UNIT 2 -Wave Optics

Content
1. Young’s double slit experiment and Fringe width
2. Newton’s ring experiment
3. Michelson’s interferometer
4. Difference between diffraction and interference.
5. Fraunhofer diffraction at single slit
6. Resolving power and Resolving power of grating
 1. Young’s Double Slit Experiment

Young’s double slit experiment uses two coherent sources of light placed at a small
distance apart. Usually, only a few orders of magnitude greater than the wavelength of
light are used. Young’s double slit experiment helped in understanding the wave theory
of light, which is explained with the help of a diagram. As shown, a screen or
photodetector is placed at a large distance, ‘D’, away from the slits.
The original Young’s double slit experiment used diffracted light from a single source
passed into two more slits to be used as coherent sources. Lasers are commonly used as
coherent sources in modern-day experiments.

Fringe width

Consider a narrow monochromatic source S of two parallel narrow slits S1 and S2

close together and equidistant from S.
Let d= separation between S1 and S2
D= distance between screen and coherent source.
At point O on screen, intensity is maximum as the path difference between the
waves is zero.
Consider a point at a distance x from O.
𝑑 𝑑
So, 𝑃𝑄 = 𝑥 − 2 𝑃𝑅 = 𝑥 + 2
 𝑑 2 𝑑 2
So, 𝑆2 𝑃2 − 𝑆1 𝑃2 = [𝐷2 + (𝑥 + 2 ) ] − [𝐷2 + (𝑥 − 2 ) ]
= 2𝑥𝑑
2𝑥𝑑
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑆2 𝑃 − 𝑆1 𝑃 =
𝑆2 𝑃 − 𝑆1 𝑃
But 𝑆2 𝑃 ≅ 𝑆1 𝑃 = 𝐷
2𝑥𝑑 𝑥𝑑
Therefore path difference 𝑆2 𝑃 − 𝑆1 𝑃 = =
2𝐷 𝐷
2𝜋 𝑥𝑑
And phase difference 𝛿 = (𝐷)
𝜆

Fringe width of dark fringe

𝑥𝑑
Path difference = = 𝑛𝜆
𝐷
𝑛𝜆𝐷
So, 𝑥 = 𝑑
𝜆𝐷
Now distance of 1st dark fringe from center is 𝑥1 = 𝑑

2𝜆𝐷
𝑥2 =
𝑑
𝜆𝐷
So, 𝑥2 − 𝑥1 = 𝑑

𝜆𝐷
So the distance between the consecutive bright fringes is 𝑑

Fringe width of bright fringe

𝑥𝑑 𝜆
Path difference = = (2𝑛 + 1)
𝐷 2
(2𝑛+1)𝜆𝐷
So, 𝑥= 2𝑑
3𝜆𝐷
Now distance of 1st dark fringe from center is 𝑥1 = 2𝑑

5𝜆𝐷
𝑥2 = 2𝑑
𝝀𝑫
So, 𝒙𝟐 − 𝒙𝟏 = 𝒅

𝜆𝐷
So the distance between the consecutive dark fringes is 𝑑
𝜆𝐷
Hence fringe width 𝛽 = 𝑑

So, 𝛽 𝛼 𝜆
𝛽𝛼𝐷
1
𝛽𝛼
𝑑
 2. Newton’s Ring

When a plano convex lens of large radius of curvature is placed with its convex
surface in contact with a plane glass plate, an air film of gradually increasing
thickness from the point of contact is formed between the upper surface of the plate
and lower surface of the lens.
If monochromatic light is allowed to fall normally on this film then alternate bright
and dark concentric rings with their center dark are formed. These rings are called
“Newton’s ring”.
Rings are obtained in reflected light, so effective path difference is
𝜆
∆= 2𝜇𝑡 cos 𝑟 + − − − − − −[1]
2
For normal incidence i=0=r. So cos r=1 and for air film 𝜇 = 1
So equation 1 becomes
𝜆
∆= 2𝑡 + − − − − − [2]
2
For maxima,
𝜆
2𝑡 + = 𝑛𝜆
2
Or
 𝜆
2𝑡 = (2𝑛 − 1) − − − − − − − [3]
2
For minima
𝜆 𝜆
2𝑡 + = (2𝑛 + 1)
2 2
2𝑡 = 𝑛𝜆 − − − − − −[4]
Diameter of rings
In In ∆𝑂𝐵𝐷
𝑂𝐵 2 = 𝑂𝐷2 + 𝐷𝐵 2
𝑅 2 = (𝑅 − 𝑡)2 + 𝑟 2
𝑅 2 = 𝑅 2 + 𝑡 2 − 2𝑅𝑡 + 𝑟 2
(Since𝑡 ≪ 𝑅 𝑠𝑜 𝑛𝑒𝑔𝑙𝑒𝑐𝑡 𝑡 2 )
2𝑅𝑡 = 𝑟 2
𝒓𝟐
𝟐𝒕 = − − − −[5]
𝑹

