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 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 :

[d] \[1\times \sin 90=\frac{2}{\sqrt{3}}\sin (90-\alpha )\]

\[\Rightarrow \]\[\cos \alpha =\frac{\sqrt{3}}{2}\]

So        \[sin\alpha =\sqrt{1-\frac{3}{4}}=\frac{1}{2}\]

Now,    \[1\times \sin \theta =\frac{2}{\sqrt{3}}\sin \alpha \]

\[=\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}\]

\[\Rightarrow \]\[\theta ={{\sin }^{-1}}\frac{1}{\sqrt{3}}\]