Interference
Superposition Principle
For all linear systems, the net response at a given place and time caused by two
or more stimuli is the sum of the responses which would have been caused by
each stimulus individually. So if input A produces response X and input B
produces Y than input (A+B) produces response (X+Y).
Interference
When the light from two different sources with same frequency and having a
constant phase difference move in the same direction, then these light wave trains
superimpose upon each other. This results in the modification of distribution of
intensity of light. This modification of the intensity of light resulting from the
superposition of wave is called interference.
Coherent Sources
Two sources are said to be coherent if they emit light of same frequency and
always having a constant phase difference between them.
At some points the resultant intensity is greater than the sum of the intensities of
the waves.
Destructive Interference
Constructive Interference
At some points the resultant intensity is smaller than the sum of the intensities of
the waves.
21 III
21 III
This is called Constructive Interference.
This is called Destructive Interference.
The wave front originating from a common source is divided into two parts by
using mirrors, prisms or lenses and the two wave fronts thus separated travels and
finally brought together to produce interference.
In this type sources are small like a point source.
Classification of Interference
1. Division of Wave front
2. Division of Amplitude
The amplitude of the incoming beam is divided into two parts either by partial
reflection or refraction. These two parts travel in different paths and finally brought
together to produce interference.
In this type broad sources are required.
Young’s Double Slit Experiment:
crest
Trough
S
(Coherent Source)
S1
S2
(Slit)
(Screen)
Let S be a narrow slit illuminated by a monochromatic light of wavelength λ. S1 and S2
are two narrow slits close to each other and equidistant from S. Suppose is the
frequency of the waves. Let a1 and a2 be the amplitudes of the two wave coming out of
S1 and S2.
Analytical treatment of Interference
tay sin11
The displacement y1 due to one wave from S1 at any instant t is
The displacement y2 due to other wave from S2 at any instant t is
)sin(22 tay
Where Φ is the constant phase difference between the two waves.
The resultant displacement at P is the algebraic sum of the individual
displacements
21 yyy
)sin(sin 21 tatay
Squaring (i) and (ii) and then adding
222222
221
22
2
2
1 sincossincos2cos AAaaaaa
sincoscossinsin 221 tatatay
tataay cossinsin)cos( 221
coscos21 Aaa
sinsin2 Aa
tAtAy cossinsincos
(i)
(ii)
Let
For maximum intensity
1cos
2
2
2
1max aaI
or
IAaaaa 2
21
2
2
2
1 cos2
n2
2
2121
2
2
2
1max )(2 aaaaaaI
Phase Difference Path Difference
2
Path Difference = nλ
For minimum intensity
1cos
2
2
2
1min aaI
or
IAaaaa 2
21
2
2
2
1 cos2
)12( n
2
2121
2
2
2
1min )(2 aaaaaaI
Phase Difference Path Difference
2
Path Difference
For good contrast
21 aa 2
max 4aI 0minI and
2)12(
n
Average intensity
2
0
2
0
d
Id
Iav
When a1 = a2
2
0
2
0
21
2
2
2
1 )cos2(
d
daaaa
2
0
2
021
2
2
2
1 sin2 aaaa
2
2
2
1 aaIav
22aIav
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.
Calculation of the fringe width:
To determine the spacing between the bands/ fringes and
the intensity at point P.
D
S1
2d
S2
S
(Coherent Source)
Slit
Screen
O
P
d
d
N
M
x
Path difference (Δ) = S2P-S1P
To calculate S2P, consider the ∆S2NP
222
22 NPNSPS
222
2 dxDPS
2
1
2
2
2 1
D
dxDPSor
Expending by binomial theorem
2
2
22
11
D
dxDPS
Therefore, higher power term of D can be neglected. Then we get
2
2
22
1D
dxDPS
D
dxDPS
2
2
2or
dxD Here
Similarly, we can calculate S1P, consider the ∆S1MP
D
dxDPS
2
2
1
Then the path difference is (Δ) = S2P - S1P D
xd2
For the nth fringe the path difference = D
dxn2
(a)Bright Fringes:
The path difference should be equal to n
.
nD
dxn 2
nd
Dxn
2 where n = 0, 1, 2, 3, 4, ----------------
nd
Dn
d
Dxx nn
21
21
nd
D
d
Dn
d
D
222
d
D
2
d
D
2
The distance between two consecutive fringes is also known as fringe width.
