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`.

Answer

`2.121 xx 10^8 ms^-1`

Explanation

`mu = (sin(A + delta_m)//2)/(sin A//2) = (sin(60^@ + 30^@)//2)/(sin 60^@//2)`
=`(sin 45^@)/(sin 30^@) = (0.7070)/(0.5000) = 1.414`
`v = (c)/(mu) = (3 xx 10^8)/(1.414) = 2.121 xx 10^8 ms^-1`.