question_answer

A ray incident at a point as an angle of incidence of \[60{}^\circ \], enters a glass sphere ofrefractiveindex, \[n=\sqrt{3}\] and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is

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,\[r_{2}^{'}={{r}_{2}}={{30}^{o}}\] \[\therefore \]    \[\alpha ={{180}^{o}}-(r_{2}^{'}+{{i}_{2}})\] \[={{180}^{o}}-({{30}^{o}}+{{60}^{o}})={{90}^{o}}\]