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 $A=6\sqrt{3\widehat{i}}+8\sqrt{3\widehat{j}}-10\widehat{k}$ 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 $A=6\sqrt{3\widehat{i}}+8\sqrt{3\widehat{j}}-10\widehat{k}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, =(6\sqrt{3\widehat{i}}+8\sqrt{3\widehat{j}})+(-10\widehat{k})$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, = 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
$i=cos^{-1}\Bigg\{\frac{10}{\sqrt{(10)^2+(6\sqrt{3})^2+(8\sqrt{3})^2}}\Bigg\}=cos^{-1}\Bigg\{\frac{1}{2}\Bigg\}$
$i=60^\circ$
Unit vector in the direction of QOQ' will be
$\widehat{q}=\frac{6\sqrt{3}\widehat{i}+8\sqrt{3}\widehat{j}}{\sqrt{(6\sqrt{3})^2+(8\sqrt{3})^2}}$
$=\frac{1}{2}(3\widehat{i}+4\widehat{j})$
Snell's law gives
$\frac{\sqrt{3}}{\sqrt{2}}=\frac{sin\, i}{sin\, r}=\frac{sin\, 60^\circ}{sin\, r}$
$\therefore sin\, r=\frac{\sqrt{3} / 2}{\sqrt{3}/ \sqrt{2}}=\frac{1}{\sqrt{2}}$
$\therefore r=45^\circ$
Now, we have to find a unit vector in refracted ray's direction
OR. Say it is $\widehat{r}$ whose magnitude is 1. Thus,
$ \widehat{r}=(1\, sin\, r)\widehat{q}-(1\, cos\, r)\widehat{k}$
$ =\frac{1}{\sqrt{2}}[\widehat{q}-\widehat{k}]=\frac{1}{\sqrt{2}}\Bigg[\frac{1}{5}(3\widehat{i}+4\widehat{j})-\widehat{k}\Bigg]$
$ \widehat{r}=\frac{1}{5\sqrt{2}}(3\widehat{i}+4\widehat{j}-5\widehat{k})$