In a Young's double slit experiment, the ratio of the widths of the two slits is 4 : 1 . If the maximum intensity in the interference pattern is I₀ , the intensity at a point on the screen where the phase difference between the interfering waves is 3 will be

Step-by-step solution

The intensity of light from a slit is directly proportional to its width. Let the widths be w₁ and w₂ . Given w₁ w₂ = 4 1 , the ratio of their intensities is I₁ I₂ = 4 1 . Let I₂ = I and I₁ = 4I . The maximum intensity in the interference pattern is given by: I_ max = ( I₁ + I₂ )^2 I₀ = ( 4I + I )^2 = (2 I + I )^2 = 9I This gives I = I₀ 9 . The resultant intensity I_R at a point with phase difference is: I_R = I₁ + I₂ + 2 I₁ I₂ For = 3 : I_R = 4I + I + 2 (4I)(I) ( 3 ) I_R = 5I + 2(2I) ( 1 2 ) = 5I + 2I = 7I Substit