For a prism of refractive index `1.732`, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

3

Explanation

The refractive index of prism angle A and angle of minimum deviation `delta_(m)` is given by `mu = (sin [(A+delta_(m))/2]/(sin(A/2)))`
here we are given, `delta_(m)` = A
Substituting the value, we have `mu = (sinA)/(sin A/2)`
`mu = (2 sin A/2 cos A/2)/(sin A/2) = 2 cos A/2`
`mu = (sin A)/(sin A/2) = (a sin A/2 cos A/2)/(sin A/2) = 2 cos A/2 `
For the given value of refractive index, we have, `cos A/2 = sqrt3/2 implies A/2 = 30^(@)` or `A=60^(@)`
This is the required value of prism angle.