The refractive index

By Snell's Law , we know that

$n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}$

In this case,

$n \sin x = \sin 2x$

$n \sin x = 2 \sin x \cos x$

$2 \sin x \cos x - n \sin x = 0$

$\sin x ( 2 \cos x - n ) = 0$

(The solutions of $\sin x = 0$ can be neglected in this case)

Therefore we know that

$2 \cos x = n$

$\cos x = \frac {n} {2}$

$x = \cos ^{-1} (\frac {n}{2})$

Since the angle of incidence is $2x$ , we have

$2x = 2 \cos ^{-1} (\frac {n}{2}) =$ Angle of Incidence.