In a Young's double slit experiment, the ratio of the widths of the two slits is 9 : 1 . Let P be a point on the screen where the path difference between the interfering waves is 3 , and Q be a point on the screen where the intensity is minimum. The ratio of the intensity at point P to the intensity at point Q is

Step-by-step solution

The intensity of light emerging from a slit is directly proportional to its width. Therefore, the ratio of the intensities of the two sources is: I₁ I₂ = w₁ w₂ = 9 1 Let I₁ = 9I₀ and I₂ = I₀ . The minimum intensity at point Q is: I_Q = ( I₁ - I₂ )^2 = ( 9I₀ - I₀ )^2 = (3 I₀ - I₀ )^2 = (2 I₀ )^2 = 4I₀ For point P , the path difference is x = 3 . The corresponding phase difference is: = 2 x = 2 3 = 2 3 The intensity at point P is given by the general interference formula: I_P = I₁ + I₂ + 2 I₁ I₂ I_P = 9I₀ + I₀ + 2 9I