A) \[30{}^\circ \]
B) \[60{}^\circ \]
C) \[90{}^\circ \]
D) \[45{}^\circ \]
Correct Answer: D
Solution :
The ray retraces its path after reflection from the silvered face AC. Therefore,\[\angle ARQ=90{}^\circ \]. As is clear from the figure, the angle of reflection RQN at face AB is\[30{}^\circ \]. Therefore, by Snells law, we have \[n=\frac{\sin i}{\sin r}\] Given, \[n=\sqrt{2},r=30{}^\circ \] \[\therefore \] \[\sin i=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[i=45{}^\circ \]You need to login to perform this action.
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