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.)

Question

Accepted Answer

(1) 25 μm. Half angular width is 30°: a sin 30° = λ ⇒ λ = 0.5 μm. Young's fringes: β = (λ D)/d ⇒ d = (λ D)/β. d = (0.5×10⁻⁶×0.5)/(10⁻²) = 25×10⁻⁶ m = 25 μm.

Step-by-step solution

Why the other options are wrong