Question

A ray of light falling at an angle of `50^@` is refracted through a prism and suffers minimum deviation. The angle of the prism is `60^@`. Find the angle of minimum deviation and refraction index of the material of the prism.

Answer 1

Gives that, `A = 60^(@) , delta_(m)= ? , mu = ?`
In position of minimum deviation, `i_(1) = i_(2) = 50^(@)`
From, `A + delta_(m) = i_(1) + i_(2) = 50 + 50 = 100^(@)`
`delta_(m) = 100 - A = 100 - 60 = 40^(@)`
`mu = (sin.((A+delta_(m)))/(2))/(sin.(A)/(2))`
`(sin.((60^(@)+40^(@)))/(2))/(sin.(60^(@))/(2))`
`= (sin 50^(@))/(sin 30^(@))`
`= (0.7660)/(0.5000)`
`= 1.532`