Question

Let the x-z plane be the boundary between transparent media. Medium 1 m in $Z\ge 0$ refractive index of $\sqrt{2}$ and medium 2 with has a refractive index of $\sqrt{3}$.
A ray of light in medium 1 given by the $\overline{A}=6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k}$ is incident on the separation. The angle of refraction in meter is :

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

Answer 1

Answer: (B)

${45}^{\circ }$

Given,

The x-y plane is boundary between the two media

The refractive index for $z>0$ and $Z⩽0$ is different

The angle of incidence along the z-axis can be given as angle of incidence $i$, hence

$\cos i=\left|\frac{A_z}{\sqrt{{A_x}^2+{A_y}^2+{A_z}^2}}\right|\Rightarrow i={\cos }^{-1}\left(\left|\frac{-10}{\sqrt{{\left(6\sqrt{3}\right)}^2+{\left(8\sqrt{3}\right)}^2+{\left(-10\right)}^2}}\right|\right)$

$i={\cos }^{-1}\left(\left|\frac{-10}{\sqrt{400}}\right|\right)\Rightarrow i={\cos }^{-1}\left(\frac{1}{2}\right)$

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

Now using Snell’s law

${\mu }_1\sin i={\mu }_2\sin r$

Where ${\mu }_1=\sqrt{2}$ and ${\mu }_2=\sqrt{3}$

Hence,

$\sqrt{2}\sin {60}^{\circ }=\sqrt{3}\sin r\Rightarrow \sin r=\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{\sqrt{3}}$

$\sin r=\frac{1}{\sqrt{2}}\Rightarrow r={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

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

Hence the angle of refraction of medium $2$ is $r={45}^{\circ }$.