Question

In Young’s double slit experiment, the separation of slits is doubled and the distance of the slits and screen is halved. How will it affect the fringe width?

Answer 1

In interference pattern, the fringe width ß is given by

ß = \(\frac{\lambda D}{d}\) ............(1)

Let ß' be new fringe width, when d' = 2d and D' = \(\frac{D}{2}\)

∴ ß' = \(\frac{\lambda D'}{d'} = \frac{\lambda D}{2.2d} = \frac{\lambda D}{4d}\)

Using Eq. (1), we get

ß' = \(\frac{\beta }{4}\)

So the fringe width is reduced to one-fourth of its previous value.