\(\mu=\frac{\sin 45^o}{\sin r} \Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\)  ..... \((i)\)

At point \(B\), for total internal reflection,

\(\sin {i_1} = \frac{1}{\mu }\)

From figure, \(\mathrm{i}_{1}=90^{\circ}-\mathrm{r}\)

\(\therefore \) \(\left(\sin 90^o-r\right)=\frac{1}{\mu}\)

\(\Rightarrow \cos r=\frac{1}{\mu}\)              ...... \((ii)\)

Now \(cos\,r\,=\,\sqrt{1-\sin ^{2} r}=\sqrt{1-\frac{1}{2 \mu^{2}}}\)

\(=\sqrt{\frac{2 \mu^{2}-1}{2 \mu^{2}}}\)        ...... \((iii)\)

From eqs \((ii)\) and \((iii)\)

\(\frac{1}{\mu}=\sqrt{\frac{2 \mu^{2}-1}{2 \mu^{2}}}\)

Squaring both sides and then solving, we get

\(\mu=\sqrt{\frac{3}{2}}\)