To find diameter of bright rings compare equation 3 and equation 5

𝑟2 𝜆
= (2𝑛 − 1)
𝑅 2
𝐷2 𝜆𝑅
= (2𝑛 − 1)
4 2
𝑫 = √𝟐(𝟐𝒏 − 𝟏)𝝀𝑹

𝑫 𝜶 √(𝟐𝒏 − 𝟏)
So diameter of the bright rings are proportional to the square root of odd natural
numbers.
To find the diameter of dark rings compare equation 4 and 5.
𝑟2
= 𝑛𝜆
𝑅
𝐷2
= 𝑛𝜆𝑅
4
𝑫 = √𝟒𝒏𝝀𝑹

𝑫 𝜶 √𝒏
So diameter of dark rings is proportional to the square root of natural numbers.
3. Michelson’s Interferometer

https://www.youtube.com/watch?v=UA1qG7Fjc2A
1. Michelson interferometer designed to determine the wavelength of light, thickness
of the strips and meter standardization.
2. Principle: - In this beam of light from source is divided into two parts by partial
refraction and reflection. After travelling in mutually perpendicular directions, they
reunite and overlap to produce interference fringes.
3. Construction: -
a. C is beam splitter of simple glass.
b. D is compensating plate of same thickness as C.
c. Both C and D are held parallel to each other and at 45 0 to M2.
d. M2 is a mirror mounted on movable carriage. Distance of movement of M2 can
be read on graduated drum.
e. M1 and M2 are made perpendicular to each other with screw attached to them.
f. Fringes are observed in the field of view of telescope.
4. Working: -
a. Monochromatic light from source S is rendered parallel by lens L and is incident
on the beam splitter C. Then ray is partly reflected along L2.
b. Beam L2 travels normally towards mirror M1 and reflected along same path and
come along eyepiece.
c. The transmitted beam travel towards M2 and is reflected along same path
towards back surface of C and proceeds towards eyepiece.
d. Two beams received along the eyepiece were produced from single source by
division of amplitude are coherent.
e. The superposition leads to interference and interference fringes.
5. Role of compensating plate:-
Light rays from source S undergoing reflection at M1passes through C twice.
In absence of D the ray reflected at M2 travels through C only once.
 To compensate this path difference a compensating plate D of same thickness is
placed into the path L1 and is kept parallel to C.

6. Applications of Michelson’s Interferometer:

1. Determination of wavelength of monochromatic light
First of all interferometer is set for circular fringes with central bright spot.
1. If r is thickness of the air film enclosed between two mirrors and n is the order of
the spot obtain, then for normal incidence cos r=1
𝜆
then 2𝑡 + 2 = 𝑛𝜆 ------------(1)
𝜆 𝜆
2. If mirror M1 is moved 2 away from M2’, then additional part difference of 2 will
be introduced and hence (n + 1)th bright spot appears at the centre of the field.
3. If N be the number of the fringes that cross the centre of the field when a mirror
M1 is move from initial position X1 to final position X2 then
𝜆
𝑁 = 𝑋2 − 𝑋1
2

𝟐(𝑿𝟐 − 𝑿𝟏 )
Or, 𝝀= 𝑵
𝑋2 − 𝑋1 is measured with micrometer screw and N is counted.

2. Determination of difference in wavelength:

Interferometer is adjusted to obtain circular fringes.

1. Let the source have to wavelength 𝜆1 𝑎𝑛𝑑 𝜆2 very close to each other.
2. The two wavelengths form separate fringe pattern, but as𝜆1 𝑎𝑛𝑑 𝜆2 to are very
close and thickness of film is very small the two patterns practically coincide.
3. Mirror M1 is moved till the two patterns separate and thickness of the film is such
that the fringe of 𝜆1 falls on the bright fringe of 𝜆2 and the result is maximum
indistinctness.
4. Now the mirror M1 is further move through a distance X till the next indistinctness
position is reached. In this if N fringes of 𝜆1 appears at the centre then n + 1 fringes
of 𝜆2 appear at the centre of the field of view. So

2𝑋 = 𝑛𝜆1 and 2𝑋 = (𝑛 + 1)𝜆2

2𝑋 2𝑋
𝑛= and (𝑛 + 1) =
𝜆1 𝜆2

1 1
𝑛 + 1 − 𝑛 = 2𝑋 ( − )
𝜆1 𝜆2
𝜆2 − 𝜆1
1 = 2𝑋 ( )
𝜆1 𝜆2
𝝀𝟏 𝝀𝟐 𝝀𝟐𝒂𝒗𝒈
𝒐𝒓 𝝀𝟐 − 𝝀𝟏 = 𝚫𝝀 = =
𝟐𝑿 𝟐𝑿
3. Determination of thickness of the film

Interferometer is set for localised fringes of white light.

