Question

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

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${75}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

Answer: (A)

${45}^{\circ }$

From the fugure, it can be seen that normal to the boundary separating the two media will be in z-direction, as incident ray, refracted ray and normal are in the same plane.

The angle made by the incidnet ray with the normal is given as:

$Cosi=\frac{A_z}{A}=\frac{10}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2+{100}^2}}=\frac{10}{\sqrt{108+192+100}}=\frac{10}{20}$

$\Rightarrow Cosi=\frac{-1}{2}\Rightarrow i=Co{s}^1\left(\frac{1}{2}\right)$

$\Rightarrow i=60{}^{\circ }$

Applying Snell's law of refraction at the interface of two media

$n_{1}Sini=n_{2}Sinr$

$\sqrt{2}Sin60{}^{\circ }=\sqrt{3}Sinr$

$Sinr=\frac{\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times 2}=\frac{1}{\sqrt{2}}$

$\Rightarrow r=Si{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)\Rightarrow r=45{}^{\circ }$

As the X, Y components remain unchanged, so unit vector along direction of refracted ray is

$\hat{R}=\frac{6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2}}Sin45+Cos45(-\hat{k})$

$\hat{R}=\frac{3}{5\sqrt{2}}\hat{i}+\frac{4}{5\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}$

$\hat{R}=\frac{1}{5\sqrt{2}}(3\hat{i}+4\hat{j}-5\hat{k})$

Hence , the answer is $\frac{1}{5\sqrt{2}}(3\hat{i}+4\hat{j}-5\hat{k})$.