Let x-y plane be the boundary between two transparent media. Medium 1 in z ≥ 0 has a refractive index of √(2) and medium 2 with z < 0 has a refractive index of √(3). A ray of light in medium 1 given by the vector A = 6√(3)i + 8√(3)j - 10k is incident on the plane of separation. The angle of refraction in medium 2 is: (a) 45° (b) 60° (c) 75° (d) 30°

Answer: (a). The normal to the x - y plane is along k, so cos i = (|Az|)/(|A|). |A| = √(108 + 192 + 100) = 20, so cos i = 10/20 = 1/2 ⇒ i = 60°. √(2)sin 60° = √(3)sin r ⇒ sin r = (√(2)·√(3)/2)/(√(3)) = 1/√(2). So the answer is (a) 45°.

Let x-y plane be the boundary between two transparent media. Medium 1… | JEE Main 2011 (AIEEE)

Q74JEE Main 2011 · AIEEEMedium

Refraction, Glass Slab & Apparent Depth

Let $x\text{-}y$ plane be the boundary between two transparent media. Medium $1$ in $z \ge 0$ has a refractive index of $\sqrt{2}$ and medium $2$ with $z < 0$ has a refractive index of $\sqrt{3}.$ A ray of light in medium $1$ given by the vector $\vec{A} = 6\sqrt{3}\hat{i} + 8\sqrt{3}\hat{j} - 10\hat{k}$ is incident on the plane of separation. The angle of refraction in medium $2$ is:

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Answer: (a)

The normal to the $x$ - $y$ plane is along $\hat{k}$ , so $\cos i = \dfrac{|A_z|}{|\vec{A}|}$ .
$|\vec{A}| = \sqrt{108 + 192 + 100} = 20$ , so $\cos i = \dfrac{10}{20} = \dfrac{1}{2} \Rightarrow i = 60^\circ$ .
$\sqrt{2}\sin 60^\circ = \sqrt{3}\sin r \Rightarrow \sin r = \dfrac{\sqrt{2}\cdot\sqrt{3}/2}{\sqrt{3}} = \dfrac{1}{\sqrt{2}}$ .
So the answer is (a) $45^\circ$ .

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