Question

A light ray is incident on an equilateral prism of refractive index $\sqrt{2}$. What should be the angle of incidence such that the ray gazes on the second surface?

Answer 1

Let's trace a path of incident light ray:


$\Rightarrow$Anlge of prism $(\angle A$) = ${60}^o$
$\Rightarrow$$\angle A$ = $\angle r_1$ + $\angle r_2$$..\left(1\right)$
$\Rightarrow$Emergent ray gazes at second surface means that ray incident at critical angle on the second surface.
$\Rightarrow$Angle of emergent = ${90}^o$
$\Rightarrow$Critical angle = $r_2$

$\frac{sin{r}_2}{sine}=\frac{Refractiveindexofair\left(1\right)}{Refractiveindexofprism}$$..\left(2\right)$

For the first surface;
$\frac{sin\theta }{sin{r}_1}=\frac{\sqrt{2}}{1}..\left(3\right)$

For the second surface;
Usind 2nd equation
$\frac{sin{r}_2}{sin90}=\frac{1}{2}$
$\Rightarrow sin{r}_2=\frac{1}{\sqrt{2}}$
$\Rightarrow r_2={45}^{\circ }$

By using 1st equation:
$\Rightarrow r_1={60}^{\circ }-{45}^{\circ }={15}^{\circ }$

$\Rightarrow \frac{sin\theta }{sin{15}^{\circ }}=\frac{\sqrt{2}}{1}$$..\left(using3rdequation\right)$
$sin\theta =\frac{\sqrt{2}}{1}\times sin{15}^{\circ }$
$\Rightarrow \theta =si{n}^{-1}(\sqrt{2}\times sin{15}^{\circ })$