A right prism is to be made by selecting a proper material and the angles A and B `(BleA)` as shown in figure. It is desired that a ray of light incident normally on AB emerges parallel to the incident direction after two internal reflection.a. What should be the minimum refractive index `mu` for this to be possible? b. `For mu=5/3` is it possible to achieve this with the angle A equl to 60 degrees?

Answer

`sqrt(2)` ; No

Explanation

In Fig. the incident ray is normal to `AB`.
Angle of incidence on `AC = theta` and angle of incidence on `BC` is `theta' = (90^(@) - theta)`.
For total internal reflection, critical angle `(C )` must be less than smaller of `theta` and `(90^(@) - theta)`.
`:. c le 45^(@)` or `sin C le (1)/(sqrt(2))`
or `(1)/(mu) le (1)/(sqrt(2))` or `mu ge sqrt(2)`
`:.` Min. value of `mu = sqrt(2)`
(b) For `mu = 5//3, C = sin^(-1)((3)/(5)) = 37^(@)`
If `/_A` were `60^(@)`, angle of incidence on `CB`would be `90^(@) - 60^(@) = 30^(@)`, which is less than critical angle `(37^(@))`. Therefore, total internal reflection cannot take place from the surface `CB`.