(c) Critical angle $= \sin^{- 1}\left( \frac{1}{\mu} \right)$
$\therefore\theta = \sin^{- 1}\left( \frac{1}{\mu_{\lambda_{1}}} \right)$ and $\theta^{'} = \sin^{- 1}\left( \frac{1}{{\mu_{\lambda}}_{2}} \right)$
Since $\mu_{\lambda_{2}} > {\mu_{\lambda}}_{1}$, hence $\theta^{'} < \theta$