The angle of incidence for a ray of light at a refracting surface of a prism is `45^(@)`. The angle of prism is `60^(@)`. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are `:`

Options

A. `30^@, sqrt(2)`
B. `45^@, sqrt(2)`
C. `30^@, (1)/(sqrt(2))`
D. `45^@, (1)/(sqrt(2))`

Answer

A

Explanation

Here, `i_1 = 45^@, A = 60^@ , delta_m = ? Mu = ?`
When the ray suffers minimum deviation,
`i_1 = i_2. As A + delta_m i_1 + i_2 = 90^@`
`delta_m = 90^@ - A = 90^@ - 60^@ = 30^@`
`mu = (sin(A + delta_m)//2)/(sin A//2) = (sin(60^@ + 30^@)//2)/(sin 60^@//2)`
=`(sin 45^@)/(sin 30^@) = (1//sqrt(2))/(1//2) = sqrt(2)`.