Engineering Physics-Chapter 1

1. INTERFERENCE OF LIGHT
by Dr. Vishal Jain
CHAPTER 1

2. Syllabus
PY-101 ENGINEERING PHYSICS CHAPTER 1
• Interference of light:
• Michelson’s Interferometer: Production of
circular & straight line fringes; Determination of
wavelength of light; Determination of wavelength
separation of two nearby wavelengths.
• Optical technology: Elementary idea of anti-
reflection coating and interference filters.

3. What is Light?
Light is an electromagnetic radiation refers to visible
regions of electromagnetic spectrum corresponding to
the wavelength range of 400nm to 760nm which has
transverse vibrations.
Wave

4. General Definitions
The Wavelength of a sin wave,
λ, can be measured between any
two points with the same phase,
such as between crests, or
troughs, as shown.
The frequency, f, of a wave is
the number of waves passing a
point in a certain time. We
normally use a time of one
second, so this gives frequency
the unit hertz (Hz), since one
hertz is equal to one wave per
second.

6. Principle of Superposition
“Whenever two or more waves superimpose in a medium, the
total displacement at any point is equal to the vector sum of
individual displacement of waves at that point”
If Y1, Y2, Y3…are different
displacement vector due to
the waves 1,2,3 …acting
separately then according to
the principle of superposition
the resultant displacement is
given by
Y=Y1+Y2+Y3+……

7. INTERFERENCE is the process in which two or more
waves of the same frequency - be it light, sound, or other electromagnetic
waves - either reinforce or cancel each other, the amplitude of the resulting
wave being equal to the sum of the amplitudes of the combining waves.
For example, if at a given instant in time and location along the medium, the
crest of one wave meets the crest of a second wave, they will interfere in such
a manner as to produce a "super-crest." Similarly, the interference of a trough
and a trough interfere constructively to produce a "super-trough." This is called
constructive interference. If the two amplitudes have opposite signs, they will
subtract to form a combined wave with a lower amplitude. For example, the
interaction of a crest with a trough is an example of destructive interference.
Destructive interference has the tendency to decrease the resulting amount of
displacement of the medium. The bright bands show constructive interference
of light. The dark bands show destructive interference of light.

9. Young’s Double Slit Experiment

10. MICHELSON’S INTERFERROMETER
The Michelson interferometer is a common
configuration for optical interferometer and was
invented by Albert Abraham Michelson in 1887.
Using a beam splitter, a light source is split into two
arms.

11. Principle:- The MI works on the principle of division of amplitude.
When the incident beam of light falls on a beam splitter which divided
light wave in two part in different directions. These two light beams
after traveling different optical paths, are superimposed to each other
and due to superposition interferences fringes formed.

12. Construction:- It consists of two highly polished plane mirror M1 and M2, with
two optically plane glass plate G1 and G2 which are of same material and same
thickness. The mirror M1 and M2 are adjusted in such a way that they are
mutually perpendicular to each other. The plate G1 and G2 are exactly parallel to
each other and placed at 45° to mirror M1 and M2. Plate G1 is half silvered from
its back while G2 is plane and act as compensating plate. Plate G1 is known as
beam-splitter plate.
The mirror M2 with screw on its back can slightly titled about vertical and
horizontal direction to make it exactly perpendicular to mirror M1. The mirror
M1 can be moved forward or backward with the help of micrometer screw and
this movement can be measured very accurately.
Working: Light from a broad source is made paralied by using a convex lens L.
Light from lens L is made to fall on glass plate G1 which is half silver polished
from its back. This plate divides the incident beam into two light rays by the
partial reflection and partial transmission, known as Beam splitter plate. The
reflected ray travels towards mirror M1 and transmitted ray towards mirror M2.
These rays after reflection from their respective mirrors meet again at 'O' and
superpose to each other to produce interference fringes. This firings pattern is
observed by using telescope.

