question_answer

A ray incident at a point at an angle of incidence of \[60{}^\circ \]enters a glass sphere of refractive index \[\mu =\sqrt{3}\] 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 :

[c] :  For 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\[{{r}_{2}}={{r}_{1}}\therefore {{r}_{2}}={{30}^{o}}\] For refraction at \[Q,\frac{\sin {{r}_{2}}}{\sin {{i}_{2}}}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \]\[\frac{\sin {{30}^{o}}}{\sin {{i}_{2}}}=\frac{1}{\sqrt{3}}\Rightarrow \sin {{i}_{2}}=\frac{\sqrt{3}}{2}\]\[\Rightarrow \]\[{{i}_{2}}={{60}^{o}}\] For refraction at \[Q,r_{2}^{'}={{r}_{2}}={{30}^{o}}\] \[\therefore \]\[\alpha ={{180}^{o}}-(r_{2}^{'}+{{i}_{2}})={{180}^{o}}-({{30}^{o}}+{{60}^{o}})={{90}^{o}}\]