Interference of Light:

Due to superposition of two light waves emitted from coherent sources intensity of light
increases at some points and decreases at other points. As a result alternate bright and dark state
is produced on a plane. The alternate variation of intensity of light from point to point on a plane
is called the interference of light.

Coherent Sources:
When two sources emit light waves of the same colour (wavelength), of the same frequency,
nearly the same amplitude and always in phase with each other, then the two sources are called
coherent sources.

It is not possible to have two independent sources which are coherent. But for experimental
purpose, two virtual sources formed from a single source can act as coherent sources.

Condition for interference of light:

1. The waves from two sources must be of the same frequency.
2. The two light waves must be coherent.
3. The path difference between the overlapping waves must be less than the coherence
length of the waves
4. The wavelengths of two interfering waves must be same and their amplitudes are either
equal or nearly equal to each other.
5. The interfering waves must be produced in the same direction.
6. The light waves should be reach at the interfering points at the same time.
7. The sources must be monochromatic.
8. The two coherent sources must lie to each other in order to discern the fringe pattern.
9. The distance of the screen from the two sources must be large.
Young’s Double Slit Experiment:
Consider a narrow monochromatic source of light S and two pinholes S1 and S2, equidistance
from S.

S1 and S2 acts as two coherent sources separated by a distance d.

Let a screen XY be placed at a distance D from the coherent sources. The point C is the centre
point of XY which is equidistance from the coherent sources. So the path difference between
waves at C is Zero. Thus the point C has maximum intensity.

Consider a point P at a distance x from C on the screen XY. The waves reach at the point P from
S1 and S 2.

S1 Q

C
d
S

S2 R

Now, from figure,

PC = x

S1S2 = QR = d
QR = CR =d/2

Hence,

PQ = PC – QC = x – d/2 …………………………..(1)

PR = PC + CR = x + d/2 …………………………..(2)

Now, From ∆ S1PQ,

S1P2 = S1Q2 + PQ2

= D2 + (x – d/2)2 …………………………….(3)

From ∆ S2PR,

S2P2 = S2R2 + PR2

= D2 + (x + d/2)2 …………………………….(4)

Subtracting equation (3) from equation (4),

S2P2 - S 1P2 = D2 + (x + d/2)2 - D2 - (x – d/2)2

or, S2P - S1P = 2xd / (S2P + S1P)

But S2P = S1P = D ( approximately)

So,

S2P - S1P = 2xd/2D = xd/D

Where, S2P - S1P is not equal to zero.

Therefore, Path difference,

x’ = xd/D ……………(5)

So, The phase difference,

δ = 2πx’ / λ = 2πxd / λD……………………(6)

Case 1:

For Bright Frings,

If the path difference is a whole number multiple of wavelength λ, then the point P is bright.

Hence, For bright,

x’ = nλ [where, n = 0,1,2,3,…….]

or, xn d/D = nλ

or, xn = nλD/d

When , n = o, x0 = 0

n = 1, x1 = λD/d

n = 2, x2 = 2λD/d

n = 3, x3 = 3λD/d

…………………………………………………..

…………………………………………………

Now,

x1- x0 = λD/d

x2 - x1 = λD/d

x3 - x2 = λD/d

…………….

……………

xn - xn-1 = λD/d

Therefore, The distance between any two consecutive bright frings is λD/d.

Case –II:

If the path difference is an odd number multiple of half wavelength then the point P is dark.

So, for dark fringe,

x’ = (2n +1) λ/2 [where, n = 0,1,2,3,……………]

or, , xn d/D = (2n + 1)λ/2

or, , xn = (2n + 1) λD /2d

When , n = o, x0 = λD/2d

n = 1, x1 = 3λD/2d
 n = 2, x2 = 5λD/2d

n = 3, x3 = 7λD/2d

…………………………………………………..

…………………………………………………

Now,

x1- x0 = λD/d

x2 - x1 = λD/d

x3 - x2 = λD/d

…………….

……………

xn - xn-1 = λD/d

Therefore, the distance between any two consecutive dark frings is λD/d.

Fringe width:

The distance between any two consecutive bright or dark fringes is known as fringe width.