13. Functioning of Compensating Plate: In absence of plate G2 the reflected ray passes
the plate G1 twice, whereas the transmitted ray does not passes even once.
Therefore, the optical paths of the two rays are not equal. To equalize this path the
plate G2 which is exactly same as the plate G1 is introduced in path of the ray
proceeding towards mirror M2 that is why this plate is called compensating plate
because it compensate the additional path difference.
Formation of fringes in MI

14. The shape of fringes in MI depends on inclination of mirror M1 and M2. Circular fringes are
produced with monochromatic light, if the mirror M1 and M2 are perfectly perpendicular to
each other. The virtual image of mirror M2 and the mirror M1 must be parallel. Therefore it is
assumed that an imaginary air film is formed in between mirror M1 and virtual image mirror
M'2. Therefore, the interference pattern will be obtained due to imaginary air film enclosed
between M1 andM'2.
From Fig. if the distance M1 and M2 and M'2 is'd', the distance between S'1 and S'2 will be
2D. These circular fringes which are due to interference with a phase difference with a phase
difference determined by the inclination '∂' are known as fringes of equal inclination. Circular
fringes can be seen by telescope because they are formed at infinity because they are formed
due to two parallel interfering rays. When d becomes zero, the whole pattern becomes dark.
Since a circular fringe is formed at the same inclination so they are called fringe of equal
inclination and also called Haidinger's fringes.

16. Radius of Fringes
The Condition for maxima and minima in MI is given by
 






2
1
cos2 nd  nd cos2
It is clear that on moving away from center the value of angle θ increases and the value of
cos θ decreases hence the order of fringe also decrease so n maximum at center, The
condition for nth dark ring at center is
nd 2
On moving m number of rings away from the center, the
order of mth ring will be ( n-m). If mth ring make an angle
θm with the axis of telescope then from equation
 )(cos2 mnd m 
For maxima For minima
……………Eq 1
……………Eq 2
D
θm
rm
n n-1
n-2
m
n-m
On Subtracting eq 1 and 2
 md m  )cos1(2
d
m
m
2
)cos1(

 
d
m
m
2
1cos,

  …Eq 3
Here
22
cos
Dr
D
m
m

 …Eq 4

17. By eq 3 and 4
d
n
Dr
D
m
2
1
22



22
2
1
Dr
d
m
D
m 


22
2
2
1
Dr
d
m
D
m 






























1
2
1
2
22
d
m
Drm

d
m
D
d
m
Drm
 222
)1...(
2
2
1 




d
m
Drm


This equation gives the radius of mth ring
……………Eq 5
……………Eq 6
……………Eq 7
……………Eq 8
……………Eq 9
……………Eq 10

18. Applications of MI
(1) Measurement of the wavelength of monochromatic light : The mirror M1 and
M2 adjusted such that circular fringes are formed. For this purpose mirror M1 and
M2 are made exactly perpendicular to each other.
Now set the telescope at the center of fringe and move the mirror M1 in any
direction, number of fringes shifted in field of view of telescope is counted.
Let on moving mirror M1 through x distance number of fringes shifted is N So the
path difference
By using both equations we will calculate wavelength corresponding to distance
and number of fringes shifted through telescope.
(2) Determination of the difference in between two nearby wavelengths :- Suppose
a source has two nearby wavelengths λ1 and λ2. Each wavelength gives rise its
own fringe pattern in MI. By adjusting the position of the mirror M1, aposition will
be found where fringes from both wavelength will coincide and form highly
contrast fringes.
Nx 2   Nxx  122

19. So the condition is given by
When a mirror M1 has been moved through a certain distance, the bright fringe due to
wavelength λ1 coincide with dark fringe due to wavelength λ2 and no fringe will be
seen. On further movies mirror M1 the bright fringes again distinct, this is the position
where n1+m order coincide with n2+m+1.
So the condition given by
Subtracting eq 2 by eq 1
So by eq 4 and 3
2211  nn  ……………Eq 1
2211 )1()(2   mnmnx ……………Eq 2
21 )1(2   mmx
)(
,)(
21
2
221




 mm
……………Eq 3
……………Eq 4












 21
21
21
1
21
2
2x
21
2
21
,
22


  where
xx

20. Problems & Solution
Q.1. In MI 200 fringes cross the field of view when the movable mirror is displaced through
0.05896mm. Calculate the wavelength of the monochromatic light used.
Solution:- Given
N=200
x= 0.05896mm = 0.05896 X 10-3 m
So the wavelength 5896105896
200
1005896.022 10
3


