Question

A monchromatic and linearly polarised light is used in a Youngs double slit experiment.A linear polariser, whose pass axis is at an angle of 45 to the polarization of the incident wave, is placed in front of one of the slits. If $_{Imax}$ and $_{Imin}$ respectively, denote the maximum and minimum intesities of the interference on the screen, the visibility defined as the ratio $\frac{_{Imax-_{Imin}}}{_{Imax+Imin}}$ is

Answer 1

Assume intensity before passing through polarizer is I I_max = (squareroot I_1 + squareroot I_2)^2 I_1 = I_0 cos^2 (45) = I_)/2 I_2 = I_0 I_min = (squareroot I_2 - squareroot I_1)^2 = (squareroot I_) - squareroot I_)/2)^2 I_min = I_0 + I_0/2 - 2 I_0/squareroot 2 I_min = 3 I_0/2 - squareroot 2 I_0 I_max = I_0 + I_0/2 + squareroot 2 I_0 = 3 I_0/2 + squareroot 2 I_0 r = I_max - I_min/I_max + I_min r = (3 I_)/2 + squareroot 2 I_0) - (3 I_0/2 - squareroot 2 I_0)/(3 I_0/2 + squareroot 2 I_0) + (3 I_)/2 - squareroot 2 I_0) r = 2 squareroot 2 I_0/3 I_0 r = 2 squareroot 2/3