A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. By rotating the prism,the angle of minimum deviation is measured to be `40^(@)` . What is the refractive index of the material of the prism? If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. Refracting angle of the prism is `60^(@)` .

Explanation

Refractive index of the material of the prism,
`mu = (sin""(A + delta_(m))/(2))/(sin""(A)/(2)) = (sin""(60^(@) + 40^(@))/(2))/(sin""(60^(@))/(2) ) = (sin50^(@))/(sin30^(@)) = 1.53`
When placed in water,
`w^(mu)g = (a^(mu)g)/(a^(mu)g) = (1.53)/(1.33) = 1.15 " " therefore w^(mu)g = (sin""(A + delta._(m))/(2))/(sin""(A)/(2))`
`or, 0.15 sin""(60^(@))/(2) = sin""(60^(@) + delta._(m))/(2)`
`therefore " " delta._(m) = 10.2^(@) 10^(@)12.`