If the width of the fringe is represented by β,

Then, β = λD /d

Which means that, all the fringes are equal in width and are independent of the order of the
fringe.

β
Newton’s Rings:
If we place a plano-convex lens on a glass plate a thin air film is formed between the curved
surface of lens and plane glass plate. If we allow monochromatic light to fall on the curved
surface of the lens, then the frings are product in the air film and contour lens will be circular.
The ring shaped frings thus produced wave analyze by Newton and is hence known as Newton’s
ring. When viewed with white light, the fringes are coloured.

Construction:

A B
P Q

Let, AOB be a plano-convex lens on the glass plate POQ. A thin air film is formed between the
AOB and POQ. The thickness of the air film is zero at the point of contact O and increases as
one moves away from the point of contact.

If we allow a monochromatic source of light S to fall on the air film. To do this a glass plate G is
placed at 45 with the incident light. This placed plate G reflects the light down on the plate.

After reflection from this surface POQ, it is transmitted through G and observed rings by a
microscope M.

For near normal incidence the optical path difference between two waves is nearly equal to 2μt
(where μ is the refractive index and t is the thickness).
Thus, whenever, the thickness of the air film satisfies the condition,

2μt = (2n +1) λ/2 …………………….. (1), where, n =0,1,2,3,………………………

we will have the maxima.

And similarly the condition,

2μt = nλ ……………………………(2)

will correspond to minima.

Radius of the nth dark ring:

Suppose the radius of curvature of the lens is R and thickness of the air film is t at a distance of
OQ = r, from the point of contact O.

Here, the interference is due to reflected light.

R
Let, the radius of nth ring is rn. 2R-t
C
For bright ring,
R
A B
2μtcosθ = (2n + 1) λ /2 H E F

Here, θ is small. So, cosθ = 1 P O Q

Then, 2μt = (2n + 1) λ /2 ……………………………………….(1)

For dark rings,

2μt cosθ = nλ

But, θ is very small. So, cosθ = 1

Then, 2μt = nλ …………………… (2)

From figure,

EF × HE = OE (2R –OE) …………..(3)

Here, EF = HE = r and OE = FQ = t

So, r2 = t (2R –t) ………………(4)

Generally, R>>t; So, we may neglect t in comparism to 2R.

So r2 = t (2R) =2Rt

or, 2t = r2 / R ……………….(5)

From equation (1), for bright rings,

μ r2 / R = (2n + 1) λ /2
(2 + 1)
or, r = 2

(2 + 1)
 r= 2
[for air, μ =1]

From equation (2), for dark rings,

μ r2 / R = n λ

or, r =

 r= , [for air, μ =1]

For nth ring,

rn = ………………………….(6)

Equation (6) implies that the radii of the rings vary at square root of natural numbers. Thus, the
rings will become close to each other as the radius increases.

Dark Centre:

When, n = 0 the radius of the dark ring is zero and the radius of the bright ring is 2 . So the
centre is dark. Alternately dark and bright fringe are produced.

Wavelength:

If the diameter of the nth dark ring is Dn and radius r n, then

rn2= nλR

  Dn2 = 4nλR …………………….. (7)

And also the diameter of the (n + m)th ring is,

Dn+m2 = 4(n+m)λR……………….(8)

Subtracting equation (7) from (8) we get,

Dn+m2 - Dn2 = 4(n+m-n)λR

  λ = (Dn+m2 - D n2) / 4 mR …………………. (9)

Where, m is the difference between the higher and lower order of the ring.

Now, by knowing the value of radius of curvature of the lens with the help of a spherometer the
wavelength of a given monochromatic source of light can be determined.

