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A boy is in a pool and shines a flashlight toward the level of it at a \(35^{\circ}\) angle to the vertical. At what angle does the flashlight beam leave the pool? (the index of refraction of glass is 1.33).
in Physics & Chemistry by Platinum (131,394 points) | 563 views

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As before, a beam of light strikes the surface boundary of two different media so to find the refracted angle in the air we must use Snell's law.

snell law 2

Contrary to the previous problems, here the light beam initially traveling in glass and entering the air with a lower index of refraction so we expect that the angle of refraction in the air bent away from the vertical.

In other words, we expect that the angle of refraction is greater than angle of incidence i.e. \(\theta_{r}>\theta_{i}\). Now, we calculate it by applying Snell's law formula as below
\(n_{i} \sin \theta_{i}=n_{r} \sin \theta_{r}\)
\((1.33) \sin 35^{\circ}=(1.00) \sin \theta_{r}\)
\[
\begin{aligned}
\Rightarrow \sin \theta_{r} &=\frac{1.33}{1.00} \sin 35^{\circ} \\
&=0.7629
\end{aligned}
\]

The angle whose sine is \(0.7629\) is found as below
\[
\begin{aligned}
\sin \theta_{r} &=0.7629 \\
\Rightarrow \theta_{r} &=\sin ^{-1}(0.7629) \\
&=49.72^{\circ}
\end{aligned}
\]
As expected.

 

by Platinum (131,394 points)

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