# Snell's Law

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Snell's Law
Field: Geometry
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Snell's Law

This is a picture of a spoon in a glass of water that seems to be bent. Snell's Law is a mathematical formula that predicts the amount of bend seen in the image.

# Basic Description

In order to understand the concept of Snell's Law, it is important to understand the behavior of light. Optics is the study of light and its behavior. We will narrow the topic of optics to geometrical optics where light travels in a straight line called a ray.

It is essential to know that there are other ways to understand light, even though we will only focus on the nature of light based on geometrical optics. For instance, in physical optics light behaves more so as a wave. Physical optics will not be discussed on this page. Light can act like a particle. Light exhibits properties that are very interesting, but still puzzling.

Geometrical optics requires basic knowledge of geometry and triangles. It is clear that optics and Snell's Law is a topic usually found in physics and electrical engineering, but these topics involve phenomena that can be interpreted through mathematical relationships.

## Reflection of Light

Again, for geometrical optics we treat light as a ray. When a ray strikes a surface it is an incident ray. The angle that the incident ray makes with the normal is called the angle of incidence.

The ray that is reflected by that surface is simply called the reflected ray. The reflected ray also makes an angle of reflection with the normal.

The angle of reflection that the reflected ray makes is the same as the angle of incidence made by the indicated ray. Refer to Image1. Rays can reflect in two ways. First, rays can perform specular reflection and this occurs when a ray strikes a smooth surface and is reflected without distortion (look at Image 2). Second, rays can perform diffuse reflection where a ray strikes a rough surface and reflects in different directions (look at Image 3).

A commonality between Image 1 and Image 2 is the fact that each set of ingoing and outgoing rays are similar to one another. In Image 2 the ingoing and outgoing rays are all the same but their starting point has shifted. In Image 3 this is not the case. In diffuse reflection individual incident rays reflect at different angles from one another. However, each ray still exhibits the fact that the angle of reflection is the same as the angle of incidence.

 Image 1. Angle of incidence and reflection are equal. Image 2. Specular Reflection Image 3. Diffuse Reflection

## Refraction of Light

Light has a specific speed in a vacuum. It is 3.0 x 10 8 m/s and is labeled as $c$. The speed of light differs when it travels through another medium, a substance that is not a vacuum. The ratio of the speed of light in a vacuum, $c$, and the speed of light in a medium, $v$, is the index of refraction , $n$.

So, $n = \frac {c} {v}$.

When light travels from one medium with a given index of refraction to another medium at an angle, the speed of light changes. This change in speed causes a change in direction of the ray of light. The wavelength shortens in second medium, which is not a vacuum. However, the cycle of waves per unit time (frequency) does not change.

Refraction is the change the incident ray undergoes from its original direction. It results from the change of speed. From refraction, a refracted ray is produced as well as an angle of refraction. Snell's law tell us what will happen at a surface given the indices of refraction and media.

Here are some indices of refraction. Substances that are gases, such as air and carbon dioxide, have an index of refraction,$n$, of about 1. Water has an index refractive of 1.33 and glass has an index of refraction of 1.5 (There are many types of glass and because of this glass ranges from 1.5-1.9. We will let glass be 1.5). A diamond has an index refraction of 2.4.

Snell's law works when light goes from a medium of lower to higher refractive index or conversely. We can actually tell based on how a ray bends if light goes from a medium of lower to higher refractive index or vice versa. For instance, if the ray bends toward the normal then it was coming from a medium with a lower refractive index and going to a medium with a higher refractive index. If the ray bends away from the normal, then the ray was coming from a medium with a higher refractive index and going to a medium with a lower refractive index. Image 4 on the left shows the second medium being higher than the first and as a result the ray bends toward the normal.

Image 4.

## Definition of Snell's Law

Snell's Law allows us to have a sense of what will happen to light at a interface given the indices of refraction and two media. Snell's law gives the relationship between the angle of incidence, angle of refraction, and the indices of refraction for different media.

Given two media with specific indices of refraction, $n_1$ and $n_2$, an angle of incidence $\theta_1$ and an angle of refraction $\theta_2$, Snell's Law states:

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

### Snell's Law Applet

See the effect of Snell's Law. Change the refraction of index for the top or bottom surface. Click on the letter A or E and move it around. The angle of incidence and refraction will be given. The concept of total internal refraction (this will be discussed in the Why it's Interesting section) will be introduced.

Alternatively -- there is a flash version of this applet for you to work with:

 If you can see this message, you do not have the Java software required to view the applet.

# A More Mathematical Explanation

Given below are two methods for deriving Snell's Law, additional information, and problems that you c [...]

Given below are two methods for deriving Snell's Law, additional information, and problems that you can solve. The first derivation requires basic geometry and trigonometry. The second derivation requires knowledge in calculus especially practice with derivatives.

## Deriving Snell's Law with Refraction

When light enters a new medium, the wave speed changes. Speed is based on how much distance is traveled in an amount of time. Think about a car, we usually say the speed is 60 miles (distance) per hour (time). Remember that "per" in math means division.

For a wave, we specify the distance as the wavelength $\lambda$ and the time as a period $p$. So the wave speed is the ratio of the wavelength and the period.

$v = \frac{\lambda} {p}$.

The period is the duration of one cycle. Frequency on the other hand is the number of how often the cycles repeat per unit time. The period is the reciprocal of frequency:

$p = \frac{1} {f}$.

So we can say that a wave speed is the product of the wavelength and the frequency. The wave speed is:

$v = \lambda f$

Wave frequency does not change as the wave goes from one medium to another, but the wavelength does change.

Refer to Image 5.

Image 5. Frequency stays the same. Light slows down because wavelength changes.

