Question

A transparent solid cylindrical rod has refractive index of $\frac{2}{\sqrt{3}}.$ It is surrounded by air. A light ray is incident on the axis at one end of the rod as shown in figure.

The incident angle $\theta$ for which the light ray grazes along the wall of the rod is

• A. ${\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• 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

The correct option is D ${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Given,
Refractive index of material of the cylinder, ${\mu }_2=\frac{2}{\sqrt{3}}$
Refractive index of air, ${\mu }_1=1$

If the light ray grazes along the wall, the angle of refraction along the wall at point $\text{Q}$ is $r_2={90}^{\circ }$

Applying Snell's at point $\text{Q}$,
${\mu }_2\sin i_2={\mu }_1\sin r_2$

Substituting the value of ${\mu }_1$, ${\mu }_2$ and $r_2$ we have,
$\frac{2}{\sqrt{3}}\sin i_2=1\sin {90}^{\circ }$

$\Rightarrow \sin i_2=\frac{\sqrt{3}}{2}$

$\therefore i_2={60}^{\circ }$

From the $\Delta \text{PQR}$,
$r_1+i_2+{90}^{\circ }={180}^{\circ }$
$\Rightarrow r_1+{60}^{\circ }+{90}^{\circ }={180}^{\circ }$
$\therefore r_1={30}^{\circ }$

Now, applying Snell's law at point $\text{P}$, we have
${\mu }_1\sin \theta ={\mu }_2\sin r_1$
Substituting the known data,
$\Rightarrow 1\sin \theta =\frac{2}{\sqrt{3}}\sin {30}^{\circ }$

$\Rightarrow \sin \theta =\frac{2}{\sqrt{3}}\times \frac{1}{2}$

$\therefore \theta ={\sin }^{-1}\frac{1}{\sqrt{3}}$

Thus, option $\left(d\right)$ is the correct answer.

Why this question : This question checks the application of $\text{TIR}$ and grazing incidence.