Refractive Index:

If this arrangement of figure is kept in a metal container C. The liquid is poured in the container
C without disturbing the arrangement. The air film between the lower surface of the lens and
upper surface of the plate is replaced by the liquid. Then for the liquid the nth order of the dark
rings from equation (6) is given by,

D’n2 = 4nλR/μ

Similarly for (n+m)th dark ring,

D’ n+m2 = 4(n+m)λR/μ

Therefore,

D’n+m2 – D’n2 = 4mλR/μ

or, μ = 4mλR / (D’n+m2 – D’n2) [for known wavelength]

  μ = (Dn+m2 - D n2) / (D’n+m2 – D’n2) [for unknown wavelength]

Interferometers:
Interferometry is a family of techniques in which waves, usually electromagnetic waves,
are superimposed causing the phenomenon of interference in order to extract information.
Interferometry is an important investigative technique in the fields of astronomy, fiber
optics, engineering metrology, optical metrology, oceanography, seismology, spectroscopy (and
its applications to chemistry), quantum mechanics, nuclear and particle physics, plasma
physics, remote sensing, biomolecular interactions, surface profiling, microfluidics, mechanical
stress/strain measurement, velocimetry, and optometry.
Interferometers are widely used in science and industry for the measurement of small
displacements, refractive index changes and surface irregularities. In an interferometer, light
from a single source is split into two beams that travel different optical paths, then combined
again to produce interference. The resulting interference fringes give information about the
difference in optical path length. In analytical science, interferometers are used to measure
lengths and the shape of optical components with nanometer precision; they are the highest
precision length measuring instruments existing. In Fourier transform spectroscopy they are used
to analyze light containing features of absorption or emission associated with a substance or
mixture. An astronomical interferometer consists of two or more separate telescopes that
combine their signals, offering a resolution equivalent to that of a telescope of diameter equal to
the largest separation between its individual elements.

Types of interferometers:

 Michelson Interferometer
 Twyman-Green Interferometer
 Fabry-Perot Etalon
 Scanning Fabry-Perot Interferometer
 Mach-Zehnder Interferometer etc.

Diffraction of light:
When a beam of light passes through a narrow slit of aperture to a screen, then the geometrical
shadow of the obstacle is not distinct uniform and a completely dark. This phenomenon is called
diffraction of light.

In other word, when a beam of light passes through a narrow slit, it spreads out to a certain
extend into a region of geometrical shadow of an object is known as diffraction.

Classification of diffraction of light:

Diffraction phenomena are conveniently divided into two general classes-

1) Fresnel Diffraction
2) Fraunhofer Diffraction

The difference between Fresnel and Fraunhofer diffraction are as follows –

Fresnel diffraction Fraunhofer diffraction

1) Either source of light or screen or both 1) Source and screen on which the
are effectively at finite distance from diffraction phenomenon observed are at
the aperture causing the diffraction of infinite distance from the aperture.
light.
2) No lenses are required to occur the 2) More than one lens is required.
diffraction phenomenon.
3) Incident wave front will be spherical or 3) Incident wave front will always plane.
cylindrical.
4) It is difficult to treat theoretically. 4) It is easy to treat theoretically.
5) It is easy to treat experimentally. 5) It is difficult to treat experimentally.
Diffraction by a Single slit:
 b

Let us consider a section of a slit of width b, illuminated parallel light from the left. Let ds be an
element of the wave front in the plane of the slit. The parts of each secondary wave which travel
normal to the plane will be focused at Po and through an angle θ will reach at P.

Consider the wavelet emitted by the element ds situated at origin whose amplitude will be
directly proportional to the ds and inversely proportional to the distance x.

The displacement of spherical wave at P be expressed as,

a ds
dy0 = x sin (ꙍ ― )

If ds is above the origin,

a ds
dy+s = x sin (ꙍt ― kx ― kssinθ)

If ds is below the origin,

a ds
dy-s = x sin (ꙍt ― kx + kssinθ)

So, the resultant displacement is given by,

a ds a ds
dy = dy+s + dy-s = x sin (ꙍt ― kx ― kssinθ) + x sin (ꙍt ― kx + kssinθ)
2a ds
= x {sin (ꙍt ― kx) cos (kssinθ)}
Integrating s = 0 to b/2 we get,

2ads
y = ∫2 {sin (ꙍt ― kx) cos (kssinθ)}
0

2a
= x sin (ꙍt ― kx) ∫2 cos (kssinθ)ds
0

2a sin (ks sinθ)

= x sin (ꙍt ― kx) ksinθ
b
2a sin (k2 sinθ)
= x sin (ꙍt ― kx) ksinθ
b
ab sin (k2 sinθ)
= x sin (ꙍt ― kx) b
k2sinθ

  y = A sin ( t – kx)

Where,
b
ab sin (k2 sinθ)
A= x b
is the resultant amplitude.
k2sinθ

b
ab sin (k2 sinθ)
Let, A0 = x and β = b
k2sinθ

sin β
So, the amplitude, A = A0 β

Thus the intensity of diffraction pattern on screen is given by,

I = A2 = A02 (sin2β / β2 )

Which is the intensity of diffraction pattern by single slit.

