Question

In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. The angular position, of this point, is

(1) sin^-1 (λ/d)

(2) sin^-1 (λ/2d)

(3) sin^-1 (λ/3d)

(4) sin^-1 (λ/4d)

Answer 1

(3) sin^-1 (λ/3d)

In Fig, P is the point where intensity = I = I₀/4 (I₀ maximum intensity). Let δ be the phase difference, between the two disturbances, at P. Then

Let Δ be the path difference between the two disturbances, at P. We know δ = 2π/λ (Δ).