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 $\overrightarrow{A}=6\sqrt{3i}+8\sqrt{3j}-10\hat{k}$ is incident on the plane of separation. If the unit vector in the direction of refracted ray in medium - 2 is $\frac{1}{5}(a\hat{i}+b\hat{j}-\frac{5}{\sqrt{2}\hat{k}})$, then find the value of ab.___

Answer 1

As incident ray, normal and refracted ray lie in same plane.
So, $\frac{a}{b}=\frac{6\sqrt{3}}{8\sqrt{3}}$
$\frac{a}{b}=\frac{3}{4}\dots \dots \left(i\right)$
and $|\hat{R}|=1$
$\sqrt{{\left(\frac{1}{5}\right)}^2(a^2+b^2+\frac{25}{2})}=1$
$a^2+b^2=\frac{25}{2}$
$a=\frac{3}{\sqrt{2}}$ and $b=\frac{4}{\sqrt{2}}$