Explanation

(d) From Snell’s law, refractive index $(\mu)$ is given by
$\mu = \frac{\sin i}{\sin r}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Where $i$ is angle of incidence and $r$ of refraction.
Also, $\mu = \frac{v_{1}}{v_{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Equating Eqs, (i) and (ii), we get
$\mu = \frac{\sin i}{\sin r} = \frac{v_{1}}{v_{2}}$
$\Rightarrow \sin{r = \ \frac{v_{2}}{v_{1}}.\sin i}$
Given, $v_{2}\ = 2u,v_{1} = u,i = 30{^\circ},\sin{30{^\circ}} = \frac{1}{2}$
$\therefore\sin{r = 2 \times \frac{1}{2} = 1}$
$\Rightarrow r = 90{^\circ}$