Question

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

Answer 1

Incident ray $A=6\sqrt{3\hat{i}}+8\sqrt{3\hat{j}}-10\hat{k}$
$=(6\sqrt{3\hat{i}}+8\sqrt{3\hat{j}})+(-10\hat{k})$
$=QO+PQ$ (As shown in figure)
Note that QO is lying on x-y plane.
Now, Q Q ' and Z-axis are mutually perpendicular. Hence, we
can show them in two-dimensional figure as below.
Note that QO is lying on x-y plane.
Now, Q Q ' and Z-axis are mutually perpendicular. Hence, we
can show them in two-dimensional figure as below.
Vector A makes an angle i with z-axis, given by
$i=co{s}^{-1}\left\{\frac{10}{\sqrt{(10{)}^2+(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2}}\right\}=co{s}^{-1}\left\{\frac{1}{2}\right\}$
$i=60^{\circ }$
Unit vector in the direction of QOQ' will be
$\hat{q}=\frac{6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2}}$
$=\frac{1}{2}(3\hat{i}+4\hat{j})$
Snell's law gives
$\frac{\sqrt{3}}{\sqrt{2}}=\frac{sini}{sinr}=\frac{sin60^{\circ }}{sinr}$
$\therefore sinr=\frac{\sqrt{3}/2}{\sqrt{3}/\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\therefore r=45^{\circ }$
Now, we have to find a unit vector in refracted ray's direction
OR. Say it is $\hat{r}$ whose magnitude is 1. Thus,
$\hat{r}=(1sinr)\hat{q}-(1cosr)\hat{k}$
$=\frac{1}{\sqrt{2}}[\hat{q}-\hat{k}]=\frac{1}{\sqrt{2}}[\frac{1}{5}(3\hat{i}+4\hat{j})-\hat{k}]$
$\hat{r}=\frac{1}{5\sqrt{2}}(3\hat{i}+4\hat{j}-5\hat{k})$