Question

A beam of flashlight traveling in air incident on a surface of a thin glass at an angle of \(38^{\circ}\) with the normal. The index of refraction of the glass is \(1.56\). What is the angle of refraction?

Answer 1

When a beam of light strikes the boundary of two different media such as air-glass, part of it is reflected, and another part is refracted. That part that enters on the other side of the boundary is called refracted ray. The angle that this ray makes with the vertical to the boundary is also called the angle of refraction.

In this problem, the light is initially in the air with an index of refraction \(n_{i}=1.00\) and strikes the boundary surface separating air and glass at \(\theta_{i}=38^{\circ}\). This is the angle of incidence. The subscript \(i\) denotes the incident.

Another different medium is glass with \(n=1.56\). The refracted ray lies in it with an unknown angle \(\theta_{r}=?\) which should be found using Snell's law of refraction.

Before going further and solve the problem, we expect that since the light beam enters from a low index of refraction medium into a one with a high index of refraction so the refracted ray should be bent toward the normal.

By applying Snell's law, we will check this claim.
\[
n_{i} \sin \theta_{i}=n_{r} \sin \theta_{r}
\]
\[
\begin{aligned}
(1.00) \sin 38^{\circ} &=(1.56) \sin \theta_{r} \\
\Rightarrow \sin \theta_{r} &=\frac{1.00}{1.56} \sin 38^{\circ} \\
&=0.3947
\end{aligned}
\]
Now find the angle whose sine is \(0.3947\) as below
\[
\begin{aligned}
\sin \theta_{r} &=0.3947 \\
\Rightarrow \theta_{r} &=\sin ^{-1}(0.3947) \\
&=23.25^{\circ}
\end{aligned}
\]
As expected.