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`.

Options

A. `2.12 xx 10^(8) ms^(-1)`
B. `1.12 xx 10^(8) ms^(-1)`
C. `4.12 xx 10^(8) ms^(-1)`
D. `5.12 xx 10^(8) m s^(-1)`

Answer

A

Explanation

Using, `mu = (sin(A+delta_(m))//2)/(sinA//2)`
Here, `A = (pi)/(3) = 60^(@), delta_(m)= (pi)/(6) = 30^(@), c = 3 xx 10^(8) m s^(-1)`
`:. Mu = (sin(60^(@) + 30^(@))//2)/(sin60^(@)//2) = (0.7071)/(0.50)= 1.414`
Therefore, `v = c/mu = (3 xx 10^(8))/(1.414)` or `v = 2.12 xx 10^(8) ms^(-1)`