Question

Let the $X-Y$ plane be the boundary between two transparent media. Medium 1 in $Z\ge 0$ has a refractive index of $\sqrt{2}$ and medium 2 in $Z<0$ 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. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

As refractive index for $Z>0$ and $Z\le 0$ is different. $X-Y$ plane should be boundary between two media.

Angle of incidence $i$ can be calculated using,
$\cos i=\left|\begin{array}{c}\frac{A_z}{\sqrt{A_x^2+A_y^2+A_z^2}}\end{array}\right|=\frac{10}{\sqrt{(36\times 3)+(64\times 3)+100}}$

$\cos i=\frac{10}{\sqrt{400}}=\frac{10}{20}=\frac{1}{2}i={60}^{\circ }$

From Snell's law, ${}_1{\mu }_2=\frac{\sin i}{\sin r}$

$\frac{{\mu }_2}{{\mu }_1}=\frac{\sin {60}^{\circ }}{\sin r}\Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\sin r}\Rightarrow \sin r=\frac{1}{\sqrt{2}}$

$\therefore r={45}^{\circ }$