Question

A laser beam in air is incident on a liquid at an angle of ${53}^{\circ }$ with respect to the normal. The laser beam's angle in the liquid is ${35}^{\circ }$. What is the liquid's index of refraction?

Short answer

Liquid's index of refraction,$n_2=1.37$

Step-by-step solution

Snell's law

It connects the incident angle ${\theta }_1$to the refraction angle ${\theta }_2$, moreover because the incident medium's $n_1$and refracted medium's $n_2$indexes of refraction.

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Angles of incidence and refraction are measured in regard to the conventional line where the ray intersects the two media's interface. Medium $2$is more optically thick (has a better refractive index) than medium $1$if the refracted ray is closer to the traditional than the incident ray. Within the same plane because the incident ray, the refracted ray, and also the normal line.

Given data

The angle of incidence of the beam is ${\theta }_1=37°$

${\theta }_2=26°$is that the laser beam's angle of refraction

The index of refraction of air is $n_1=1.00$

To obtain n2, rearrange Snell's law from equation.

$n_2=\frac{n_1\sin {\theta }_1}{\sin {\theta }_2}$

Put the values,

$n_2=\frac{(1.00)\left(\sin 37°\right)}{\sin 26°}$

$=1.37$