Q.
A beam of monochromatic light reflects and refracts at point $A$, as shown in the diagram. Find the angle of refraction at point $A$.
AMUAMU 2014Ray Optics and Optical Instruments
Solution:
By Snells law
$\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}} $
${[\because i=(90-30)]} $
$\frac{\sin 60^{\circ}}{\sin r}=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}$
$\frac{\sqrt{3}}{2 \sin r}=\frac{\sqrt{3}}{1} $
$\sin r=\frac{1}{2}$
$\sin r=\sin 30^{\circ}$
$r=30^{\circ}$
So, the angle of refraction at point $A$ is $30^{\circ}$.
