Q. In Young’s double slit experiment the fringe pattern is observed on a screen placed at a distance D. The slits are illuminated by light of wavelength λ . The distance from the central point where the intensity falls to half the maximum is

(a) λD/3d

(b) λD/2d

(c) λD/d

(d) λD/4d

Ans: (d)

Sol: I_{R} = 2I(1+cosφ ) = 2I.2cos^{2}(φ/2) = 4Icos^{2}(φ/2)

I_{R} = I_{0}cos^{2}(φ/2) , Where 4I = I_{0}

I_{0} /2 = I_{0}cos^{2}(φ/2) => cos^{2}(φ/2) = 1/2

=> φ/2 = π/4 => φ = π/2 , Hence path difference = λ/4

Path diff .Δx = dsinθ

Δx = d.y/D

=> y = ΔxD/d

=( λ/4) .(D/d) = λD/4d