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 √(2) and M₂ with Z < 0 has a refractive index of √(3). A ray of light travelling in M₁ along the direction given by the vector P = 4√(3)i - 3√(3)j - 5k, is incident on the plane of separation. The value of difference between the angle of incidence in M₁ and the angle of refraction in M₂ will be __________ degree. Numerical answer type. Work it out, then open the answer.

Answer: 15. Normal is along k: |P| = √(48 + 27 + 25) = 10, so cos i = 5/10 ⇒ i = 60°. √(2)sin 60° = √(3)sin r ⇒ sin r = 1/√(2) ⇒ r = 45°. i - r = 60° - 45° = 15°. Answer: 15 degree.

The X-Y plane be taken as the boundary between two transparent media… | JEE Main 2022 (29 Jul, Shift 1)

Q46JEE Main 2022 · 29 JUL · SHIFT 1Medium

Refraction, Glass Slab & Apparent Depth

The X-Y plane be taken as the boundary between two transparent media $M_1$ and $M_2.\ M_1$ in $Z \geqslant 0$ has a refractive index of $\sqrt{2}$ and $M_2$ with $Z < 0$ has a refractive index of $\sqrt{3}$ . A ray of light travelling in $M_1$ along the direction given by the vector $\vec{P} = 4\sqrt{3}\hat{i} - 3\sqrt{3}\hat{j} - 5\hat{k}$ , is incident on the plane of separation. The value of difference between the angle of incidence in $M_1$ and the angle of refraction in $M_2$ will be __________ degree.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

Normal is along $\hat{k}$ : $|\vec{P}| = \sqrt{48 + 27 + 25} = 10$ , so $\cos i = \dfrac{5}{10} \Rightarrow i = 60^\circ$ .
$\sqrt{2}\sin 60^\circ = \sqrt{3}\sin r \Rightarrow \sin r = \dfrac{1}{\sqrt{2}} \Rightarrow r = 45^\circ$ .
$i - r = 60^\circ - 45^\circ = 15^\circ$ .
Answer: 15 degree.

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