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^(8)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`