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)`