A thin paper of thickness 0.02 mm with a refractive index of 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the incident light energy. The wavelength of the light used is 600 nm . Determine the ratio of the maximum intensity to the minimum intensity in the fringe pattern.

Step-by-step solution

Let the intensity of light from the unblocked slit be I₀ . Since the paper transmits 4/9 of the incident light energy, the intensity of light from the blocked slit is I₁ = 4 9 I₀ . The maximum intensity in the interference pattern is given by: I_ max = ( I₀ + I₁ )^2 I_ max = ( I₀ + 4 9 I₀ )^2 = (1 + 2 3 )^2 I₀ = 25 9 I₀ The minimum intensity in the interference pattern is given by: I_ min = ( I₀ - I₁ )^2 I_ min = ( I₀ - 4 9 I₀ )^2 = (1 - 2 3 )^2 I₀ = 1 9 I₀ The ratio of the maximum intensity to the minimum intensit