In a Young’s double slit experiment, the intensities at two points, for the path difference λ 4 and λ 3 ( λ being the wavelength of light used) are I 1 and I 2 respectively. If I 0 denotes the intensity produced by each one of the individual slits, then I 1 + I 2 I 0 = _ _ _ _ _ _ .

Step-by-step solution

As we know, resultant intensity is given by, I = 4 I 0   cos 2 Δ ϕ 2 and phase difference can be written as ∆ ϕ = 2 π λ ∆ x . For first case: I 1 = 4 I 0   cos 2 2 π λ × λ 4 = 4 I 0   cos 2 π 4 = 2 I 0 For second case: I 2 = 4 I 0   cos 2 2 π λ × λ 3 = 4 I 0   cos 2 2 π 3 = I 0 Therefore, required value of I 1 + I 2 I 0 = 3 .