A monochromatic light of `lambda = 500 Å` is incident on two indentical slits separated by a distance of `5 xx 10^(-4)` m. The interference pattern is seen on a screen placed at a distance of 1 m from the plane of slits. A thin glass plate of thickness `1.5 xx 10^(-6)` m and refractive index `mu = 1.5` is placed between one of the slits and the screen . Find the intensity at the center of the screen.

Answer

0

Explanation

Take a point at a distance y from the center of the slit. Path differece between waves reaching this point:
`Delta x = (d y)/(D) - (mu - 1) t`
For center of slit: `y = 0`
So `Delta x = -(mu - 1) t`
Phase difference : `phi = (2pi)/(lambda) Delta x = - (2 pi)/(lambda) (mu - 1) t`
`I = 4 I_(0) cos^(2) ((phi)/(2)) - 4I_(0) cos^(2) ((pi)/(lambda) (mu - 1)t)`
`= 4 I_(0) cos^(2) [(pi (1.5 - 1) 1.5 xx 10^(-6))/(500 xx 10^(-10))] = 4 I_(0) cos^(2) ((3pi)/(2)) = 0`