Condition for maximum and minimum intensity:

Since intensity depends on amplitude. So here,

sin β
A = A0 β will define the maxima and minima.

For central maxima, β→0

 sinβ
  lim β
=1
→0

So, I = A2 = A02 = I0

Maximum intensity falls to zero at β = ± , ± 2 , ± 3 , …………………

= mπ

Condition for minima:

Intensity will be minimum if amplitude is minimum

  Imin = 0

When sinβ = 0

or , sin β = sin nπ, [where, n= 0,1,2,…..]

 β = nπ

Hence, (kb/2) sin θ = n π

or, (b/2) (2π /λ) sin θ = nπ [ where, k = (2π /λ)]

  bsinθ = nλ
Diffraction due to a double slit :

d c

Let us consider two equal slits of width b, separated by an opaque space of width c, the origin
may be choosen at the center of c. So, the separation of the slits is d = b+c.

The parallel rays are reflected at an angle θ to the normal of the slit. Consider for small element
ds due to the emitted wavelet by the element, the infinitesimal displacement may written as,
a ds
dy0 = x sin (ꙍ ― )

If ds is above the origin,

a ds
dy+s = x sin (ꙍt ― kx ― kssinθ)

If ds is below the origin,

a ds
dy-s = x sin (ꙍt ― kx + kssinθ)

So, the resultant displacement is given by,

a ds a ds
dy = dy+s + dy-s = x sin (ꙍt ― kx ― kssinθ) + x sin (ꙍt ― kx + kssinθ)
2a ds
= x {sin (ꙍt ― kx) cos (kssinθ)}
For total integrating by taking limit,

s = (d-b) /2 to (d+b)/2

we get,
+
2ads
y = ∫d ―2 b {sin (ꙍt ― kx) cos (kssinθ)}
2

+
2a
= x sin (ꙍt ― kx) ∫ 2
― cos (kssinθ)ds
2

2a k (d + b)sinθ k (d ― b)sinθ
= x sin (ꙍt ― kx) sin{ 2
} - sin{ 2
}

2 sin ( t ― kx) k dsinθ k bsinθ k dsinθ k bsinθ

= kxsinθ
sin( 2
+ 2
) - sin( 2
- 2
)

2 sin ( t ― kx) k dsinθ k bsinθ

= kxsinθ
2 cos ( 2 ) sin ( 2 )

Let,
k bsinθ 2 bsinθ
β = 2
= λ 2
= λ bsinθ

k dsinθ 2 dsinθ
γ = 2
= λ 2
= λ dsinθ

Therefore,
2 sin ( t ― kx)
y= kxsinθ
2 cosγ sinβ

2 sinβ
= cosγ sin ( t ― kx)
2

2 sinβ
= cosγ sin ( t ― kx)

sinβ
= 2A0 cosγ sin ( t ― kx) [where, A0 = ]

= A sin ( t ― kx)

sinβ
Where, Amplitude, A = 2A0 cosγ
  Intensity,
sinβ sinβ
I = A2 = (2A0 cosγ)2 = 4 A02 ( cosγ)2

Condition for minima:

The resultant intensity will zero when either (sin2β )/β2 or cos 2γ is zero.

That is,

β = π, 2π, 3π,……………,mπ

  (kb/2) sin θ = mπ,

So, bsinθ = mλ

This is the condition no 1 for minima

or,

γ = π/2, 3π/2, 5π/2,……………,(2m +1)π/2

so, dsinθ = (2m +1)λ/2

this is the condition no 2 for minima.

Condition for maxima:

For maximum,

γ = 0, π, 2π, 3π, ……………………..nπ

  dsinθ = mλ

This is the condition for maima.

Resolving power of optical instrument:
The ability of an optical instrument expressed in numerical measure, to resolve the images of two
nearby points is termed as its resolving power.

