question_answer

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

[c] Refraction at P,

\[\frac{\sin 60{}^\circ }{\sin {{r}_{1}}}=\sqrt{3}\]\[\Rightarrow \]\[\sin {{r}_{1}}=\frac{1}{2}\]\[\Rightarrow \]\[{{r}_{1}}=30{}^\circ \]

Since, \[{{r}_{2}}={{r}_{1}}\]

\[\therefore \]    \[{{r}_{2}}=30{}^\circ \]

Refraction at Q, \[\frac{\sin {{r}_{2}}}{\sin {{i}_{2}}}=\frac{1}{\sqrt{3}}\]

Putting \[{{r}_{2}}={{30}^{o}},\] we obtain \[{{i}_{2}}={{60}^{o}}\]

Reflection at Q,

\[r{{'}_{2}}={{r}_{2}}=30{}^\circ \]

\[\therefore \]    \[\alpha =180{}^\circ -(r{{'}_{2}}+{{i}_{2}})\]

\[=180{}^\circ -(30{}^\circ +60{}^\circ )=90{}^\circ \]