Explanation

(b) When a ray of light passes from glycerine (denser, $\mu = 1.47$) to water (rarer, $\mu = 1.33$) the angle of refraction $(r)$ is greater than angle of incidence ($i$), then from Snell’s law
$\frac{\sin i}{\sin r} = \frac{\mu_{2}}{\mu_{1}} < 1$
When $r = 90{^\circ}$, corresponding angle of incidence is known as critical angel, $ie,\ i = \ \theta_{C}$
$\therefore\frac{\sin\theta_{C}}{\sin{90{^\circ}}} = \frac{\mu_{2}}{\mu_{1}}$
$\Rightarrow \sin\theta_{C} = \frac{\mu_{2}}{\mu_{1}}$
$\Rightarrow \ \theta_{C} = \sin^{- 1}\left( \frac{\mu_{2}}{\mu_{1}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \sin^{- 1}\left( \frac{1.33}{1.47} \right)$
$\theta_{C} = \ 64{^\circ}48'$