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Snell's law practice problems with answers

In this article, we are going to answer some problems about Snell's law (helpful in AP physics 2) to learn how to solve such problems in geometric optics.

Problem (1): 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? 

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

A beam of light incident of the air-glass boundary

In this example 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. \begin{align*} n_i \sin\theta_i&=n_r\sin\theta_r \\ \\(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{align*} Now find the angle whose sine is $0.3947$ as below \begin{align*} \sin\theta_r&=0.3947 \\ \Rightarrow \theta_r&=\sin^{-1}(0.3947)\\&=23.25^\circ\end{align*} As expected. 


 

Problem (2): 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 $1.33$).

Solution: 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. 

A ray traveling from water into air

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 \begin{align*} n_i \sin\theta_i&=n_r\sin\theta_r \\ \\(1.33)\sin 35^\circ&=(1.00)\sin\theta_r \\ \\\Rightarrow \sin\theta_r&=\frac{1.33}{1.00}\sin 35^\circ\\\\&=0.7629\end{align*} The angle whose sine is $0.7629$ is found as below \begin{align*} \sin\theta_r&=0.7629 \\ \Rightarrow \theta_r&=\sin^{-1}(0.7629)\\&=49.72^\circ\end{align*} As expected. 



Problem (3): A slab of glass has an index of refraction of 1.5 and is submerged in water with $n=1.33$. A beam of monochrome light is incident on the slab and is refracted. 
(a) Find the angle of refraction if the angle of incidence is $30^\circ$. 
(b) Now, assume that the light is initially in the glass and incident on the glass-water surface. What is the refraction of the light? 

Solution: (a) Light strikes at an angle of $\theta_i=30^\circ$ in the water $n_i=1.33$ on the glass at point $A$ with $n_r=1.5$. Applying Snell's law of refraction, $n_i \sin \theta_i=n_r \sin \theta_r$, at the surface that the light enters the glass and solving for unknown refracted angle we get \begin{align*} \sin\theta_r&=\frac {n_i}{n_r}\sin \theta_i\\\\ &=\frac{1.33}{1.5}\sin 30^\circ\\\\&=0.443\end{align*} And therefore, $\theta_r=26.3^\circ$. 

(b) In the second case, the light originates from the glass with $n_i=1.5$ and strikes at an angle of $30^\circ$ on the surface of glass-water. The refracted ray would be in the water $n_r=1.33$. Using Snell's law equation, we have \begin{align*} n_i \sin\theta_i&=n_r\sin\theta_r \\ \\(1.5)\sin 30^\circ&=(1.33)\sin\theta_r \\ \\\Rightarrow \sin\theta_r&=\frac{1.5}{1.33}\sin 30^\circ\\\\&=0.563\end{align*} Thus, the angle of refraction in the water is $34.3^\circ$. 



Problem (4): A beam of light traveling in the air, strikes a flat slab of glass at an incident angle of $30^\circ$. The index of refraction of the glass is $1.5$. At the moment of leaving the glass, what is the angle of refraction? 

Solution: In this problem, we must apply Snell's law twice. One for the entering stage at the air-glass interface and the second for leaving at the glass-air surface. 

A ray through two mediums with deferent refractive indices

In the first stage, the incident beam is in the air $n_i=1.00$ and strike at point $A$ with angle of incidence $45^\circ$ and the refracted beam lies inside the glass with $n_r=1.5$ and unknown refracted angle $\theta_r$ which should be determined as below \begin{align*} n_i \sin\theta_i&=n_r\sin\theta_r \\ \\(1.00)\sin 45^\circ&=(1.5)\sin\theta_r \\ \\\Rightarrow \sin\theta_r&=\frac{1.00}{1.5}\sin 45^\circ\\\\&=0.4715\end{align*} Therefore, the first refracted angle which occurs in the slab of glass is found $\theta_r=28.13^\circ$. 

For the second stage, the beam of light inside the glass incident on the glass-air interface at point $B$ with angle of incidence of $28.13^\circ$ and refracted into the air with unknown angle $\theta'_r$ which is found as below \begin{align*} n_i \sin\theta_i&=n_r\sin\theta'_r \\ \\(1.5)\sin 28.13^\circ&=(1.00)\sin\theta'_r \\ \\\Rightarrow \sin\theta'_r&=\frac{1.5}{1.00}\sin 28.13^\circ\\\\&=0.7072\end{align*} Finally, The beam leaves the glass with an angle of refraction of $45^\circ$ which is the same angle of entering. 

