The xy plane is the boundary between two transparent media. Medium 1 with $z \geq 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z \leq 0$ has refractive index of $\sqrt{3}$ . A ray of light in medium 1 given by the vector $6 \sqrt{3} \hat{i} + 8 \sqrt{3} \hat{j} - 10 \hat{k}$ is incident on the plane of separation. Find the angle of refraction in medium 2 vector in the direction of the refracted ray in medium 2

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30⁰

45⁰

60⁰

75⁰

answer is B.

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Detailed Solution

Taking boundary as x−y plane, the angle of incidence is given as

$cosθ = \frac{(- \hat{k}) \cdot (6 \sqrt{3} \hat{i} + 8 \sqrt{3} \hat{j} - 10 \hat{k})}{\sqrt{108 + 192 + 100}} = \frac{1}{2}$

$\Rightarrow θ = 60^{o}$

Now, $\sqrt{2} \sin 60 ° = \sqrt{3} sinr \Rightarrow sinr = \frac{1}{\sqrt{2}}$

So the angle of refraction is 45^o.

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