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

Angle of incidence, `i=60^(@)`
Refractive index of glass, `mu=3//2`
If r is the angle of refraction then
`(sini)/(sinr)=mu`
`rArr" "sinr=(sini)/(mu)=(sin60^(@))/(3//2)=(sqrt3)/(2)xx(2)/(3)`
`=(sqrt3)/(3)=0.5773`
`rArr" "r=35^(@)15.`
`"Given, thickness of slab "=10cm=0.1m`
Lateral shift produced is
`x=(t)/(cosr)sin(i-r)`
`=(0.1)/(cos35.26^(@))sin(60^(@)-35^(@)15.)`
`=(0.1xx0.4185)/(0.8165)=0.05m`