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