Question

The speed of light in medium $b$ is $\frac{c}{3}$, where $c=$ speed of light in vacuum and light refracts from medium $a$ to medium $b$ making angles ${45}^{\circ }$ and ${60}^{\circ }$ respectively with normal. The speed of light in medium $a$ is

Answer 1

Applying Snell's law at the interface,
${\mu }_a\sin i={\mu }_b\sin r$
$\Rightarrow \left(\frac{c}{v_a}\right)\sin i=\left(\frac{c}{v_b}\right)\sin r$
$\Rightarrow \frac{c}{v_a}\sin {45}^{\circ }=\frac{c}{v_b}\sin {60}^{\circ }$
$\Rightarrow \frac{c}{\sqrt{2}{v}_a}=\frac{c}{\left(\frac{c}{3}\right)}\times \frac{\sqrt{3}}{2}$
$\Rightarrow \frac{c}{v_a}=\frac{3\sqrt{3}}{2}\times \sqrt{2}$
or, $v_a=\frac{2c}{\left(3\sqrt{3}\right)\left(\sqrt{2}\right)}$
$\therefore v_a=\frac{\sqrt{2}c}{3\sqrt{3}}$

Tip: Apply Snell's law and replace the $\mu$ with $\frac{c}{\upsilon }$