A ray of light is incident at an angle of `60^(@)` on one face of a rectangular glass slab of thickness `0.1 m`, and refractive index `1.5`.Calculate the lateral shift produced.

Explanation

Given `I = 60^(@) , t = 0.1 m , n = 1. 5, S _(L) = ?`
From Snell.s law,` n = ( sin i)/( sin r)`
`sin r = ( sin i)/(n) = ( sin 60^(@))/(1.5)`
`therefore r = sin ^(-1) (0.5773)`
` = 35 ^(@) 15.`
Leteral shift,
`S _(L) = ( t tan (i - r))/( cos r) = (0.1 sin ( 24 ^(@) 45.))/( cos (35 ^(@) 15.)) = 0. 0 51 m`