Explanation

(c) From law of reflection, $\angle\ i = \ \angle\ r\ \ \ \ \ \ \ldots(i)$
And $\frac{\sin r^{'}}{\sin i} = \frac{\mu_{d}}{\mu_{r}}\ \ \ \ \ \ \ \ldots(ii)$
From the figure,
$r + r' + 90{^\circ} = 180{^\circ}$
$\Rightarrow \ r + r' = 90{^\circ}$
Or $i + r' = 90{^\circ}$ $r^{'} = (90{^\circ} - i)\ \ \ \ \ \ \ \ldots(iii)$
From Eq. (ii), $\frac{sin(90{^\circ}\ - \ i)}{\sin i} = \frac{\mu_{d}}{\mu_{r}}$
Or $\frac{\cos i}{\sin i} = \frac{\mu_{d}}{\mu_{r}} \Rightarrow \cot{i = \ \frac{\mu_{d}}{\mu_{r}}}$
But $\frac{\mu_{d}}{\mu_{r}} = \sin C(where\ C\ is\ critical\ angle)$
$\therefore\cot{i = \sin{C \Rightarrow C =}}\sin^{- 1}\ (\cot{\ i})$