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

Correct Answer: 15

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°.