When light travels from one medium into another, direction of motion changes if the index of refraction of the two materials differ. Snell's Law defines how the change of direction relates to the index of refraction of the two materials.

The eye uses refraction to see. Check it out!

https://www.youtube.com/watch?v=SOqdetf1Izg

Pre-lecture Study Resources

Watch the pre-lecture videos and read through the OpenStax text before doing the pre-lecture homework or attending class.

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BoxSand Introduction


Ray Optics  |  Snell's Law of Diffraction

We've learned that light slows down as it enters a material with a higher index of refraction. For continuity of the wave, the frequency must remain constant. If $v=f \lambda$, and the frequency is constant but the speed decreases, the wavelength must also decrease.

This is a gif of the law of refraction. It shows the incident ray traveling through some medium at some angle to the vertical entering another medium as the refracted ray at some different angle to the vertical.                      This is an image of an incident ray moving through some medium at some angle theta one to the horizontal and the ray that enters the second medium is called the refracted ray that is refracted at a different angle to the vertical theta two. Some of the light is reflected back and does not enter medium two which is bounced off at some angle theta one to the vertical.

Geometrically the only way for the wavelength to decrease but the wave to remain continuous is if the angle of propagation changes. Snell's law defines the relationship between the incident angle $\theta_1$ and the refracted angle $\theta_2$ (sometimes called the transmitted angle $\theta_t$).

*NOTE: All angles are measured with respect to the normal to the surface.*

$n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$

With this relationship if light travels into a medium with a higher index of refraction (transmission), like from air into water, then $\theta_1>\theta_2$ and the light bends towards the normal. If the light instead traveled in the opposite direction, so from a higher to a lower index of refraction, the light would bend away from the normal.

Total Internal Refraction (TIR)

In the right geometry, with the right relationship of index of refraction, light can actually be forced to stay within a medium and not be refracted at all. Imagine traveling from a higher to a lower index material.

This is an image three different kinds of refraction depending on the angle of the light that enters the second medium. In all scenarios, the light travels first through water labeled n one and then through air labeled n two. The first shows the incident ray coming at some angle theta one to the vertical and is refracted at some different angle theta two into the air. The second shows the incident ray coming int at some angle theta c and the refracted light is perfectly perpendicular to the vertical and travels between the boundary. Theta c is called the critical angle. The third shows the incident ray coming in at some angle theta one to the vertical and is reflected inward back into the medium at some angle theta two. This is called total internal reflection. This is to show that at different angles, there are different kinds of reflection or refraction that can occur.

The refracted (transmitted) angle will always be larger than the incident. As $\theta_1$ increases and approaches 90º, $\theta_2$ will reach 90º before $\theta_1$ does. At the critical angle ($\theta_c$), the refracted angle is 90° and $\sin(90)=1$, so all the light is reflected and none of it is refracted.

if $n_1>n_2$,     $\sin(\theta_{c}) = \frac{n_1}{n_2}$

Fiber opticsHAM radio waves that travel around the Earth, and Lightboards work on the principle of Total Internal Refraction.


Key Equations and Infographics

A representation with the words Snell’s law of refraction on the top. There is an equation that shows that the product of the index of refraction of medium one and the sine of the incident angle is equal to the product of the index of refraction of medium two and the sine of the refracted angle. This is also written in words below.


A representation with the words total internal reflection on the top. There is an equation that shows that the sine of the critical angle is equal to the ratio the indexes of refraction. This is also written in words below.

Now, take a look at the pre-lecture reading and videos below.

OpenStax Reading


OpenStax Section 25.3  |  The Law of Refraction

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OpenStax Section 25.4  |  Total Internal Reflection

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OpenStax Section 25.6  |  Dispersion: The Rainbow and Prisms

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Additional Study Resources

Use the supplemental resources below to support your post-lecture study.

YouTube Videos


Here is a good introduction to Snell's Law and its application.

https://www.youtube.com/watch?v=Pq07YCs1R8Q

 Doc Schuster provides a geometric derivation of Snell's Law. This will be a very helpful video.

https://www.youtube.com/watch?v=twURUE075y8

Here is a quick introduction to Snell's Law

https://www.youtube.com/watch?v=K2kiSnHCax8

Here is an apparent depth example of Snell's Law

https://www.youtube.com/watch?v=30FCqf46TK8

Other Resources

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Simulations


 

Bending of light simulation by PhET to examine Snell's Law.

