Question

A light ray moving in medium - I (of refractive index $n_1$) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index $n_2$ of medium - II is decreased, then:

Answer 1

As $n_2$ decreases, $i_C={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$ also decreases.
So, condition $i>i_c$ is still satisfied and there will be still total internal reflection at interface. If angle of incidence is increases, then ray will be still totally internally reflected at interface because $i>i_C$.