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: IR = 2I(1+cosφ ) = 2I.2cos2(φ/2) = 4Icos2(φ/2)
IR = I0cos2(φ/2) , Where 4I = I0
I0 /2 = I0cos2(φ/2) => cos2(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