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Here is an easy simulation that looks at Snell's law in an easier form.

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One more Snell's law simulation to check out. This one changes the physical direction of the beam for another perspective.

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For additional simulations on this subject, visit the simulations repository.

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Demos


Kitty learns about refraction through experiment.

Kitty learns about refraction through experiment.

For additional demos involving this subject, visit the demo repository

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History


Oh no, we haven't been able to write up a history overview for this topic. If you'd like to contribute, contact the director of BoxSand, KC Walsh (walshke@oregonstate.edu).

Physics Fun

OMG it's a double rainbow!

https://www.youtube.com/watch?v=MX0D4oZwCsA

Other Resources


This resource is very direct. Consider looking through it at least once.

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The Hyperphysics reference for Snell's Law gives a quick two sentence description with a nice diagram.

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Resource Repository

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Problem Solving Guide

Use the Tips and Tricks below to support your post-lecture study.

Assumptions

 

Checklist

(1) Draw a horizontal line. This represents the boundary between the two media with different indices of refraction. Draw a dotted vertical line perpendicular to the horizontal line that bisects it. This represents the line normal to the surface of the boundary between the media.

(2) Sketch the incoming ray. Remember that the angle in Snell's Law is relative to the normal, so make sure you read the problem carefully to determine if you were given that angle or an angle relative to some other line.

(3) Use Snell's Law $$n_1 sin \left( \theta_1 \right) = n_2 sin \left( \theta_2 \right) $$ and solve for the unknown varaible (you will be given some combination of 3 of the 4 variables $n_1$, $n_2$, $\theta_1$, or $\theta_2$, and asked to solve for the fourth).

More challenging Snell's Law problems will involve solving for the index of refraction or angle of refraction for some third medium. If you are given three media instead of 2, just split the problem into two stages and do the 3 steps above twice: first for $n_1$ and $n_2$, and then for $n_2$ and $n_3$ (similar to using $v_f$ from one stage of a problem as $v_i$ in the next stage).

Misconceptions & Mistakes

 

Pro Tips

 

Multiple Representations

Multiple Representations is the concept that a physical phenomena can be expressed in different ways.

Physical

Physical Representations describes the physical phenomena of the situation in a visual way.

Mathematical

Mathematical Representation uses equation(s) to describe and analyze the situation.

A representation with the words Snell’s law of refraction on the top. There is an equation that shows that the product of the index of refraction of medium one and the sine of the incident angle is equal to the product of the index of refraction of medium two and the sine of the refracted angle. This is also written in words below.


A representation with the words total internal reflection on the top. There is an equation that shows that the sine of the critical angle is equal to the ratio the indexes of refraction. This is also written in words below.

Graphical

Graphical Representation describes the situation through use of plots and graphs.

 

Descriptive

Descriptive Representation describes the physical phenomena with words and annotations.

 

Experimental

Experimental Representation examines a physical phenomena through observations and data measurement.

Here is an excellent experiment to verify the laws of refraction and determine the refractive index of glass using Snell's law.

https://www.youtube.com/watch?v=CdjjDsb5Fz8

Practice

Use the practice problem sets below to strengthen your knowledge of this topic.

Fundamental examples

 (1) Light is traveling through air ($n = 1$) and is incident on a plastic medium at an angle of incidence of $\theta_{inc} = 40 ° $. It enters the plastic medium at an angle of inidence of $\theta_{refr} =  30 °$. What is the index of refraction of the second medium?

(2) Light travels from a medium with index of refraction $n = 2$ into a medium with an index of refraction of $n = 46$. The light crosses the boundary between mediums at an angle of $\theta =30 °$ from the horizontal. What is the angle of refraction?

(3) A piece of glass is floating on a liquid in a beaker that is open to the air. Light is incident on the piece of glass at an angle $60 °$ relative to the normal. The index of refraction of the glass is $n_g = 2.5$ and the index of refraction of the liquid is $n_l = 3$. At what angle relative to the normal does the light enter the liquid?

(4) You are standing above a glass water tank that is 1 meter deep with a red laser pointer, and you aim it down at the tank at a very shallow angle - 80 degrees from the normal. The laser exits the far wall of the tank before it hits the bottom of the tank. The horizontal distance between the laser's entry into the water and its collision with the far wall of the tank is $50$ cm. At what depth does the laser exit the tank? 

Solutions found HERE.

Short foundation building questions, often used as clicker questions, can be found in the clicker questions repository for this subject.

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