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 equal to 60 degrees?

Explanation

a. Consider the ray incident normally oN/AB ure. The asngle of reflectin at the surface AC is theta. It is clear from the ure that the angle of incidence ast th secon surface CB is `90^@-theta`. The emergent ray will be paralel to the incidnet ray after two total inteN/Al refelection.The critical angle `theta_c` should be less than theta as well as `90^@-theta`. Thus `theta_c` should be smaller than or equal to the smaler of `theta and 90^@-theta, i.e,`
theta_clemin(theta,90^@-theta)`
As min (theta, 90^@-theta)le45^@, theta_cle5^@`
or `sintheta_cle1/sqrt2 or 1/mule1/sqrt2`
or `muge1/sqrt2`
Thus the refrctive index of the material of hte prism should be greater than or equal to `sqrt2.` In this case teh gien ray can undrgo tow inteN/Al reflections for a suitable theta.
b. For `mu=5/3` the critical angle `theta_c` is
`sin^-1(3/5)=37^@`
As the ure suggests we consider the lightincident normally on the face AB. The angle of incidence `theta` on the surface AC is equal to `theta=60^@`. As this is larger than the cirtical angle `37^@`, total internla reflection takes place here. The angle of incident at the surface CB is `90^@-theta=30^@`. As this is less than teh critial ange, total interN/Al reflection does not take place at this surface.