Explanation

(b) Here the requirement is that $i > c$
$\Rightarrow \sin{i > \sin c} \Rightarrow \sin{i >}\frac{\mu_{2}}{\mu_{1}}\ \ \ \ \ \ \ldots(i)$
From Snell’s law $\mu_{1} = \frac{\sin\alpha}{\sin r}$ …(ii)
Also in $\mathrm{\Delta}OBA$
$r + i = 90{^\circ} \Rightarrow r = (90 - i)$
Hence from equation (ii)
$\sin\alpha = \mu_{1}\sin{(90 - i)}$
$\Rightarrow \cos i = \frac{\sin\alpha}{\mu_{1}}$
$\sin i = \sqrt{1 - \cos^{2}i} = \sqrt{1 - \left( \frac{\sin\alpha}{\mu_{1}} \right)^{2}}\ \ \ldots(iii)$
From equation (i) and (iii) $\sqrt{1 - \left( \frac{\sin\alpha}{\mu_{1}} \right)^{2}} > \frac{\mu_{2}}{\mu_{1}}$
$\Rightarrow \sin^{2}{\alpha < (\mu_{1}^{2} - \mu_{2}^{2}}) \Rightarrow \sin\alpha < \sqrt{\mu_{1}^{2} - \mu_{2}^{2}}$
$\alpha_{\max} = \sin^{- 1}\sqrt{\mu_{1}^{2} - \mu_{2}^{2}}$