Question

The x-y plane is the boundary between two transparent media. Medium-1 with z > 0 has refractive index $\sqrt{2}$ and medium -2 with z < 0 has a refractive index $\sqrt{3}$. A ray of light in medium-1 given by the vector $\overrightarrow{A}=6\sqrt{3}\overrightarrow{i}+8\sqrt{3}\overrightarrow{j}-10\overrightarrow{k}$ is incident on the plane of separation. If the unit vector in the direction of refracted ray in medium-2 is $\frac{1}{5}(a\overrightarrow{i}+b\overrightarrow{j}-\frac{5}{\sqrt{2}}\overrightarrow{k})$ then find the value of ab.

Answer 1

Given: The x-y plane is the boundary between two transparent media. Medium-1 with z > 0 has refractive index $\sqrt{2}$ and medium -2 with z < 0 has a refractive index $\sqrt{3}$. A ray of light in medium-1 given by the vector $\overrightarrow{A}=6\sqrt{3}\overrightarrow{i}+8\sqrt{3}\overrightarrow{j}-10\overrightarrow{k}$ is incident on the plane of separation. If the unit vector in the direction of refracted ray in medium-2 is $\frac{1}{5}(a\overrightarrow{i}+b\overrightarrow{j}-\frac{5}{\sqrt{2}}\overrightarrow{k})$

To find the value of $a,b$

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 refracted angle.

Now,

$\cos {45}^{\circ }=\left|\frac{A_z}{\sqrt{A_x^2+A_y^2+A_z^2}}\right|$

For unit vector, $\left|\sqrt{A_x^2+A_y^2+A_z^2}\right|=1$

So

$\cos {45}^{\circ }=\left|\frac{A_z}{1}\right|⟹A_z=\frac{1}{\sqrt{2}}$

The other two components must be adjusted by a factor r to result in a unity magnitude:

$\left|\sqrt{r^2\left(\right(6\sqrt{3}{)}^2+\left(8\sqrt{3}{)}^2\right)+{\left(\frac{1}{\sqrt{2}}\right)}^2}\right|=1$

Take square on both sides,

$⟹r^2\left(\right(6\sqrt{3}{)}^2+\left(8\sqrt{3}{)}^2\right)+0.5=1⟹r^2\left(\right(6\sqrt{3}{)}^2+\left(8\sqrt{3}{)}^2\right)=1-0.5⟹r^2\left(300\right)=0.5⟹r^2=\frac{0.5}{300}⟹r=\frac{1}{\sqrt{600}}$

Hence the unit vector in the direction of refracted ray in medium-2 is

$(6\sqrt{3}\times \frac{1}{\sqrt{600}})\overrightarrow{i}+(8\sqrt{3}\times \frac{1}{\sqrt{600}})\overrightarrow{j}-\frac{1}{\sqrt{2}}\overrightarrow{k}$

But given this is equal to

$\frac{1}{5}(a\overrightarrow{i}+b\overrightarrow{j}-\frac{5}{\sqrt{2}}\overrightarrow{k})$

Equating i component of the two we get

$\frac{a}{5}=6\sqrt{3}\times \frac{1}{\sqrt{600}}⟹a=\frac{5\times 6\times \sqrt{3}}{\sqrt{600}}$

Multiply both numerator and denominator by $\sqrt{3}$, we get

$a=\frac{5\times 6\times 3}{\sqrt{1800}}⟹a=\frac{5\times 6\times 3}{\sqrt{9\times 2\times 100}}⟹a=\frac{5\times 6\times 3}{3\times 10\times \sqrt{2}}⟹a=\frac{3}{\sqrt{2}}$

Now, equating j component of the two we get

$\frac{b}{5}=8\sqrt{3}\times \frac{1}{\sqrt{600}}⟹b=\frac{5\times 8\times \sqrt{3}}{\sqrt{600}}$

Multiply both numerator and denominator by $\sqrt{3}$, we get

$b=\frac{5\times 8\times 3}{\sqrt{1800}}⟹b=\frac{5\times 8\times 3}{\sqrt{9\times 2\times 100}}⟹b=\frac{5\times 8\times 3}{3\times 10\times \sqrt{2}}⟹b=\frac{4}{\sqrt{2}}⟹b=2\sqrt{2}$

So the value of a and b are $\frac{3}{\sqrt{2}},2\sqrt{2}$ respectively