1. The cross wire is set on central fringe.
2. Now introduce thin plate in one of the interfering rays.
3. With this plate of thickness t and refractive index 𝜇 path is increased by
2(𝜇 − 1)𝑡. This leads to shift of fringes from their position.
4. M1 is moved till again central fringes coincides with cross wire.
5. This distance is noted by micrometer screw. Hence,
𝜆
Since 𝑖𝑛 𝑎𝑖𝑟 Δ = 2𝑡𝑐𝑜𝑠𝑟 + 2

𝜆
& 𝑖𝑛 𝑝𝑙𝑎𝑡𝑒 Δ = 2𝜇𝑡𝑐𝑜𝑠𝑟 +
2
So net path difference Δ = 2(𝜇 − 1)𝑡 = 𝑛𝜆
𝒏𝝀
𝐭=
𝟐(𝝁 − 𝟏)
 4. Difference between interference and diffraction

Interference Diffraction

Interference may be defined as waves Diffraction, on the other hand, can be termed
emerging from two different sources, as secondary waves that emerge from the
producing different wave fronts. different parts of the same wave.

The intensity of all the points on maxima In diffraction, there is a variance of the
is of similar intensity in interference. intensity of positions.

It is absolutely dark in the region of We see a variance in the intensity of

minimum intensity, in the case of interference in diffraction.
interference.

The width of the fringes in interference is The width of the fringes is not equal in
equal in interference. interference.
 5. Fraunhofer Diffraction at single slit

Let b is the width of single slit.

Let he disturbance caused at P by the wavelets from unit width of the slit at centre of
the slit be
𝑦 = 𝐴 cos 𝜔𝑡 − − − −[1]
The wavelet at small width 𝑑𝑥 at centre of the slit when it reaches at P has amplitude
2𝜋
𝐴𝑑𝑥 and phase (𝜔𝑡 + 𝑥 𝑠𝑖𝑛 𝜃)
𝜆

Let the small disturbance be 𝑑𝑦, we have

2𝜋
𝑑𝑦 = 𝐴𝑑𝑥 𝑐𝑜𝑠 {𝜔𝑡 + 𝑥 𝑠𝑖𝑛 𝜃}--------------[2]
𝜆

To get total disturbance at P at an angle 𝜃

𝑦 +𝑏/2
2𝜋
𝑦 = ∫ 𝑑𝑦 = ∫ 𝐴 cos (𝜔𝑡 + 𝑥 sin 𝜃) 𝑑𝑥
0 −𝑏/2 𝜆

𝜋𝑏 𝑠𝑖𝑛𝜃
𝑠𝑖𝑛 [ ]
𝑦 = {𝐴 𝜆 } 𝑐𝑜𝑠𝜔𝑡
𝜋 𝑠𝑖𝑛𝜃
𝜆
𝜋𝑏 𝑠𝑖𝑛𝜃
𝑠𝑖𝑛 [ ]
𝑦 = {𝐴𝑏 𝜆 } 𝑐𝑜𝑠𝜔𝑡
𝜋𝑏 𝑠𝑖𝑛𝜃
𝜆
𝜋𝑏 𝑠𝑖𝑛𝜃
𝑠𝑖𝑛 [ ]
𝑦 = {𝐴0 𝜆 } 𝑐𝑜𝑠𝜔𝑡 − − − −[3]
𝜋𝑏 𝑠𝑖𝑛𝜃
𝜆
𝐴0 = 𝐴𝑏 𝑖𝑠 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑓𝑜𝑟 𝜃 = 0
𝜋𝑏 sin 𝜃
Let =𝛼
𝜆
𝑠𝑖𝑛𝛼
Therefore 𝑦 = 𝐴0 𝑐𝑜𝑠𝜔𝑡----------[4]
𝛼
𝑠𝑖𝑛𝛼
So resultant amplitude is 𝑅 = 𝐴0 𝛼
Also we know 𝐼 = 𝑅 2

