question_answer

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

A) \[{{30}^{o}}\]         

B) \[{{45}^{o}}\]

C) \[{{60}^{o}}\]  

D) \[{{75}^{o}}\]

Correct Answer: B

Solution :

[b] \[{{\mu }_{1}}\,\sin {{\theta }_{1}}={{\mu }_{2}}\,\sin {{\theta }_{2}}\]

\[\cos {{\theta }_{1}}=\frac{10}{\sqrt{{{(6\sqrt{3})}^{2}}+{{(8\sqrt{3})}^{2}}+100}}\]

\[=\frac{10}{\sqrt{400}}=\frac{10}{20}\]

\[\Rightarrow \]            \[{{\theta }_{1}}={{60}^{o}}\]

\[\sqrt{2}\,\sin {{60}^{o}}\,=\sqrt{3}\,\sin {{\theta }_{2}}\]

\[\Rightarrow \]            \[\sqrt{2}\times \frac{\sqrt{3}}{2}=\sqrt{3}\sin {{\theta }_{2}}\]

\[\Rightarrow \]            \[\sin {{\theta }_{2}}=\frac{1}{\sqrt{2}}\,\Rightarrow \,\,\,{{\theta }_{2}}={{45}^{o}}\]