A monochromatic light is incident on the plane interface AB between two media of refractive indices `mu_(1)` and `(mu_(2)gtmu_(1))` at an angle of incidence `theta` as shown in Fig.
The angle `theta` is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now, if a transparent slab DEFG of uniform thickness and of refractive index `mu_(3)` is introduced on the interface (as shown in the figure), show that for any value of `mu_(3)` all light will ultimately be reflected back into medium II.

Explanation

When the slab is not inserted,

`theta ge C = sin^(-1)((mu_(1))/(mu_(2))) :. Sin theta ge (mu_(1))/(mu_(2))` … (i)
When the slab is inserted, two cases arise.
(i) `mu_(3) le mu_(1)`
We have `sin theta ge (mu_(1))/(mu_(2)) ge (mu_(3))/(mu_(2))`
`:.` light incident on `AB` at an angle greater than critical angle `("sin"^(-1) (mu_(3))/(mu_(2)))` will suffer total internal reflection and go back to medium II.
(ii) `mu_(3) gt mu_(2)`
`sin theta ge (mu_(1))/(mu_(2)) lt (mu_(3))/(mu_(2))`
As angle of incidence is less than critical angle, light enters into medium III.
Angle of refraction `theta'` is given by `(sin theta)/(sin theta') = (mu_(3))/(mu_(2))`
`sin theta' = (mu_(2))/(mu_(3)) sin theta ge (mu_(2))/(mu_(3)).(mu_(1))/(mu_(2))`
`sin theta' ge (mu_(1))/(mu_(3))`
i.e., is `theta`' is greater than critical angle. So total internal reflection occurs from `DE` and the entire light is refracted back to medium II.