 

m
m
N
x
 Å
Q.2. The initial and final readings of MI screw are 10.7347 mm and 10.7057mm
respectively, when 100 fringes pass trough the field of view. Calculate the wavelength of
light used.
Solution:- Given
N=100
x=x2-x1= 10.7347-107057=0.029mm=0.029 X 10-3m
So the wavelength
5800105800
100
10029.022 10
3


 

m
m
N
x
 Å

21. Problems & Solution
Q.3. MI is set to form circular fringes with light of wavelength 5000Å. By Changing the
path length of movable mirror slowly, 50 fringes cross the center of view How much path
length has been changed?
Solution:- Given
N=50
λ= 5000 X 10-10m
So the path length m
mn
d 6
10
105.12
2
10500050
2






Q.4. In a Michelson Interferometer, when 200 fringes are shifted, the final reading of the
screw was found to be 5.3675mm. If the wavelength of light was 5.92 X 10-7 m, What was
the critical reading of the screw?
Solution:- Given
N=200
x=x2-x1= 5.3675 X10-3m - ?
and wavelength λ = 5.92 X 10-2m
So the wavelength
N
x2
 m
mN
x 5
7
1092.5
2
1092.5200
2
, 





Now initial reading of screw d1=d2 ± x = 5.3675 x 10-3m + 0.0592 x10-3m =5.4267 x 10-3m

22. Interference in Optical Technology
Antireflection Coating:- In 1935, Smaulka discovered that reflection losses can be
reduced by coating the surface with a thin transparent dielectric film known as
antireflection coating.
Principle:-
The schematic diagram of a single layer
antireflection coating is outline in Fig. 1. This
coating consists of a layer of transparent
material of thickness equal to one quarter the
wavelength of the light in the coating material.
The index of air and that of the optical part to be
coated (µ2=(µ1µ3)1/2 for minimum reflection) .
Reflection occurs at each surface of the coating.
The two reflected waves are 180 degree out of
phase and cancel through destructive
interference; no energy, therefore, can appear in
the reflected beam, and all the light is
transmitted.
Hence a film can acts as antireflection coating if it satisfies the following two conditions
(1) Phase Condition, (2) Amplitude condition
n1=1.0
n2=1.3
n2=1.5
λ/4
Fig. 1. Antireflection coating

23. Incident Ray Ray 1 Ray 2
t
Air (µa)
Film (µf)
Glass (µg)
Top Surface
Bottom Surface
(1) Phase Condition:- The wave reflected from the top and bottom surface of thin
film should be in opposite phase so that they cancel each other due to destructive
interference. It enable us to determine the required thickness of the thin film.
From the fig 2. the optical path difference between
ray 1 and ray 2 is
Fig. 2. Antireflection coating
22
cos2

  rtf
Here the first λ/2, is due to π phase change at air to
film surface and the second λ/2 is due to π phase
change at thin film/glass surface since both the
reflection occurs from denser medium with respect
to rare medium.
For normal incidence cos r = 1
  t2
Addition of the λ in pat difference does not effect
anything so
t2
For destructive interference 

 






2
1
2
)12(2 nntf
For minimum thickness of thin film n=0
f
f tt



4
,4 m inm in 

24. (1) Amplitude Condition:- The wave reflected from the top and bottom surface of
thin film should have equal amplitude. Let the amplitude of reflected ray one and
ray two are A1 and A2 respectively.
A1=A2
As the light is striking from air to thin film/glass boundaries. Suppose the refractive
index’s of air, film and glass medium are μa, μf and μg respectively
When a beam of light travel from one medium μ1 to another medium μ2, a part of
light is reflected and the optical reflectivity of surface given by
2
12
12











