JEE Advanced · Physics · 24. Ray Optics
A ray \(O P\) of monochromatic light is incident on the face \(A B\) of prism \(A B C D\) near vertex \(B\) at an incident angle of \(60^{\circ}\) (see figure). If the refractive index of the material of the prism is \(\sqrt{3}\), which of the following is (are) correct?

- A The ray gets totally internally reflected at face \(C D\)
- B The ray comes out through face \(A D\)
- C The angle between the incident ray and the emergent ray is \(90^{\circ}\)
- D The angle between the incident ray and the emergent ray is \(120^{\circ}\)
Answer & Solution
Correct Answer
(A) The ray gets totally internally reflected at face \(C D\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}& \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r} \\& \therefore r=30^{\circ}\end{aligned}\)

\(\theta_C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
or \(\sin \theta_C=\frac{1}{\sqrt{3}}=0.577\)
At point \(Q\), angle of incidence inside the prism is \(i=45^{\circ}\).
Since \(\sin i=\frac{1}{\sqrt{2}}\) is greater than \(\sin \theta_C=\frac{1}{\sqrt{2}}\), ray gets totally internally reflected at face \(C D\). Path of ray of light after point \(Q\) is shown in figure.
From the figure, we can see that angle between incident ray \(O P\) and emergent ray \(R S\) is \(90^{\circ}\).
Therefore, correct options are (a), (b) and (c).

\(\theta_C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
or \(\sin \theta_C=\frac{1}{\sqrt{3}}=0.577\)
At point \(Q\), angle of incidence inside the prism is \(i=45^{\circ}\).
Since \(\sin i=\frac{1}{\sqrt{2}}\) is greater than \(\sin \theta_C=\frac{1}{\sqrt{2}}\), ray gets totally internally reflected at face \(C D\). Path of ray of light after point \(Q\) is shown in figure.
From the figure, we can see that angle between incident ray \(O P\) and emergent ray \(R S\) is \(90^{\circ}\).
Therefore, correct options are (a), (b) and (c).
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