A) \[0{}^\circ \]
B) \[30{}^\circ \]
C) \[60{}^\circ \]
D) \[45{}^\circ \]
Correct Answer: D
Solution :
At second surface, there is no refraction so \[{{r}_{2}}=0\] \[\therefore \] \[{{r}_{1}}=A={{30}^{o}}\] From Snells law \[n=\frac{\sin {{i}_{1}}}{\sin \,{{r}_{1}}}\] \[n=\frac{\sin {{i}_{1}}}{\sin {{30}^{o}}}\] \[\sin {{i}_{1}}=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\] \[\sin {{i}_{1}}=\frac{1}{\sqrt{2}}\] \[{{i}_{1}}={{45}^{o}}\]
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