Question

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

Answer 1

${\mu }_1\sin {\theta }_1={\mu }_2\sin {\theta }_2$

$\cos {\theta }_1\frac{10}{\sqrt{(6\sqrt{3}{)}^2}}+(8\sqrt{3}{)}^2+100=\frac{10}{\sqrt{400}}=\frac{10}{20}$

$\cos {\theta }_1=\frac{1}{2}$

${\theta }_1={60}^o$

${\mu }_1\sin {\theta }_1={\mu }_2\sin {\theta }_2$

$\sqrt{2}\sin 60=\sqrt{3}\sin {\theta }_2$

$\sqrt{2}\frac{\sqrt{3}}{2}=\sqrt{3}\sin {\theta }_2$

$\frac{\sqrt{2\times 3}}{2}=\sqrt{3}\sin {\theta }_2$

$\sin {\theta }_2=\frac{1}{\sqrt{2}}$

${\theta }_2={45}^o$