A light rays is incident on a prism of prism angle A and refractive index `mu`. Show that the smallest value of the angle of incidence for which the reflected ray does not suffer total internal reflection at the second face, i.e. just comes out of it, is given by
`i=sin^-1[(mu^2-1)^(1//2)sinA-cosA]`.

Explanation

`i^'=90^@` i.e. `r^'=C`, `sinC=1//mu`
`r+r^'=A`
`r+C=Aimpliesr=A-C`
Refraction on PQ:
`1xxsini=musinr=musin(A-C)`
`=mu(sinAcosC-cosAsinC)`
`=mu[sinAsqrt(1-sin^2C)-cosAsinC]`
`=mu[sinAsqrt(1-1/mu^2)-cosA*1/mu]`
`=mu[(sinAsqrt(mu^2-1)-cosA)/(mu)]`
`=sinAsqrt(mu^2-1)-cosA`
`i=sin^-1[sinAsqrt(mu^2-1)-cosA]`