The redistribution of light energy due to
superposition of two or more waves with
same phase or constant phase difference
is called interference of light.
Consider two light waves represented by
Y1  A1 sin t and Y2  A2 sint    …(1)
Interference (superposition) of these two
waves at a point leads to the resultant
displacement given by, Y=Y1+Y2
Y  Y1  Y2 ..........................................................(2)
 A1 sin t  A2 sin t   
 A1 sin t  A2 sin t cos  A2 sin  cost
 sin t  A1  A2 cos   A2 sin  cost..........(3)
Define,
A1  A2 cos  R cos ...................................(4)
A2 sin   R sin  ....................................(5)
(4) and (5)in(3),gives
Y  R sin t cos   R sin  cos t
 R sint   ....................(6)
R is the amplitude of the resultant wave and

is the phase difference between the first wave and resultant wave.
Squaring and adding equations (4) and (5), it can be shown that
R  A1  A2  2 A1 A2 cos  ...............(7)
2
2
dividing (5) by (4), we get,
A2 sin 
tan 
.........(8)
A1  A2 cos
Constructive interference is said to be taken place, when two or more
waves interfere at a point in phase leading to the maximum intensity
(Brightness) at that point.
Destructive interference is said to be taken place when two or more
waves interfere at a point out of phase leading to the minimum
intensity (Darkness) at that point.
We have
R
A1  A2  2 A1 A2 cos 
2
2
For Constructive interference, R must be maximum, i.e., when cos(fi)=1
i.e.,  cos1 1  0,2 ,4 ,......2n .
Wheren  0,1,2,3,....etc.
T husfor constructive interference, thephasediffernecemust be
an integralmultipleof  . T hatis   2n .


In t ermsof P at hdifference,  

2n  n.
2
2
T hus we get const ructve
i int erference when t hepat hdifference,
  n.
When cos   1, Rmax 
or ,
Rmax 
A12  A22  2 A1 A2 (1)
 A1  A2 
Rmax  A1  A2
If
A  A1  A2
Rmax  A1  A2  2 A
2
We have
R
A1  A2  2 A1 A2 cos 
2
2
For Destructive interference, R must be minimum, i.e., when cos(fi)=-1
i.e.,  cos1 (1)   ,3 ,5 ......,(2n  1) .
Wheren  0,1,2,3,....etc.
T husfor destructive interference, thephasediffernecemust be
an odd multipleof  . T hatis   (2n  1) .


In termsof P athdifference,  

(2n  1)  2n  1  .
2
2
2
T hus we get destructive interference when thepathdifference,
  2n  1  2 .
When cos   1, Rmin 
or ,
Rmin 
A12  A22  2 A1 A2 (1)
 A1  A2 
Rmin  A1  A2
If
A  A1  A2
Rmin  A1  A2  0
2
Experimental Setup:
A parallel beam of monochromatic light passes through a single slit and
advances towards closely separated double slits S1 and S2 as a
single wave front. It is divided into two separate wave fronts by S1
and S2. The two wave fronts, which are coherent, advance towards
the screen Y.
While moving towards Y, the two wave fronts superpose according to
the principle of superposition of waves. At all points on the screen
where the crest of one wave falls on the crest of the other wave and
the trough of one wave falls on the trough of the other wave,
constructive interference takes place – giving rise to bright fringes.
At all the points on the screen where the crest of one wave falls on
the trough of the other wave, destructive interference takes placegiving rise to dark fringes. A pattern consisting of alternate bright and
dark fringes of equal width is obtained.
The path difference between the two waves
interfering at P is given by
  S 2 P  S1 P
From the right angled triangle S1QP
d

S1P  S1Q  PQ  D   X  
2

2
2
2
2
2
From the right angled triangle S2RP
d

S2 P  S2 R  PR  D   X  
2

2
2
2
2
2
2
2




d
d




2
2
2
2
S 2 P  S1 P   D   X      D   X   
2   
2  



2
2
d 
d

  X    X  
2 
2

d2
d 22
2
2
 X 2 Xd  2  
X  Xd d
 
2
2
S1P  S1Q  PQ  D 4   X  4
 2 Xd
2

or, S 2 P  S1 P S 2 P  S1 P   2 Xd
But, d  D hence P is very close to O and S1P= S2P=D
D  D S 2 P  S1 P   2 Xd
d