𝑨𝟐𝜽 𝒔𝒊𝒏𝟐 𝜶
𝑰𝜽 =
𝜶𝟐
𝜋𝑏 sin 𝜃 𝑠𝑖𝑛2 𝛼
As = 𝛼 so 𝛼 depends on 𝜃 and gives intensity on screen at different values of
𝜆 𝛼2
𝜃.
Position of central maxima
For central point o 𝛼=0
Therefore
sin 𝛼
lim =1
𝛼→0 𝛼

Hence Intensity at o,
𝑠𝑖𝑛2 𝛼
𝐼 = 𝐼0 = 𝐼0
𝛼2
This is maximum as all waves reach O in phase.
𝛼=0
𝜋𝑏 sin 𝜃
=0
𝜆
𝜃=0
Thus all waves travelling normal to slit gives position of central maxima.
Position of minima
sin 𝛼
I=0, when =0
𝛼

Therefore 𝑠𝑖𝑛𝛼 = 0 (𝑤ℎ𝑒𝑟𝑒 𝛼 ≠ 0)

𝛼 = ±𝑚𝜋
𝜋𝑏𝑠𝑖𝑛𝜃
= ±𝑚𝜋
𝜆
So 𝒃 𝐬𝐢𝐧 𝜽 = ±𝒎𝝀

Position of secondary maxima

nth secondary maxima is given by
1
𝑏 sin 𝜃 = ± (𝑚 + 2)λ
𝜋𝑏𝑠𝑖𝑛𝜃 1 𝜋
As 𝛼 = = ± (𝑚 + 2) 2
𝜆

𝟏 𝝀
𝒃 𝒔𝒊𝒏𝜽 = ± (𝒎 + )
𝟐 𝟐
 6. Resolving power
1. The ability of the instrument to produce separate patterns is known as “resolving
power”.
2. The limit of resolution of an optical instrument is defined as the smallest angle
subtended at its objective by two-point objects which just can be distinguished as
separate.
3. The reciprocal of limit of resolution is called resolving power.
4. According to Rayleigh Criterion, two sources are resolvable by optical instrument
when the central maxima in diffraction pattern of one falls over the first minimum in
the diffraction of the other and vice versa.
5. The illustrated above criterion consider intensity curve of diffraction pattern of two
wavelength λ1 & λ2 . In figure central maxima of λ1 falls over secondary minimum of
λ2 . This makes principal maximum separately visible. Thus two wavelengths are
resolved.

6. When Central maxima of λ1 falls near the central maxima of λ2 then two images
cannot be resolved.

7. When Central maxima of λ1 falls on primary minima of λ2 then two images are just
resolved.
 8. When Central maxima of λ1 falls on secondary minima of λ2 then two images are
completely resolved.

Resolving power of grating

The ability of grating to form separate spectral lines of two wavelengths very
𝜆
close together is called resolving power. It is denoted by 𝑑𝜆

Let a parallel beam of light of wavelength 𝜆 𝑎𝑛𝑑 𝜆 + 𝑑𝜆 be incident normally on

grating.
Let nth principal maxima of 𝜆 is formed in direction θ. Then
(𝑒 + 𝑑) sin 𝜃 = 𝑛𝜆 − − − [1]
Let 1st minima adjacent to nth maxima be formed in direction (𝜃 + 𝑑𝜃). Then
grating equation for minima is
𝑁(𝑒 + 𝑑) sin 𝜃 = 𝑚𝜆 − − − [2]
N= no. of lines on grating
m=1,2,3,……….except 0,N,2N………………nN as these values gives zero 1st,
2nd,…..order principal maxima.
1st minima adjacent to nth principle maxima will be obtained for m=nN+1
Put m=nN+1 in eq [2]
N(e+d)sin(θ+dθ)=(Nn+1)λ
𝑁𝑛+1
(e+d)sin(θ+dθ)= 𝜆----------[3]
𝑁

According Rayleigh criterion the wavelength 𝜆 𝑎𝑛𝑑 𝜆 + 𝑑𝜆 are just resolved by

grating when maxima of 𝜆 + 𝑑𝜆 is obtained in the direction (θ+dθ). Then
(e+d)sin(θ+dθ)=n(𝜆 + 𝑑𝜆)----[4]
Compare eq [3] &[4]
𝑁𝑛 + 1
𝜆 = 𝑛(𝜆 + 𝑑𝜆)
𝑁
nNλ+λ=Nnλ+Nndλ
This is resolving power of grating
𝝀
or = 𝑵𝒏
𝒅𝝀