A slab of water is on the top of a glass slab of refractive index 2. At what angle to the normal must a ray be incident on the top sureface of th eglass slab so that it is reflected from the bottom surface as shown in figure ?

Explanation

For the light ray to be reflected from the lower surface RS, the angle of incidence must be greater than the critival angle for the glass-air interface (at N).
`i gt sin^(-1)((1)/(2))` or `i gt 30^@`
At the water-glass interface,
`mu_(')=(4)/(3), mu_(2)=2`
`theta_(1)=theta=?, theta_(2)=i=30^@`
Since `mu_(1)sin theta_(1)=mu_(2)sintheta_(2),` we get
`theta=sin^(-1)[(mu_(1))/(mu_(2))sintheta_(1)]`
`rArr theta=sin^(-1)((3)/(4))`