S2 P  S2 R  PR  D   X  
2

or
2
2 Xd
22D  2
Xd
 
D
2 or
To obtain bright fringe, the condition is

Xd
nD
 n  X 
D
d
2
Or the distance of the nth bright fringe from O is
nD
Xn 
d
. Similarly the distance of the (n-1)th bright fringe from O is
( n  1)D
X n 1 
d
nD (n  1)D D
 Fringe width   X n  X n1 


d
d
d
D
 
d
Similarly if we consider the formation of a dark fringe at P, we
get the condition for dark fringe as
Xd

(2n  1)D

 (2n  1)  X 
D
2
2d
Or the distance of the
nth
dark fringe from O is
(2n  1)D
Xn 
2d
. Similarly the distance of the (n-1)th bright fringe from O is
X n 1
[2(n  1)  1]D

2d
Therefore, for dark fringes, fringe width,
  X n  X n 1 
 
D
d
D
d
Therefore, Bright and dark fringes are of equal width
Coherent Waves:
Waves having same wavelength and
frequency are called coherent waves.
Or
Waves are said to be coherent if the
phase difference between them is zero or
constant at any instant of time.
Methods to produce coherent
source:
Two coherent sources can be obtained in two
ways, one, by division of amplitude and the
other, by division of wave front. The second
technique is used to produce coherent
sources in the Young’s double slit
experiment.
[Example of obtaining coherent sources by
division of amplitude method is by using
Biprism]
Conditions to obtain sustained
and distinct interference fringes
 The two sources of light must be coherent.
 A distance as small as possible must
separate the two coherent sources.
 The two slits must be very narrow.
 The two waves must have the same (or
nearly the same) wavelength and period.
 The two waves must have the same or
nearly the same amplitude.
Interference at thin films
Maxima or constructive int.
takes place when,
2t cos r  (2n  1) 
2
Minima or destructive int.
takes place when,
2t cos r  n
Where,
ref. index of the thin film
t thickness of the film
r angle of incidence
n=0,1,2,….
Write a short note on Newton’s
rings.
Newton’s rings are circular interference fringes formed by
interference at thin air film formed between a plane glass
plate and a convex lens of large radius of curvature.
In the adjacent diagram C is a Planoconvex lens of large radius of
curvature placed on the glass plate P.
Light from a monochromatic source is
rendered parallel by a convex lens L.
Another glass plate Q reflects the
parallel beam and allows it to fall on
the convex lens C. Here some of the
light is reflected from the upper
surface of the glass plate and some
from the lower surface of the lens.
Division of amplitude takes place.
Interference fringes are produced.
They are circular because the system
is symmetrical about the center of the
lens and hens the name Newton’s
rings.
Radius of mth dark fringe is given by
rm  mR
Radius of mth bright fringe is given by
rm 
2m  1R
2
m  0,1,2,3,4,.....
R Radius of curvature of the lens surface and
Wavelength of the light
I  2 f R v
2
I R
2
2
2
Constructive Interference
Destructive Interference
Rmax  A1  A 2
Rmin  A1  A 2
I max  A1  A 2 
I max 

