Question-28 In Young’s double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is d and D is the distance between source and screen. The possible wavelengths used are proportional to
(A) \frac{1}{D},\frac{1}{5D},\frac{1}{7D},\ldots
(B) \frac{1}{D},\frac{1}{3D},\frac{1}{5D},\ldots
(C) \frac{1}{D},\frac{1}{2D},\frac{1}{3D},\ldots
(D) \frac{1}{D^2},\frac{1}{2D^2},\frac{1}{3D^2},\ldots
Answer : (B)
Explanation:
If W is the fringe width, then there will be a minimum in front of the slit if
\frac{d}{2} = \frac{W}{2}, \frac{3W}{2}, \frac{5W}{2}, \frac{7W}{2}, …. or d = W, 3W, 5W, …..
∴ W = \frac{d}{3}, \frac{d}{5}, ….
λ = \frac{Wd}{D}
∴ λ = \frac{d^2}{D}, \frac{d^2}{3D}, \frac{d^2}{5D}
∴ λ is proportional to \frac{1}{D}, \frac{1}{3D}, \frac{1}{5D}, …..