question_answer

A transparent solid cylindrical rod has a refractive index of\[\frac{2}{\sqrt{3}}\].It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle \[\theta \] for which the light ray grazes along the wall of the rod is:   AIEEE  Solved  Paper-2009

A) \[{{\sin }^{-1}}\left( \frac{1}{2} \right)\]

B)        \[{{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]  

C)        \[{{\sin }^{-1}}\left( \frac{2}{\sqrt{3}} \right)\]  

D)        \[{{\sin }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]

Correct Answer: D

Solution :

\[1\times \sin 90=\frac{2}{\sqrt{3}}\sin (90-\alpha )\] \[\Rightarrow \]\[\cos \alpha =\frac{\sqrt{3}}{2}\] So           \[sin\alpha =\sqrt{1-\frac{3}{4}}=\frac{1}{2}\] Now,     \[1\times \sin \theta =\frac{2}{\sqrt{3}}\sin \alpha \] \[=\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \]\[\theta ={{\sin }^{-1}}\frac{1}{\sqrt{3}}\]