Question

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 -

• 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. ${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

Answer 1

Answer: (D)

${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

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