Optics Question 947

Question: A ray incident at a point at an angle of incidence of $ 60^{o}, $ enters a glass sphere of refractive index $ n=\sqrt{3} $ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is

Options:

A) $ 50{}^\circ $

B) $ 60{}^\circ $

C) $ 90{}^\circ $

D) $ 40{}^\circ $

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Refraction at P, $ \frac{\sin 60{}^\circ }{\sin r _{1}}=\sqrt{3} $

$ \Rightarrow $ $ \sin r _{1}=\frac{1}{2} $

$ \Rightarrow $ $ r _{1}=30{}^\circ $ Since, $ r _{2}=r _{1} $

$ \therefore $ $ r _{2}=30{}^\circ $ Refraction at Q, $ \frac{\sin r _{2}}{\sin i _{2}}=\frac{1}{\sqrt{3}} $ Putting $ r _{2}=30^{o}, $ we obtain $ i _{2}=60^{o} $ Reflection at Q, $ r{’ _{2}}=r _{2}=30{}^\circ $

$ \therefore $ $ \alpha =180{}^\circ -(r{’ _{2}}+i _{2}) $ $ =180{}^\circ -(30{}^\circ +60{}^\circ )=90{}^\circ $



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