Question

Let the X-y plane be the boundary between two transparent media. Medium $1inz\ge 0$ has a 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 $\overrightarrow{[}A]=6\sqrt{3\hat{i}}+8\sqrt{3}\hat{j}-10\hat{k}$ is incident on the plane of separation. The angle of refraction in medium 2 is:

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

Answer 1

Answer: (D)

${45}^{\circ }$

$\text{Given}=6\sqrt{3}\hat{i}+8\sqrt{3}\hat{ȷ}-10\hat{k}$

$\begin{array}{c}{\mu }_1=\sqrt{2}\\ {\mu }_2=\sqrt{3}\\ \text{For first medium}\\ \text{According to vector}\end{array}$

$\begin{array}{c}K=\text{angle between}\\ \text{medium and}\\ \text{z axis}\end{array}$

$\begin{array}{cc}\cos k& =\frac{10}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2+(10{)}^2}}\\ & =\frac{10}{20}=\frac{1}{2}\end{array}$

$\begin{array}{c}\text{K}={60}^{\circ }=i\\ \text{Now,}\text{Using shell's law}\end{array}$

$\begin{array}{c}{\mu }_1\sin i={\mu }_2\sin r\\ \sqrt{2}-\sin {60}^{\circ }=\sqrt{3}\sin r\\ \frac{\sqrt{2}\times \sqrt{3}}{2}-\sqrt{3}\sin r\Rightarrow \frac{1}{\sqrt{2}}=\sin r\\ r={45}^{\circ }\end{array}$