Question

Light is incident on a prism with an equilateral triangular cross-section, making an angle of $\theta$ w.r.t the normal as shown in the figure. The index of refraction of the prism is $\frac{2}{\sqrt{3}}$. If the angle $\theta$ starts at ${60}^{\circ }$ and is gradually made smaller until no light exist from side BC, the final value of $\theta$ in degrees is

Answer 1

$A={60}^{\circ }$

$r^{\prime }+r=A$

So, $\sin {90}^{\circ }=\sqrt{2}\sin r$

$\Rightarrow r^{\prime }={45}^{\circ }$

$45+r=A$

$\Rightarrow r={15}^{\circ }$

Since

$\left(1\right)\sin \theta =\sqrt{2}\sin {15}^{\circ }$

$\sin \theta =\sqrt{2}\sin {15}^{\circ }$

$\theta ={\sin }^{-1}\left(\sqrt{2}\sin {15}^{\circ }\right)$