The angle of minimum deviation for prism of angle `pi//3 is pi//6`. Calculate the velocity of light in the material of the prism if the velocity of light in vacuum is `3 xx 10^8 ms^-1`.

Explanation

Here, `A = (pi)/(3) = 60^@ , delta_m = (pi)/(6) = 30^@`,
`c =3 xx 10^9 ms^-1 , v = ?`
We know, `mu = (sin (A + delta_m)//2)/(sin A//2)`
=`(sin (60^@ + 30^@)//2)/(sin 60^@ //2)`
=`(0.7071)/(0.50) = 1.414`
From `mu = ( c)/(v), v = ( c)/(mu) = (3 xx 10^8)/(1.414) = 2.12 xx 10^8 m//s`.