In a Young's double slit experiment, the ratio of the maximum intensity to the minimum intensity in the interference pattern is 25 : 9 . Let P be a point on the screen where the path difference between the interfering waves is 4 . The ratio of the intensity at the central maximum to the intensity at point P is:

Step-by-step solution

Given the ratio of maximum to minimum intensity: I_ max I_ min = 25 9 Since I_ max = (A₁ + A₂)^2 and I_ min = (A₁ - A₂)^2 , we have: A₁ + A₂ A₁ - A₂ = 5 3 Solving for the ratio of amplitudes: 3A₁ + 3A₂ = 5A₁ - 5A₂ 2A₁ = 8A₂ A₁ = 4A₂ The ratio of individual intensities is I₁ = 16 I₂ . The intensity at the central maximum is: I_ max = ( I₁ + I₂ )^2 = (4 I₂ + I₂ )^2 = 25 I₂ At point P , the path difference is x = 4 . The corresponding phase difference is: = 2 x = 2 4 = 2 The intensity at point P is given by the genera