Explanation

(b) At point $A$, by Snell’s law
$\mu = \frac{\sin 45}{\sin r} \Rightarrow \sin r = \frac{1}{\mu\sqrt{2}}\ \ \ \ \ \ \ldots(i)$
At point $B$, for total internal reflection $\sin i_{1} = \frac{1}{\mu}$
From figure, $i_{1} = 90 - r$
$\therefore\sin{(90{^\circ} - r) = \frac{1}{\mu}}$
$\Rightarrow \cos r = \frac{1}{\mu}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Now $\cos r = \sqrt{1 - \sin^{2}r} = \sqrt{1 - \frac{1}{2\mu^{2}}}$
$= \sqrt{\frac{2\mu^{2} - 1}{2\mu^{2}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
From equation (ii) and (iii), $\frac{1}{\mu} = \sqrt{\frac{2\mu^{2} - 1}{2\mu^{2}}}$
Squaring both side and then solving, we get $\mu = \sqrt{\frac{3}{2}}$