A `60^@` prism has a refractive index of `1.5`. Calculate (a) the angle of incidence for minimum deviation (b) angle of minimum deviation ( c) the angle of emergence of light at maximum deviation (d) angle of maximum deviation.

Explanation

As the prism is in the position of minimum of deviation, `r = (A//2) = (60° // 2) = 30°`, so that at either face s i nz = 1.55^30° =0 .75 or `i=sin^(-1)(0.75)49^@`
Note: In this situation angle of emergence is equal to angle of incidence = `49^@` and deviation
`delta_(m)=(2i-A)=(2xx49-60)=38^@`
b) For maximum deviation, `i_(1)=90^@` so that `r_(1)=theta_(c)=sin^(-1)(2/3)=42^@`, But as in a prism
`r_(1) + r_(2)=A` so `r_(2)=A-r_(1)=60^@ - 42^@=18^@`
Now applying Snell’s law at the second face,
`musinr_(2)=sini_(20` i.e., `3/2sin18^@=sini_(2)`
i.e `i_(2)=sin^(-1)[1.5xx0.31]=sin^(-1)(0.465)equiv28^@`