Newton.
Interference from thin films is due to
Reflected light
Transmitted light
75
Change in Phase
(Phase reversal on reflection)
Consider a light wave of amplitude E incident on
surface XY of two media of refractive indices n1
and n2 respectively.
M1
76
Let the coefficient of reflection into upper medium
be r1.
Then amplitude of reflected ray = r1E.
Let coefficient of transmission in lower medium be
t 1.
Then amplitude of transmitted wave = t1E.
Now, let reflected and refracted (transmitted) rays
be reversed.
Then amplitude of refracted ray = t1r1E.
77
Also reversed ray BA should be partially reflected
along AB’ and partially refracted along AO.
If coefficient of reflection into the lower medium
be r2.
Then amplitude of reflected ray along AB’ should
be r2t1E.
Total amplitude of ray of light along AB’ is
78
According to principle of reversibility, amplitude of
reversed ray along AO should be same as that of
incident ray, OA.
This is possible only if amplitude of ray along
AB’ = 0.
M1
But
79
Therefore
or
This is known as Stoke’s Law.
Now if a wave
is reflected into upper medium its amplitude will
be
If same wave is reflected into the lower medium,
its amplitude will be
80
Thus,
and
i.e. on reflection an additional phase difference of
π is introduced.
This additional phase difference is equal to a path
difference of
.
81
Interference of Light in Thin Films
(Reflected Light)
Consider a transparent film of thickness t and of
refractive index μ.
When a ray falls on the upper surface of the film, it
is partially reflected and partially transmitted.
The transmitted beam again partially reflect and
partially transmit from the lower surface of the film
(see Figure).
82
83
Interference of Light in Thin Films
(Reflected Light)
The interference takes place between the reflected
rays going along BR1 and DR2.
The path difference between the rays can be
calculated.
DN and BM are the normals drown on BR1 and CD.
If i is the angle of incidence and r is the angle of
refraction, the line CD meets back at point P to line
BE
84
By geometry
and
The optical path difference
But according to Snell’ law
Triangle BMP
PBM=90-r
Triangle PBD
Angle PBD=90
MBD=r
Therefore
BPD = r
BDM = 90-r
MBD= r
85
So path difference
Now since
therefore
so
In ∆ BPM,
86
hence path difference
Since, in this case light is reflected from surface of a
denser medium a path change of λ/2 occurs.
The effective path difference is then
But path difference for points of maximum intensity
is equal to mλ.
So for maximum (bright fringe)
87
This is the condition for maxima.
Similarly for minima (dark fringe):
88
The phase relationship does not change if one full
wave is added or subtracted from any of the
interfering waves. Therefore (m+1)λ can be
replaced by mλ in the above equation. Thus,
Here the interference pattern observed is not a
perfect pattern because the intensities of
interfering rays are not the same and the
amplitude depends on the amount of reflected and
transmitted rays.
89
Interference of Light in thin films
(Transmitted light)
•Consider a thin transparent film of thickness ‘t’
and of refractive index ‘μ’.
•A ray AB after refraction goes along BC.
•At the lower surface of the film, it will be partially
reflected along CD and partially transmitted along
CT1.
•The ray CD will again be partially reflected and
go along DE and partially transmitted along ET2.
•The interference occurs between the transmitted
ray CT1 and ET2.
•CN and EM are normal drawn on DE and CT1
respectively.
90
91
• The optical path difference between the
transmitted CT1 and ET2 ray is given by
• Path difference Δ = μ (CD +DE) – CM
CM
• But
sin i CE
μ

sin r EN
CE
CM  μ EN
Δ  μ (CD  DE)  μ EN
 μ (PD  DE)  μ EN
(CD  PD)
 μ (PE  EN)
 μ PN
By geometry PN  2t cosr
92
Therefore, the path difference  2 μ t cos r
The path difference for maxima = nλ
Hence for maxima, 2 μ t cos r  n λ
and for minima,
λ
2 μ t cos r  (2n  1)
2
Therefore, a phase difference of π or a path
difference of λ/2 occurs only because of reflection,
not because of refraction (transmission ).
Hence, this phase difference or path difference can
not be taken into account for transmitted pattern.
93