In a Young's double slit experiment, the widths of the two slits are in the ratio n : 1 (where n > 1 ). The visibility of the interference fringes on the screen is given as 0.8 . Assuming that the intensity of light from a slit is directly proportional to its width, what is the value of n ? (Fringe visibility V is defined as V = I_ max - I_ min I_ max + I_ min )

Step-by-step solution

Let the widths of the two slits be w₁ and w₂ . We are given w₁ w₂ = n . Since the intensity is directly proportional to the slit width ( I w ), the ratio of intensities is: I₁ I₂ = n I₁ = n I₂ The maximum and minimum intensities in the interference pattern are: I_ max = ( I₁ + I₂ )^2 = ( n + 1)^2 I₂ I_ min = ( I₁ - I₂ )^2 = ( n - 1)^2 I₂ The fringe visibility V is given by: V = I_ max - I_ min I_ max + I_ min Substitute the expressions for I_ max and I_ min : I_ max - I_ min = 4 n I₂ I_ max + I_ min = 2(n + 1) I₂ V