A light ray is incident at `45^(@)` on a glass slab. The slab is `3cm` thick, and the refractive index of the glass is `1.5`. What will the displacement of the ray be as a result of its passage through the slab? At what angle will the ray emerge formn the slab?

Answer

`3((1)/(sqrt(2)) - (1)/(sqrt(7)))cm = 9.9mm, 45^(@)`

Explanation

Emergent ray is parallel to incident Ray, angle of emergence `= 45^(@)`
`mu = (sin 45)/(sin r) = 1.5`
`sin r = (1)/(sqrt(2) xx1.5) = (sqrt(2))/(3)`

`AB = (3)/(cos r), x = AB sin(i-r)`
`= (3[(1)/(sqrt(2))xx (sqrt(7))/(3) - (1)/(sqrt(2)) xx (sqrt(2))/(3)])/((sqrt(7))/(3)) = 3[(1)/(sqrt(2)) - (1)/(sqrt(7))]cm`