In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If I_m be the maximum intensity, the resultant intensity I when they interfere at phase difference is given by

Step-by-step solution

Let A₁=A₀, A₂=2 A₀ If amplitude of resultant wave is A then A ^2= A ₁^2+ A ₂^2+2 ~A ₁ A ₂ For maximum intensity, A _ ^2= A ₁^2+ A ₂^2+2 ~A ₁ A ₂ A^2 A_ ^2 = A₁^2+A₂^2+2 A₁ A₂ A₁^2+A₂^2+2 A₁ A₂ = A₀^2+4 A₀^2+2 (A₀ ) (2 A₀ ) A₀^2+4 A₀^2+2 (A₀ ) (2 A₀ ) I I _ m = 5+4 9 = 1+8 ^2( / 2) 9