In the diagram shown, light is incident on the interface between media `1` (refractive index `n_(1)`) and `2` (refractive index `n_(2)`) at angle slightly greater than the critical angle, and is totally reflected. The light is then also totally reflected at the interface between media `1` and `3` (refractive index `n_(3)`), after which it travels in a direction opposite to its initial direction, The media must have a refractive indices such that.
.

Options

A. a. `n_(1) lt n_(2) lt n_(3)`
B. b. `n_(1)^(2) - n_(3)^(2) gt n_(2)^(2)`
C. c. `n_(1)^(2) - n_(2)^(2) lt n_(3)^(2)`
D. d. `n_(1)^(2) - n_(2)^(2) gt n_(3)^(2)`

Answer

B, D

Explanation

Let the critical angle of interface between media `1` and `2` is `C_(1)` and between `1` and `3` is `C_(2)`
Then `sin C_(1)=(n_(2))/(n_(1))` and `C_(2)=(n_(3))/(n_(1))`
From `TIR` at second interface `90 - C_(1) gt C_(2)`
taking sin of both side we get
`cos C_(1) gt C_(2)`
or `sqrt(1-((n_(2))/(n_(1)))^(2)) gt (n_(3))/(n_(1))` or `n_(1)^(2)-n_(3)^(2) gtn_(2)^(2)`.