When a beam of light passing through a slab of any material with uniform thickness the angle of entering (incident angle) and the leaving angle (refracted angle) are the same.



Problem (5): A uniform rectangular block of glass ($n=1.56$) surrounded by some fluid with index of refraction $n=1.63$. A beam of light incident strike at point $A$ at an angle of $32^\circ$ to the normal. At what angle does the light beam leave the slab? 

Solution: We must apply Snell's law formula twice at points $A$ and $B$ to find the refracted ray at the moment of leaving the slab. Snell's law equation is written as $n_i \sin\theta_i=n_r\sin\theta_r$ where $n_i$ and $\sin\theta_i$ are the index of refraction of the medium where the ray originates and angle of incidence, respectively. 

A block of glass surrounded by a fluid with a lower index of refraction

Similarly, $n_r$ and $\sin\theta_r$ are the indices of refraction of the medium where the ray leaves and angle of refraction, respectively. 

Therefore, to find the angle of refraction at point $A$, solve Snell's law for $\sin \theta_r$ as below \begin{align*} n_i \sin\theta_i&=n_r\sin\theta_r \\ \\(1.63)\sin 32^\circ&=(1.56)\sin\theta_r \\ \\\Rightarrow \sin\theta_r&=\frac{1.63}{1.56}\sin 32^\circ\\\\&=0.5537\end{align*} Now, take the inverse sine of both sides get the refracted angle at point $A$ \begin{align*}\theta_r&=\sin^{-1}(0.5537)\\&=33.62^\circ\end{align*} Now, the previous ray acts as incoming ray strike the point $B$ at an angle of incidence $33.62^\circ$ to the normal. This ray is inside the glass with $n_i=1.56$. Again, applying Snell's law of refraction at point $B$, we have \begin{align*} n_i \sin\theta_i&=n_r\sin\theta'_r \\ \\(1.56)\sin 33.62^\circ&=(1.63)\sin\theta'_r \\ \\\Rightarrow \sin\theta'_r&=\frac{1.56}{1.63}\sin 33.62^\circ\\\\&=0.5200\end{align*} Taking inverse of sine of both sides get \begin{align*}\theta'_r&=\sin^{-1}(0.5200)\\&=32.00^\circ\end{align*} Thus, the light beam leaves the slab with the same angle as the angle of entering to the normal and this is as expected. 



Problem (6): A beam of light incident on and deflected by a $30^\circ-60^\circ-90^\circ$ prism as shown in the figure. The deflected light makes an angle of $25^\circ$ with the direction of the incident ray. Using laws of refraction, find the prism's index of refraction. 

A beam of light incident on a prism with unknown refractive index

Solution: In this example problem, there are, in fact, two boundary surfaces. One is the front face, which the original ray incident on it with a normal angle so it enters without deflection into the prism. 

The second surface is the hypotenuse of the prism that the previous ray incident on it and the final deflection by the prism occur by it. 

An sketch of a beam passing through a prism

Since the final refraction occurs on the hypotenuse so first draw the normal on it and then from the geometry obtained, we can find the all ingredients necessary for solving Snell's law formula i.e. angle of incidence and refraction. 

As you can see from geometry, the ray's angle of incidence on the hypotenuse is the same as the prism's apex angle i.e. $\theta_1=30^\circ$.

The ray exits the prism at angle of $\theta_2$ which ,from geometry, is the sum of deflection angle $25^\circ$ and the angle of incidence $30^\circ$ i.e. $\theta_2=25^\circ+30^\circ=55^\circ$. Putting all these values and index of refraction for air as $n_2=1.00$ into Snell's law, we can then find $n_1$ for the prism. \begin{align*} n_1\sin\theta_1&=n_2\sin\theta_2 \\\\n_1(\sin 30^\circ)&=(1.00)(\sin 55^\circ)\\\\\Rightarrow n_1&=\frac{(1.00)(\sin 55^\circ)}{\sin 30^\circ}\\\\&=1.63 \end{align*} Note that to use Snell's law, we must find all angles with the normal to the boundary between two different mediums. 

Author: Ali Nemati
Date Published: 5/8/2021