Optics Question 960
Question: Let the $ x-y $ plane be the boundary between two transparent media. Medium 1 in $ z\ge 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} $
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Answer:
Correct Answer: B
Solution:
[b] $ {\mu _{1}}\sin {\theta _{1}}={\mu _{2}}\sin {\theta _{2}} $ $ \cos {\theta _{1}}=\frac{10}{\sqrt{{{(6\sqrt{3})}^{2}}+{{(8\sqrt{3})}^{2}}+100}} $ $ =\frac{10}{\sqrt{400}}=\frac{10}{20} $
$ \Rightarrow $ $ {\theta _{1}}=60^{o} $ $ \sqrt{2}\sin 60^{o}=\sqrt{3}\sin {\theta _{2}} $
$ \Rightarrow $ $ \sqrt{2}\times \frac{\sqrt{3}}{2}=\sqrt{3}\sin {\theta _{2}} $
$ \Rightarrow $ $ \sin {\theta _{2}}=\frac{1}{\sqrt{2}}\Rightarrow {\theta _{2}}=45^{o} $
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