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 $
BETA
