Question

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

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${75}^o$

Answer 1

The correct option is B ${45}^o$

Angle of incidence with $Z$ direction (normal)

$\cos \alpha =\frac{10}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2+(10{)}^2}}=\frac{1}{2}$

$\alpha ={60}^o$.

So, ${\mu }_1\sin \alpha ={\mu }_2\sin \beta$

$\sqrt{2}\times \sin 60=\sqrt{3}\sin \beta$

$\beta ={45}^o$