A ray of light is incident at an angle of `60^(@)` on the face of a prism with an angle of `60^(@)`. Then the refractive index of the material of the prism is (the prism is in minimum deviation position)

Explanation

Here, `A=60^(@), mu=1.632`
Now, `mu=(sin ((A+delta_(m))/2))/(sin (A/2))`
or, `1.632=(sin ((60^(@)+delta_(m))/2))/("sin"60^(@)/2)=(sin((60^(@)+delta_(m))/2))/(sin 30^(@))`
or, `sin ((60^(@)+delta_(m))/2)=1.632xxsin 30^(@)`
`=1.632xx0.5`
or, `sin ((60^(@)+delta_(m))/2)=0.816`
or, `(60^(@)+delta_(m))/2=54^(@)42'`
`delta_(m)=49^(@)27'`