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 crictical angle `(37^@)`. Therefore, total internal reflection cannot take place from the surface `CB`.