Question

The X–Y plane be taken as the boundary between two transparent media M1 and M2.M1 in Z 0 has a refractive index of $\sqrt2$ and M2 with $Z<0$ has a refractive index of $\sqrt3$. A ray of light travelling in M1 along the direction given by the vector $\mathbf{\overrightarrow{P}} = 4\sqrt{3}\mathbf{\hat{i}} - 3\sqrt{3}\mathbf{\hat{j}} - 5\hat{k}$, is incident on the plane of separation. The value of difference between the angle of incident in M1 and the angle of refraction in M2 will be ______ degree.

Answer 1

Normal will be $−\hat{k}$
so

$\cos i = \frac{\mathbf{P} \cdot \hat{n}}{\left|\mathbf{P}\right| \cdot \left|\hat{n}\right|}$
$\frac{5}{10} = \frac{1}{2}$
$⇒ i = 60°$
Using snells law
$\sqrt{2} \sin 60^\circ = \sqrt{3} \sin r$

$\frac{\sqrt{3}}{\sqrt{2}} = 3 \sin r$
$⇒ r = 45°$
So, $i – r = 60°-45°$
$= 15°$
So, the answer is 15°.