SATHEE: Optics Question 960

Optics Question 960

Question: Let the $x - y$ plane be the boundary between two transparent media. Medium 1 in $z \geq 0$ has 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

Options:

A) $30^{o}$

B) $45^{o}$

C) $60^{o}$

D) $75^{o}$

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $μ_{1} \sin θ_{1} = μ_{2} \sin θ_{2}$ $\cos θ_{1} = \frac{10}{\sqrt{{(6 \sqrt{3})}^{2} + {(8 \sqrt{3})}^{2} + 100}}$ $= \frac{10}{\sqrt{400}} = \frac{10}{20}$

$\Rightarrow$ $θ_{1} = 60^{o}$ $\sqrt{2} \sin 60^{o} = \sqrt{3} \sin θ_{2}$

$\Rightarrow$ $\sqrt{2} \times \frac{\sqrt{3}}{2} = \sqrt{3} \sin θ_{2}$

$\Rightarrow$ $\sin θ_{2} = \frac{1}{\sqrt{2}} \Rightarrow θ_{2} = 45^{o}$

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Optics Question 960

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