Diffraction grating:
An arrangement which is consists of a large number of parallel equidistance slits of the same
width is called diffraction grating. Corresponding diffraction pattern is known as grating spectra.

The sum of the width of each slit and width of the opaque space is called grating constant.

Let, b = breadth of each slits

c = width of opaque space

d = (b+c) = grating constant/ grating element

therefore, for a grating with N lines per inch,

d = (2.54 / N) cm

Polarization of light:
Ordinary light consists of transverse vibrations that is the vibrations are at right angles to the
direction of propagation of the wave. An ordinary beam of light consist million of such waves,
each with its own plane of vibration. If by some means the vibrations constituting the beam of
light are confined to one plane, the light is said to be plane polarized.

The process by which light vibrations are confined to one particular direction is known as
polarization.
Polarization by reflection or Brewster’s Law:
Brewster’s law states that,

“When the angle of incidence is equal to the angle of polarization for a transparent substance,
the tangent of the angle of polarization will be equal to the refractive index of the substance.
Moreover, the reflected and the refracted rays are perpendicular to each other.”

If φ is the polarizing angle and μ is the refractive index of the medium, then according to the
Brewster’s law,

μ = tanφ

Explanation: N

A C

Unpolarized light plane polarized

i Ф

B
r
glass refracted light

Let an uonpolarized light AB is incident at an angle equal to the polarizing angle on the glass
surface. It is reflected along BC and refracted along BD.

From Snell’s law,

sin
μ = sin
and from Brewster’s law,

μ = tanφ =

By comparing, we get,
sin
sin =

sin
or, sin = [since, i = φ ]

or, cosφ = sinr

or, cosφ = cos (2 - r)

or, φ = 2 - r

 φ+r=2

Therefore, The refracted and the reflected (polarized) rays are perpendicular to each other.

Problem 1:
An unpolarized light is incident at an angle equal to the polarizing angle on glass surface. For
a refractive index 1.54, what is the value of polarizing angle?

Solution:

Here,

Refractive index, μ = 1.54

Polarizing angle, φ =?

We know that,

μ = tanφ

or, φ = tan-1 (1.54)

   φ = 57

Problem 2:
The path difference between two points of a wave is 4λ. What is the phase difference between
these two points?

Solution:

Here,

Path difference, x = 4λ

Phase difference, δ = ?

We know that,

δ = (2π / λ) x

= (2π / λ) × 4 λ

= 8π radian

Problem -3: The straight and narrow parallel slits of 1mm apart are illuminated by
monochromatic light. Fringes formed on the screen held at a distance of 1m from the slits
are 0.50mm apart. Calculate the wavelength of light used.

Solution:

We are given,

Distance between two slits, d = 1mm

Distance of the screen from source, D = 1m =1000mm

Fringe width, β = 0.50mm

Wavelength, λ =?

We know that,

β=
 or, λ = = 5 × 10 – 4 mm = 5 × 10 – 4 × 10 6 Å = 500Å

Problem -4: A plano-convex lens of radii 3 m is placed on an optically flat glass plate and
illuminated by monochromatic light. The diameter of the 8th bright ring in the transmitted
system is 0.72cm. Calculate the wavelength of light used.

Solution:

Here,

Radius of plano-convex lens, R = 3m = 300cm

No of dark ring, n = 8

Diameter of the nth ring, D = 0.72 cm

  radius of the nth ring, r = 0.36 cm

Wavelength, λ =?

We know that,
(2 + 1)
r2 = 2

2
or, λ = (2 + 1)
= 5760 Å

Problem – 5: In Newton’s ring experiment, the diameter of the 5th ring was 0.336cm and
that of the 15th ring was 0.590cm. Find the radius of curvature of the plano-convex lens if
the wavelength of light used was 5890 Å.

Solution:

Here,

Diameter of 5th ring, D5 = 0.336cm

Diameter of 15th ring, D 15 = 0.590cm

Wavelength, λ= 5890 Å = 5890 × 10 -8cm

n = 15 – 5 =10

Radius of curvature, R =?

We know that,
R = {( Dn+m)2 – (Dn)2} / (4nλ)

= 99.83cm