Question

describe young's double slit experiment to produce interference and hence derive an expression for the width of the interference fringes​

Answer 1

Given that,

The young's double slit experiment to produce interference.

We know that,

Young's double slit experiment :

A light waves is incident on a barrier containing two narrow slits separated by a distance d. A screen placed at D distance from the slits.

The intensity is maximum At the point where two wave crests or troughs superimpose.

The intensity is minimum at the point where the crest of one wave superimpose the trough of other wave.

The variation in light intensity along the screen near the centre O shown in the figure.

In the figure,

All bright and dark fringes are of equal width.

We need to find the formula of width of fringe

According to figure

$\beta=\dfrac{\lambda D}{d}$

Where, D = distance between slit and screen

d = distance of slit

$\lambda$ = wavelength of light

Hence, The formula of width of fringe is $\dfrac{\lambda D}{d}$