2
12
n
2
122
nD
dxn
d
Dnxn
412
(b) Dark Fringes:
The path difference should be equal to
Point P to be dark
where n = 0, 1, 2, 3, 4, --------------
d
D
d
Dxx nn
24
21
d
D
2
Fringe width
D
d
1
From the above equations, it is clear that fringe width β depends on
1. It is directly proportional to the distance between two coherent sources
and screen
2. It is directly proportional to the wavelength of light
3. It is inversely proportional to the spacing between two coherent sources
.
Fresnel’s Biprism:
Fresnel’s biprism is a device to produce two coherent
sources by division of wave front.
D
b a
Overlap
region S
S2
S1
O
H
P
G
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’.
03
03
179°
(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:
U’
v u
v’
lensandobjectbetweencedis
lensandimagebetweencedis
Oobjecttheofsize
Iimagetheofsize
tan
tan
According to the linear magnification produced by the lens:
Further the lens moves towards the eyepiece and a focused image of virtual
sources S1 and S2 is visible in eyepiece again. This time the image separation of S1 and
S2 should be appear different (d2) so that:
u
v
d
d1
'
'2
u
v
d
d
(1)
(2)
d1 d2
L2 L1
S1
S2
d
1
2
d
d
d
d
212 ddd
21ddd
From equation (1) and (3), we get
or
But
'vu and
'uv
Thus equation (2) becomes
v
u
d
d2 (3)
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?
S1
2d
S2
t
x
D
P m
O
v
t
c
tPST
1
c
t
c
tPST
m
1
The time taken by light to reach P from S1 is
v
cmBut
ttPScT m 1
ttPSPS m 12
The path difference S2P and S1P will then be given by
The path difference between S2P and S1P is
m nttPSPS 12
m ntPSPS 112or
We have already calculated that PSPS 12 D
dxn2
m ntD
dxn 12
m ntd
Dxn 1
2or
Let the point P is the center of the nth bright fringe if the path difference is equal to nλ
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
(1)
td
Dx 1
20 m
m0xSince >1 so that is positive.
(2)
The fringe width is
d
Dxx nn
21
Using equation (1)
It means the fringe width is not affected by introduce of mica sheet.
d
D
2
Put these values in equation (2) we get,
1
2 0
mD
dxt
1
0
m
xtor
Thus we can calculate the thickness of mica sheet.
Light Reflection From Denser Media:
/2 Shift In Position of Wave
Inversion with /2 shift in position No inversion
Change of Phase on Reflection
When a wave of light is reflected at the surface of denser medium, it always gives a phase change
of π or path difference of λ/2
i Air
Glass
ar
at r
N M
D
C
B
A
a
Here r and t are the reflection and the transmission coefficients when wave is travelling from rarer
to denser medium.
If we reverse the direction of reflected and transmitted light then according to the Principle of
reversibility , the original wave of amplitude a is produced, provided that there is no absorption of
energy
i Air
Glass
ar
at r
N M
D
C
B
A
ar2+att’
art+atr’
Here r’ and t’ are the reflection and the transmission coefficients when wave is travelling from
denser to rarer medium.
The reversal of ar and at must reproduce the amplitude a. The sum of components along BE
should be zero.
