Monochramatic light is incident on the pLane interface AB between two media of refractive indices `mu_1` and `mu_2(mu_2gtmu_1)` at an angle of incidence theta as shown in figure. The angle theta is infinitesimally greater thannte critical angle for the two media so thast total internal reflection takes place. Now, if a transparent slab DEFG of uniform thickness and of refractive inde `mu_2` is introduced on theinterface (as shown in figure ) , show that for any value of `mu_2` 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 gt = (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_3))` will suffer total internal reflection and go back to medium `II`.
(ii) `mu_3 gt mu_1`
`sin theta ge (mu_1)/(mu_2) lt (mu_3)/(mu_2)`
As angle of incidence is less than critical angle, light entres into medium `III`. ltrgt 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., `theta'` is greater than critical angle. So total internal reflection occurs from face `DE` and the entire light is reflected back to medium II.
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