Question

What should be the value of the refractive index $n$ of a glass rod placed in air so that the light entering through flat surface of the rod does not cross the curved surface of the rod?

• A. $n=\frac{1}{\sqrt{2}}$
• B. $n\le \sqrt{2}$
• C. $n\ge \frac{1}{\sqrt{2}}$
• D. $n\ge \sqrt{2}$

Answer 1

The correct option is D $n\ge \sqrt{2}$


The critical angle for glass-air interface is,
${\theta }_c=\frac{1}{n}$


Also, $(r+r^{\prime })={90}^{\circ }$

$\Rightarrow r_{\text{min}}^{\prime }={90}^{\circ }-r_{\text{max}}$

Using snell's law,

$n=\frac{\sin (i{)}_{\text{max}}}{\sin (r{)}_{\text{max}}}=\frac{\sin {90}^{\circ }}{\sin (r{)}_{\text{max}}}$

$(\because i_{\text{max}}={90}^{\circ })$

Thus, $\sin (r{)}_{\text{max}}=\frac{1}{n}={\theta }_c$

$\Rightarrow (r{)}_{\text{max}}={\theta }_c$
$\text{Or},r_{\text{min}}^{\prime }={90}^{\circ }-{\theta }_c$

Now if the minimum value of $r^{\prime }$ is greater than ${\theta }_c$ then obviously all values of $r^{\prime }$ will be greater than the ${\theta }_c$ i.e in all cases the total internal reflection will take place at the glass-air interface.

Therefore the necessary condition for $\text{T.I.R}$ is,

$\Rightarrow r_{\text{min}}^{\prime }\ge {\theta }_c$

$\text{Or},{90}^{\circ }-{\theta }_c\ge {\theta }_c...\left(i\right)$

Taking $\sin$ both sides,

$\Rightarrow \sin ({90}^{\circ }-{\theta }_c)\ge \sin {\theta }_c$

$\Rightarrow \cos {\theta }_c\ge \sin {\theta }_c$

$\text{Or,}\cot {\theta }_c\ge 1....\left(ii\right)$

From the triangle,

$\cot {\theta }_c=\frac{\sqrt{n^2-1}}{1}$

From Eq.$\left(ii\right)$ we get,

$\sqrt{n^2-1}\ge 1$

$\Rightarrow n^2\ge 2$

$\therefore n\ge \sqrt{2}$