I1  I 2
I min  A1  A 2 
2
2

2
I min 

I1  I 2
if A1  A 2  A
if A1  A 2  A
I max  4 A
I min  0
2

2
Formulae:
R  A1  A2  2 A1 A2 cos 
2
2
A2 sin 
t an 
A1  A2 cos
nD
Xn 
d
(2n  1)D
Xn 
2d
Rmax  A1  A 2
I max  A1  A 2 

d
Rmin  A1  A 2
I1  I 2
I min  A1  A 2 
2
2
I max 

D

2
I min 

I1  I 2

2
1. Calculate the %ge change in the
fringe width when the distance
between the double slit and screen
is increased by 20% while the slit
separation is increased by 50% in a
young’s double slit experiment.
Ans:
20%
2. The distance between two coherent sources is
0.3mm and the screen is 0.9 m from the sources. The
15th dark fringe is at a distance 2mm from the central
bright fringe, find the
a. wavelength of the light used.
b. distance of the 10th bright fringe from the central
bright fringe.
c. distance of the tenth bright fringe from the central
bright fringe when the separation between the screen
and the slits is increased by 0.1 m.
Ans:
44.4 nm
1.33 mm
1.48 mm
3. In a young’s double slit experiment, the slits are at
a separation of 1.5 mm. When the slits are illuminated
by a monochromatic source and a screen is held at a
distance 1.5 m, the width of 10 fringes is found to be
5.89 mm. When the screen is moved towards the slits
through a distance 0.5 m, the width of 10 fringes has
a value 3.93 mm. Find the wavelength of the light
used.
Ans:
589 nm
4. Two slits in a young’s experiment are 0.5
mm apart. Interference fringe of width 1.3 mm
are observed on a screen placed 1.2 m away
from the slits. Calculate the wavelength of the
light used. If the entire apparatus is immersed
under water of nw=4/3, find the change in the
fringe width. Hint : 0  n
w
Ans:
w
541.67 nm
0.325 mm
5. In an young’s experiment, interference
fringes are obtained on a screen placed 1.5 m
away from the two slits which are 1.5mm apart.
If the wavelength of the light used is 650 nm,
find
a) The fringe width and
b) Change in the fringe width if the screen is
moved away by 0.5 m.
Ans:
0.65 mm
0.216 mm
6. A beam of light consisting of two wavelengths 650 nm
and 520 nm is used to obtain interference fringes in a
young’s double slit experiment.
a) Find the distance of the third bright fringe on the screen
from the central maximum for the wavelength of 650 nm.
b) What is the least distance from the central maximum
where the bright fringe due to both the wavelengths
coincide?
The distance between the slits is 2mm and distance
between the slits and the screen is 1.2m.
Ans:
1.17 mm
1.56 mm
7. When two narrow slits separated by a small
distance illuminated by a light of wavelength 5x107m, interference fringes of width 0.5 mm are formed
on a screen. What should be the wavelength of light
source to obtain fringe 0.3mm wide if the distance
between the screen and the slits is reduced to half
the initial value?
Ans:
6x10-7 m
8. In a Young’s experiment, the width of
the fringes obtained with light of
wavelength 6000 A is 2.0mm. What will
be the fringe width, if the entire
apparatus is immersed in a liquid of
refractive index 1.33?
Ans:
1.5 mm
9. If the two slits in Young’s experiment have
width in the ratio 4:9, then find the ratio of
intensity at the maximum to the intensity at
the minima in the interference pattern.
Ans:
25:1
10. In young’s experiment, the distance of screen fro
the two slits is 0.1m. When a light of wavelength 6000
A is allowed to fall on the slits, the width of the fringe
obtained on the screen is 2.0mm. Determine
a) distance between the two slits.
b) the width of the fringe, if the wavelength of incident
light is 4800 A.
Ans:
3x10-4 m
1.6x10-3 m
11. In a Young’s double slit experiment, the
slits are separated by 0.03 cm and the
screen is placed 1.5 m away. The distance
between the central fringe and the fourth
fringe is 1cm. Determine the wavelength of
the light used in the experiment. [Hint:
4=1cm]
Ans:
5000 A
12. If the two slits in young’s experiment
have widths in the ratio 1:9, deduce the ratio
of intensity at the maxima and minima in the
interference pattern.
Ans:
4:1
13. The distance between two coherent
sources is 0.02x10-2 m. If the 6th band is at a
distance of 1.2x10-2 m from the central fringe,
find the wavelength of source used. The
distance of the screen from the wavelength
of source used. The distance of the screen
from the coherent sources is 1m.
Ans:
4000 A
14. The ratio of the intensities at minima to
maxima in the interference pattern is 16:25.
Determine the ratio of the widths of the two
slits in young’s experiment.
Ans:
81:1
15. In Young’s experiment, the interference fringes are
produced on the screen placed at 1.5 m from the 2
slits separated by a distance of 0.15x10-2 and
illuminated by a light of wavelength 5x10-7.
a) Find the fringe width
b) Find the change in fringe width when the screen is
brought towards the slits by 0.5m.
Ans:
5x10-4 m
1.67x10-4 m
16. Find the distance between 15th and 25th fringe
in an interference pattern due to two narrow slits
separated by 0.25x10-3m and illuminated by light of
wavelength 6x10-7m. The distance between the
screen and slit is 1m.
Ans:
24x10-3 m
17. Two coherent sources of wavelength 5893 A are
separated by 0.6mm. Calculate the fringe width
formed on a screen 0.5 m away from the sources.
Ans:
491x10-7 m
That’s it.