Question

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

Answer 1

The correct option is A ${45}^o$

Given:

Let the x-z plane be the boundary between two transparent media. Medium 1 in $z\ge 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z<0$ has 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.

To find the angle of refraction in medium 2

Solution:

As the refractive index for $z>0$ and $z\le 0$ is different, x-y plane should be the boundary between the two media.

The incident angle is the angle the vector makes with the z-axis (that is the direction of the normal to the x-y plane).

Angle of incidence,

$\cos i=\left|\frac{A_z}{\sqrt{A_x^2+A_y^2+A_z^2}}\right|⟹\cos i=\left|\frac{-10}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2+(10{)}^2}}\right|⟹\cos i=\left|\frac{-10}{\sqrt{108+192+100}}\right|⟹\cos i=\left|\frac{-10}{\sqrt{400}}\right|⟹\cos i=\frac{1}{2}⟹i={60}^{\circ }$

Now will apply the snell's law, we get

$\frac{\sin i}{\sin r}=\frac{{\mu }_2}{{\mu }_1}$, where ${\mu }_1,{\mu }_2$ are the refractive index of medium 1 and 2 respectively and whose values are given in the question.

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

multiply the numerator and denominator by $\sqrt{2}$, we get

$\sin r=\frac{1}{\sqrt{2}}⟹r={45}^{\circ }$

is the angle of refraction in medium 2.