- JEE Mains 2015
- Hard
- MCQ
Monochromatic light is incident on a glass prism of angle \(A\). If the refractive index of the material of the prism is \(μ\), a ray, incident at an angle \(\theta\), on the face \(AB\) would get transmitted through the face \(AC\) of the prism provided

Answer & Solution
(D) \(\theta > Si{n^{ - 1}}\left[ {\mu sin\left( {A - si{n^{ - 1}}\left( {\frac{1}{\mu }} \right)} \right)} \right]\)
Step-by-step Solution
Detailed explanation
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 \(\mathrm{AC}\)
\(\theta > {\sin ^{ - 1}}\) \(\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {\frac{1}{\mu }} \right)} \right]} \right.\)
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