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^(@) , (1)/(sqrt2)`
B. `45^(@) , (1)/(sqrt2)`
C. `30^(@) , sqrt2`
D. `45^(@) , sqrt2`

Answer

C

Explanation

Since the ray suffers minimum deviation , `i= e = 45^(@)` and A` = 60^(@)`
For a prism , `i+e = A + delta_(m)`
`therefore delta_(m) = i+ e - A = 45^(@) + 45^(@) - 60^(@) = 30^(@)`
`therefore mu = (sin ((A + delta_(m))/(2)))/(sin (A//2)) = (sin ((60 ^(@) + 30^(@))/(2)))/(sin((60^(@))/(2)))`
= `(sin 45^(@))/(sin 30^(@)) = ((1)/(sqrt2))/((1)/(2)) = (2)/(sqrt2) = sqrt2`
Thus `delta_(m) = 30^(@)` and `mu = sqrt2`.