In Young's double slit experiment, the amplitudes of the two waves incident on the two slits are A and 2 A . If I 0 is the maximum intensity, then the intensity at a spot on the screen, where the phase difference between the two interfering waves is ϕ .

Step-by-step solution

Resultant intensity when two waves with phase difference ϕ interfere is I = I 1 + I 2 + 2 I 1 I 2 cos ϕ As, intensity I ∝ A 2 or I = k A 2 , where k = a constant and A is amplitude. So, I = A 1 2 + A 2 2 + 2 A 1 A 2 cos ϕ Here, A 1 = A and A 2 = 2 A ⇒    I = A 2 + 2 A 2 + 2 A 2 A cos ϕ = A 2 5 + 4 cos ϕ ........(i) Intensity is maximum I 0 , when cos ϕ = 1 ⇒ I 0 = A 2 5 + 4 × 1 = 9 A 2 ⇒ A 2 = I 0 / 9 .......(ii) So, resultant intensity, I = A 2 5