Question

Light is incident at an angle $i$ on one planar cross-sectional end of a transparent cylindrical rod of refractive index $\mu$. Find the minimum value of $\mu$, so that the light entering the rod does not emerge from the curved surface of the rod, irrespective of the value of $i$.

• A. $\sqrt{6}$
• B. $\sqrt{5}$
• C. $\sqrt{3}$
• D. $\sqrt{2}$

Answer 1

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

For the light ray to remain contained inside the cylindrical rod, it must suffer total internal reflection at the curved surface.
At the curved surface, from the geometry of the diagram,
$i^{{}^{\prime }}=({90}^{\circ }-r)\ge i_c$
In critical case,
$i^{{}^{\prime }}=({90}^{\circ }-r)=i_c$


From formula,
$\sin i_c=\frac{{\mu }_r}{{\mu }_d}$
$\Rightarrow \sin ({90}^{\circ }-r)=\frac{1}{{\mu }_d}$
$\Rightarrow \cos r=\frac{1}{{\mu }_d}$ ..............$\left(1\right)$
At the air-rod interface,
${\mu }_a\sin i={\mu }_d\sin r$
$\Rightarrow {\mu }_d\sin r=\sin i$ ..............$\left(2\right)$
For the same rod $({\mu }_d=\text{constant})$, $i^{{}^{\prime }}={90}^{\circ }-r$ is minimum when $r$ is maximum and $r$ is maximum when $i$ is maximum $[$from $\left(2\right)]$.
The maximum value of $i$ is ${90}^{\circ }$.
From $\left(2\right)$,
${\mu }_d\sin r=\sin {90}^{\circ }=1$
$\Rightarrow \sin r=\frac{1}{{\mu }_d}$ ..............$\left(3\right)$
From $\left(1\right)$ and $\left(3\right)$,
$\cos r=\sin r$
$\Rightarrow r={45}^{\circ }$
Putting the value of $r$ in $\left(1\right)$, we get,
${\mu }_d=\sqrt{2}$

Why this question? It tests your basic knowledge of trigonometry and total internal reflection.