optics2

Chapter III

OPTICS

Lecture 3.2

We shall discuss about

_{Conditions for sustained interference}

Conditions for sustained interference:

(i) Two sources must be coherent sources

(ii) Two sources must emit waves continuously

(iii) Two coherent sources of light should be close to each other

(iv) Amplitude of the waves emitted by two coherent sources of light should be equal to each other so as to get complete contrast

between dark and bright band

(v) Two coherent sources should be very narrow

Interference of light waves:

The light waves coming from two independent light sources does not produce the stationary interference pattern because these light waves do not have fixed phase relation.

Fig. Two independent light sources Does not produce interference

In 1801, Thomas Young devised an experiment to study the interference phenomenon.

The idea was to split a single wave front in two wave fronts and these two new wave fronts act as if emanated from two sources and

have constant phase relation and thus

In fig light from source falls on pinhole S which then falls on two pinholes s₁ and s₂. The light coming from these s₁ and s₂ interfere and interference pattern is produced.

Young explained the interference pattern with the help of superposition principle.

Motive was to find the Position of maxima

We consider some point P on the line LL’ and if this corresponds to Some maximum then we have

---(1)

Also we can write from figure

---(2)

Above eqn can be written as (using a2_-b2 _{= (a-b)(a+b))}

If then

Thus from eqn (3), we have

---(4) And using eqn (1), we get

---(5)

The fringe width i.e. distance between consecutive dark and bright fringe is given as

Work:

Show that the fringes produced in the above interference pattern

are almost straight lines.

We can also study the intensity distribution of the interference pattern

by considering that the light has electric and magnetic component

After superposition, we have

After few calculations we can write above eqn as

where,

and

Discuss the followings:

•If hole s1 and s2 are illuminated from different light source then we don't have interference pattern

Fresnel Bi-prism Experiment:

It is a simple arrangement for the production of sustained interference pattern.

Fresnel used a bi-prism which is a combination of two prism placed base to base and the base angles of which are extremely small (~300) as shown in fig.

Actually the prism is made from single glass plate by grinding and polishing so that it is a single prism with one of its angle about

1790 _{and two angle are very small (30}0_).

Prism is perpendicular to the plane of paper.

Light from slit S gets refracted by the prism and produces two virtual images S ₁ and S₂. These images act as coherent sources and produce interference fringes on the right

Determination of the wavelength of light using Fresnel Biprism:

The bi-prism arrangement can be used for the determination of wavelength of an almost monochromatic light such as the one coming from a sodium lamp.

Light from the sodium lamp illuminates slit S, and interference fringes can be easily viewed through the eyepiece.

The fringe width β can be determined by means of a micrometer attached to the eyepiece.

Once β is known, wavelength can be determined by using the following relation:

---(1)

In above eq., D which is the distance between the slit and screen can be find simply by taking the reading from scale fitted

on optical bench.

To find the distance between the virtual sources we proceed as below:

We place a convex lens of short focal length between the

Biprism and eyepiece.

The lens is then moved along the length of the bench so as to obtain two positions L₁ and L₂ such that the real images of slit S₁ and S₂ are seen in the eyepiece.

If u and v are the distances between the slit and eyepiece from the lens in position L₁, then from magnification formula, we have

---(2) Since the two positions L₁ and L₂ are conjugate, so we have magnification formula ,

---(3)

Multiplying (2) and (3), we get

This give us distance between virtual sources,

Thus by knowing the value of fringe width, distance between source and eyepiece (D) and the distance between virtual sources(d),

we can find the wavelength of light using eqn (1).

Problems:

•What is the effect of slit spacing d and wavelength on fringe width?

•_{What is the effect of width of slit on interference pattern.}

•If we use the white light in Fresnel biprism experiment then

we get general illumination instead of interference pattern. Discuss.

•_{The relation between the angle of Fresnel biprism and distance}

between the virtual sources is given by

d =

Displacement of the fringes and thickness of the thin transparent plate:

Here we shall find the expression for the displacement of the fringes if we introduce a thin transparent plate in the path of one of the wave.

As shown in the fig.

We consider the superposition

of light emanating from two sources S₁ and S₂.

Let t be the thickness of the plate, and let n be its refractive

The light coming to point P from source S₁ has to travel the distance t in plate and distance (S₁P-t) in air.

Therefore, time required by light to go from S₁ to P is

---(1)

where v (= c/n ) represents the speed of light in the plate.

Equation (1) shows that by introducing the thin plate the effective optical path increases by (n – 1)t. Thus, when the thin plate is introduced, the central fringe (which

corresponds to equal optical path from S₁ and S₂) is formed at point O’ where

Since

---(3) using 2, we get

_______________(4)

Thus the fringe pattern gets shifted by a distance which is given by

Above eqn can be used to find the thickness t of the plate if we know other parameters.