Path difference is Δx = nλ (for maxima) dsin θ = nλ If d = λ, we have sin θ = n. Since sin θ is maximum for θ → π/2, we get nmax = 1 (central maximum) If λ < d < 2λ; d = 2λ, we have sin θ =
n
2
That is, a total of five maxima are seen on the screen. We know that Imax = I1+I2+2√I1I2 Imin = I1+I2−2√I1I2 Initially, Imax = I + 4I + 2√I(4I) = 9I Imin = I + 4I - 2√I4I = I Now, if I1 = I2 , we get Imax′ = I + I + 2√I2 = 4I Imin′ = I + I - 2√I2 = 0 That is, the intensities of both bright and dark fringes decrease.