A rectangular glass slab ABCD of refractive index `n_(1)` is immersed in water of refractive index `n_(2)(n_(1) lt n_(2))`. A ray of light is incident at the surface AB of the slab as shwon. The maximum value of the angle of incidence `alpha_(max)` such that the ray comes out from the other surface CD is given by

Options

A. `sin^(-1)[(n_(1))/(n_(2))cos(sin^(-1)((n_(2))/(n_(1))))]`
B. `sin^(-1)[n_(1)cos(sin^(-1)((1)/(n_(2))))]`
C. `sin^(-1)((n_(1))/(n_(2)))`
D. `sin^(-1)((n_(2))/(n_(1)))`

Answer

a.

Explanation

See figure. The ray will come out form CD if it sufferes total internal reflection at surface AC, i.e., it strikes the surface AD at critical angle C (the limiting case) . Applying Snell's law at
P `, n_(1)sinC=n_(2)or sinC=(n_(2))/(n_(1))`
Applying Snell's law at Q, `n_(2)sinalpha=n_(1)cosC,`
`sinalpha=(n_(1))/(n_(2))cos{sin^(-1)((n_(2))/(n_(1)))}`
or `alpha=sin^(-1)[(n_(1))/(n_(2))cos{sin^(-1)((n_(2))/(n_(1)))}]`
`:.` (a) is the correct option.