0' atrart rr '
E
Interference due to Reflection:
Source
t
airinANpathfilminABCPath
i i
r r
R1 R2
T1 T2
A
B
C
D
M
N
(Reflected rays)
(Transmitted rays)
The path difference between the reflected rays
ANBCAB m
rAB
BMcos
r
tAB
cos
r
tBC
cos BCAB
rBM
AMtan
rBMAM tan
(1)
BM = t
and also
Now, for AN
rtAM tan
rtAC tan2
iAC
ANsin
iACAN sin
irtAN sintan2
r
ri
r
rt
sin
sinsin
cos
sin2
rr
rt sincos
sin2 m
ri sinsin m
r
rtAN
cos
sin2
2
m
AC = AM + CM
(because AM = CM)
or
r
rt
r
t
r
t
cos
sin2
coscos
2
mm
r
rt
r
t
cos
sin2
cos
2 2
mm
rr
t 2sin1cos
2
m
rr
t 2coscos
2m
rt cos2m
So that,
So that the actual path difference:
2cos2
m rt
As the ray is reflected from a denser medium, so an addition of
path difference of λ/2 will be there.
So for Maximum Intensity, path difference should be equal to
m nrt cos2
m nrt 2
cos2
n
2)12(cos2
m nrt
Where n = 0,1,2,3,4,5…………………..
So for Minimum Intensity, path difference should be equal to 2
)12(
n
2)12(
2cos2
m nrt
Where n = 0,1,2,3,4,5…………………..
Interference will not be perfect as there is difference in the amplitude of the reflected rays.
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
the following reasons
The path difference depend on 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 various
colors appear for the same angle of incidence when seen in white light.
The path difference changes with the angle r and angle r changes with
angle i. So that the films assumes various colors when viewed from
different directions with white light.
Newton’s Rings:
Source
2
cos2
m rt
2
cos2
m rt
Actually the path difference between the interfering rays is
The effective path difference for large radius of curvature or for small angle θ
m nt 2
2
2
122
m nt
2
122
2
m nt
m nnt 22
122
m nt 2
For normal incidence cosr =1, then the path difference 2
2
m t
For maxima
For minima
Why Central Ring is dark in Newton’s rings experiment
2
2
m t
At point of contact t = 0
2
Path difference for dark ring 2
12
n
For n = 0, Path difference 2
That’s why Central Ring is dark
How to calculate the radius or diameter of the nth fringe:
n
22 2 tRtn
Rt 2
Rtn 22
Rt n
2
2
Let be the radius of bright ring at point C and t is the thickness of air film at that point.
Let R be the radius of plano-convex lens. In triangle OAC
so that the higher power terms are neglected.
Therefore, we have
ρn
R
R
R-t
O
A
B
C
D
RttRR n 22222
222 tRR n
or
But t
2
122
m nt
2
122
22
m nR
n
m
2122 R
nn
For constructive interference: We have,
m
2
12nRn
m
2
1222
nRD nnDiameter
m
m
122
2
1242
nRnRDn
1m
Now, for the air film the refractive index
Therefore,
Therefore,
1222 nRDn
12 nDn
Diameter of the nth bright ring is proportional to the square root of the odd natural numbers
For Dark rings
m nt 2
m nR
pn 2
22
1mBut for air film
So,
Rnpn 2
Therefore, RnDn 42
nDn
Diameter of the nth dark ring is proportional to the square root of the natural numbers
Rnpn
Spacing between successive rings
This shows that the spacing decreases with increase in the order of the rings.
]1[1 nnkDD nn
]12[12 kDD
]23[23 kDD
]34[34 kDD
1222 pnRD pn
12212222 nRpnRDD npn
RnRRpRnR 24244
pRDD npn422
pR
DD npn
4
22
Determination of wavelength of light by Newton’s Rings method
thpn Now the diameter of bright ring,
Therefore,
1222 nRDn
For Air film,
RnD airn 4][ 2
Determination of refractive index of unknown liquid by Newton’s Rings method
For unknown liquid with refractive index m
m
RnD liquidn
4][ 2
liquidn
airn
D
D
][
][2
2
m