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, |
\[r{{'}_{2}}={{r}_{2}}=30{}^\circ \] |
\[\therefore \] \[\alpha =180{}^\circ -(r{{'}_{2}}+{{i}_{2}})\] |
\[=180{}^\circ -(30{}^\circ +60{}^\circ )=90{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec