Question

A right-angled prism is to be made by selecting a proper material and the angles $A$ and $B$ $(B\le A)$, as shown in the figure. It is desired that a ray of light incident on the face $AB$ emerges parallel to the incident direction after two internal reflections. What should be the minimum refractive index above which the internal reflections happens?

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

Answer 1

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

For the light ray to emerge parallel to the incident direction, it must satisfy total internal reflection conditions at the surface $AC$ and $BC$ as shown in figure.


At interface $AC$,
$i_A\ge i_c$
So, $\sin i_A\ge \sin i_c=\frac{1}{\mu }$ ............$\left(1\right)$
Similarly, at interface $CB$,
$\sin i_B\ge \sin i_c=\frac{1}{\mu }$ ............$\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$\sin i_A=\sin i_B$
(In critical condition)
$\Rightarrow i_A=i_B$
From the geometry,
$i_A+i_B={90}^{\circ }$
So, $i_A=i_B={45}^{\circ }$
Putting thus in $\left(1\right)$, we get,
$\mu =\sqrt{2}$

Why this question? It intends to test your knowledge of geometry and its application in total internal reflection.