In Young's double slit experiment if the slits widths are in the ratio 1 : 9, the ratio of the intensities at minima to that at maxima will be

(a) 1
(b) 1/9
(c) 1/4
(d) 1/3

Solution:
Answer: (c) 1/4
Intensity of fringe is directly proportion to width os the slits.

Therefore, I₂/I₁ = W₂/W₁ = 1:9

or I₁ = 9I₂

The maximum intensity and minimum intensity are I_max = (√I₁ + √I₂)²

and I_min = (√I₁ – √I₂)² respectively.

So, I_max = 16 I₂

and I_min = 4 I₂

Therefore, I_min/I_max = 1/4