Question

What would be the refractive index of the liquid in the given figure?

• A. $\sqrt{\frac{2}{3}}$
• B. $\sqrt{\frac{3}{4}}$
• C. $\sqrt{\frac{4}{3}}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is D $\sqrt{\frac{3}{2}}$

Angle of incident and the angle of refraction are the angles which the incident ray and the refracted ray make with the normal.
$\therefore$ Angle of incidence = ${90}^{\circ }-{30}^{\circ }={60}^{\circ }$
Angle of refraction = ${90}^{\circ }-{45}^{\circ }={45}^{\circ }$
So, $i={60}^{\circ }$ and $r={45}^{\circ }$
By Snell's law
$sini{\mu }_1=sinr{\mu }_2$
or $sin{60}^{\circ }\times 1=sin{45}^{\circ }\times {\mu }_2$ [$\because$ Refractive index of air is taken as 1]
or, ${\mu }_2=\frac{sin60}{sin45}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\sqrt{\frac{3}{2}}$