d
When \({{\text{r}}_2} = {\text{C}},\)  \(\angle {{\text{N}}_2}{\text{Rc}} = {90^o}\)

Where \(C=\) critical angle

As \(\sin C=\frac{1}{v}=\sin r_{2}\)

Applying snell's law at \('R'\)

\(\mu \,\,\sin {r_2} = 1\,\,\sin {90^o}\)   ..... \((i)\)

Applying snell's law at \('Q'\)

\(1 \times \sin \theta=\mu \sin r_{1}\)  ...... \((ii)\)

But \(r_{1}=A-r_{2}\)

So, \(\sin \theta=\mu \sin \left(A-r_{2}\right)\)

\(\sin \theta=\mu \sin A \cos r_{2}-\cos A\)      ...... \((iii)\)            [using \((i)\) ]

From \(( 1 )\)

\(\cos r_{2}=\sqrt{1-\sin ^{2} r_{2}}=\sqrt{1-\frac{1}{\mu^{2}}}\)  ..... \((iv)\)

By eq. \((iii)\) and \((iv)\) 

\(\sin \theta=\mu \sin A \sqrt{1-\frac{1}{\mu^{2}}}-\cos A\)

on further solving we can show for raynot to transmitted through face \(AC\)

\(\theta=\sin ^{-1}\left[u \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right]\right.\)

So, for transmission through face \(\mathrm{AC}\)

\(\theta  > {\sin ^{ - 1}}\) \(\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {\frac{1}{\mu }} \right)} \right]} \right.\)