R
From the equation 1 and 2 the reflectivity
…………………….Eq 1
…………………….Eq 2
22





















fg
fg
af
af

























fg
fg
af
af




……..….Eq 3
………….Eq 4
))(())(( affgfgaf  
)(
)(
2
2
fafagfg
fagafgf




agf  2
agf  
For air µa=1
gf  
…….Eq 4
…….Eq 5
…….Eq 6

25. Multilayer Antireflection Coating:- The construction and working of a dielectric
coating used for partial and total reflection is shown in fig. The coating is composed
of alternating layers of two transparent dielectric materials, one having a high index
of refraction and another has lower index of refraction. Each layer has a thickness of
λ/4 where λ is the wavelength of light in the appropriate layer and reflection occurs
at each boundary between the two materials.
When the direction of light travel is from a low index material to a high index material, there is
a 180 degree phase change in the electric field of the reflected wave this occurs at surface 1, 3
and 5. So the wave reflected are 180 degree phase change. When light travels from higher
refractive index material to lower one, there is no phase change in reflected wave.

26. Interference in Optical Technology
Interference Filter :- An interference filter is an optical system that
will transmit a very narrow range of wavelengths (monochromatic
beam of light)
Principle:-
When parallel beam of light is incidence normally on a pair of plane
parallel glass plate silvered on a inner surface and separated by a thin
transparent film, then interference occurs for all wavelengths of
white light and the maxima of different orders are formed in the
transmitted light corresponding to wavelength given by
 nrt cos2 For normal incidence, cos r = 1
 nt 2



n
t2 Here t = Thickness of thin film separating the plate
μ = Refractive index of transparent film

27. Construction and Working :- In an interference filter, a thin transparent film of
MgF2 or Cryolite is placed between two semi reflective coating. When a beam of
light is incident normally on the filter, multiple reflections take place within the film.
The condition for maxima in transmitted light is given by
Cover glass plate
Monochromatic
Cover glass plate
Reflecting metal film Transparent
dielectric material
 nt 2
2


n
t 
Hence if the optical thickness µt of the film is integral multiple
of half wavelength then the other wavelengths will be
eliminated by destructive interference and λ, 2λ.. Will be
transmitted through the filter without any obstructions.
White Light

28. If the angle of incidence is “i” , the wavelength of light passing through filter is
given by
 nrt cos2
 nrt  2
sin12
……………………………..[1]
……………………………..[2]
by using Snell’s law 
i
r
sin
sin 


 






2
0
sin
1
i
For n=1


 n
i
t 






2
sin
12 ……………………………..[3]
……………………………..[4]
Here λ0=2μt
Interference filter widely used in instrumentation for clinical chemistry, environment
testing, colorimeteric, elemental and laser line seperation and fliroescene.

29. Problems & Solution
Q.5. Find the reflectivity of a glass surface of refractive index 1.6 when a light beam is
incident normally on it.
Solution:- Given
µg=1.6 and µair= 1
So the optical reflectivity is given by
Q.6. Consider an antireflection film of refractive index 1.38.Assume that its thickness is 9 X
10-6 cm. Calculate the wavelength in the visible region for which the film will be
antireflection.
Solution:- Given
μf= 1.38
Thickness t= 9 X 10-6cm= 9 X 10-8m
So the wavelength
Angstrommht f 49681068.4938.141094 88
 

%32.50532.0
6.2
6.0
16.1
16.1
222
12
12

























R

30. Problems & Solution
Q.7. Find the minimum thickness of a layer of magnesium floride μ = 1.38 required in an
interference filter design to isolate light of wavelength 5893Å. How will peak transmittance
change if the filter in tilled by 10 degree.
Solution:- Given
µmf=1.6
Wavelength = 5893 X 10-10m
2µt = nλ so for n=1
Q.6. Two λ/4 thick layers are deposited on an opthalmic glass (μ =1.52) to reduce reflection
loss. The first layer has refractive index (μ =1.38). Find the reflective index of the material
of second layer.
Solution:- Given µg=1.52, μ1 =1.38, μ2 =?
We know that by using reflectivity amplitude condition µf= (1.38 X 1.52)1/2=1.448
Angstrom
m
h
t
mf
2135
38.12
105893
2
10






Angstromm
h
i
mf
5846
38.138.1
10sin
1105893
sin
1
2
10
2
2
0
'


 

For 10 degree tilled filter
Angstrom5058965846'
 
When the filter is tilled, transmittance always changes towards shorter wavelength.