A) \[50{}^\circ \]
B) \[60{}^\circ \]
C) \[90{}^\circ \]
D) \[40{}^\circ \]
Correct Answer: C
Solution :
Refraction at P, \[\frac{\sin \,{{60}^{o}}}{\sin \,{{r}_{1}}}=\sqrt{3}\] \[\Rightarrow \] \[\sin \,{{r}_{1}}=\frac{1}{2}\] \[\Rightarrow \] \[{{r}_{1}}={{30}^{o}}\] Since, \[{{r}_{2}}={{r}_{1}}\] \[\therefore \] \[{{r}_{2}}={{30}^{o}}\] Refraction at \[Q,\,\frac{\sin \,{{r}_{2}}}{\sin \,{{i}_{2}}}=\frac{1}{\sqrt{3}}\] Putting, \[{{r}_{2}}={{30}^{o}},\] we obtain \[{{i}_{2}}={{60}^{o}}\] Refraction at Q,\[r_{2}^{'}={{r}_{2}}={{30}^{o}}\] \[\therefore \] \[\alpha ={{180}^{o}}-(r_{2}^{'}+{{i}_{2}})\] \[={{180}^{o}}-({{30}^{o}}+{{60}^{o}})={{90}^{o}}\]You need to login to perform this action.
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