JEE Main | 2018Wave OpticsHard

Question

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)

Options

A.

50 μm

B.

75 μm

C.

100μm

D.

25μm

Solution

In diffraction
d sin 30º =λ

λ=d2

Young's fringe width
[d' – separation between two slits]

β=λ×Dd'

10-2  = d2×50×10-2d'

10-2=10-6×50×10-22×d'

 d'=25μm

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