The angle of minimum deviation for an incident light ray on an equilateral prism is equal to its refracting angle. The refractive index of its material is

Options

A. `1/(sqrt(2))`
B. `sqrt(3)`
C. `(sqrt(3))/2`
D. `3/2`

Answer

B

Explanation

For an equilateral prism,
angle of prism or refracting angle `A=60^(@)`
Angle of minimum deviation `delta_(m)=A=60^(@)` (given)
`mu=(sin((A+delta_(m))/2))/(sin(A/2))=(sin((A+A)/2))/(sin(A/2))=(sin(A))/(sin(A/2))`
`=(2sin(A/2)cos(A/2))/(sin(A/2))=2cos(A/2)`
`=2cos ((60^(@))/2)` (As `A=60^(@)`)
`mu=2cos(30^(@))=sqrt(3)`