A) \[45{}^\circ \]
B) \[60{}^\circ \]
C) \[0{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: A
Solution :
According to the given condition, the beam of light will retrace its path after reflection from BC. So, \[\angle CPQ=90{}^\circ \] Thus, angle of refraction at surface AC \[\angle PQN=\angle r=90{}^\circ -60{}^\circ =30{}^\circ \] By Snells law \[\mu =\frac{\sin i}{\sin r}\] \[\Rightarrow \] \[\sqrt{2}=\frac{\sin i}{\sin 30{}^\circ }\] \[\therefore \] \[\sqrt{2}\times \sin 30{}^\circ =\sin i\] \[\Rightarrow \] \[\sqrt{2}\times \frac{1}{2}=\sin i\] \[\Rightarrow \] \[\sin i=\frac{1}{\sqrt{2}}=\sin 45{}^\circ \] \[\therefore \] \[i=45{}^\circ \]You need to login to perform this action.
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