The wavelength in a medium is:

$\lambda = \frac{v} {f}$.

It was mentioned before that index of refraction was equal to the ratio of the speed of light in a vacuum and the velocity in a particular medium.

$n = \frac{c} {v}$.

So, the wavelength is:

$v = \lambda f = \frac{c} {nf}$.

The image below shows how refraction occurs from a change in wave speed by the wavelength. Medium 2 has a greater index of refraction than medium 1. As a result the wavelength is shorter.

Image 6.

Image 6 shows two triangles that share the hypotenuse (the common side shaded red). The short sides are labeled as the wavelength in the media. In each case the hypotenuse is given by the appropriate wavelength over the sine of the angle of incidence and refraction.

In general the hypotenuse is: $\frac{\lambda} {\sin \theta}$.

Remember there are two triangles in different media so we will label one with the subscript of one and the other with a subscript of two.

Solving for the common hypotenuse gives:

$\frac{\lambda_1} {\sin \theta_1} = \frac{\lambda_2} {\sin \theta_2}$

Rewrite the expression using multiplication.

$\lambda_1 * \frac {1} {\sin \theta_1} = \lambda_2 * \frac{1} {\sin \theta_2}$

Substitute $\lambda = \frac{c} {nf}$

$\frac{c} {n_1 f} * \frac{1} {\sin \theta_1} = \frac{c} {n_2 f} * \frac{1} {\sin \theta_2}$

Multiply both side by $\frac{f} {c}$.

$\frac {1} {n_1 \sin \theta_1} = \frac {1} {n_2 \sin \theta_2}$

We can rewrite the expression as such:

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

This is Snell's Law.

## Snell Law's on the Beach

This is a famous problem that relies on Snell's Law. The following problem will help introduce an important concept to dervide Snell's Law and it is not presented as a computational problem for you to solve. It is a real world situation that Snell's Law can solve. Problems, that you might work on to develop and test your understanding will be provided later on.

It's summertime and you work as a lifeguard. On your shift you become aware of a drowning swimmer in the water. You must save the swimmer by reaching the swimmer in as little time as possible. Since you can run faster on sand than you can swim in water, it makes sense that you should cover more distance in the sand than in the water. So you decide that you won't directly run at the drowning swimmer. Instead, you think of the best path that will get you to swimmer in the least amount of time.

Snell's Law can help you determine the optimal entry point into the water so you can reach the drowning swimmer in the least amount of time. Pierre de Fermat a French mathematician in the 17th century figured that light seeks to minimize time of travel between two points rather than distance. This came to be known as Fermat's Principle which can be starting point in deriving Snell's Law (see below). Fermat's principle is another way to understand the solution to the lifeguard problem.

The image below maps out the most ideal paths one would take to reach the drowning swimmer.

We can get Snell's Law by using Fermat's Time Principle. Snell's Law will be derived based on Image 7.

Image 7

Recall that distance equals rate multiplied by time $\left ( d = r t \right )$. Therefore time is the ratio of distance and rate. In our case we rename rate as velocity, "v", and rewrite the relationship so that we are solving for time. Time is equal to distance over velocity:

$t = \frac{d} {v}$

To get the time needed to travel from point A to point B, we need the total distance and the velocity.

Velocity though is easy. We explained before that velocity relates the speed of light and the index of refraction in any particular medium. $v = \frac{c} {n}$. The velocity above the horizontal line is different from the velocity below the line. Let $v_1$ be the velocity above the line and $v_2$ be the velocity below the line.

Let's focus on the distance. Using the two right triangles that can be made with the light ray and the medium surface, we can calculate the distance for the path from point A to point B. The total distance will include the distance from point A to the medium interface and the distance from the medium interface to point B. We look at the hypotenuse of the triangle.

The distance from point A to the medium interface is:

$\sqrt{x^2 + h_1 ^2}$

The distance from the medium interface to point B is:

$\sqrt{h_2 ^2 + \left ( w-x \right ) ^2}$

Using these expressions as the distances in the equation for time, we can find the total time from point A to point B. It is:

$t \left ( x \right ) = \frac{\sqrt{x^2 + h_1 ^2}} {v_1} + \frac{\sqrt{h_2 ^2 + \left ( w-x \right ) ^2}} {v_2}$

To minimize the time we take the first derivative and set it equal to 0. We need to specify that we are taking the derivative with respect to the variable $x$.

$t ' \left ( x \right ) = \frac{1} {2} \frac{x} {v_1 \sqrt{x^2 + h_1 ^2}} - \frac{1} {2} \frac{w-x} {v_2 \sqrt{h_2 ^2 + \left ( w-x \right ) ^2}} = 0$

Multiply both sides by $2$ to get rid of the $\frac {1} {2}$.

$t' \left ( x \right ) = \frac{x} {v_1 \sqrt{x^2 + h_1 ^2}} - \frac{w-x} {v_2 \sqrt{h_2 ^2 + \left ( w-x \right ) ^2}} = 0$

If we look at Image 7 (take another look below) we notice that in the expressions above is the ratio of the opposite side and the hypotenuse, which we know is sine. Sine is the ratio of the opposite side and the hypotenuse.

Image 7

Given the angle, $\theta_1$ found near the point $A$, the opposite side length is $x$. The length of the hypotenuse equals the square root of the sum of the squares of the lengths of the other two sides. This turns out to be $\sqrt{x^2 + h_1 ^2}$. From all the given information we can rewrite all the expression using sine and $\theta_1$. We get:

$\frac{x} {\sqrt{x^2 + h_1 ^2}} = \sin \theta_1$

Given the angle, $\theta_2$ found near the point $B$, the opposite side length is the difference between $w$ and $x$ or written as such: $w-x$. The length of the hypotenuse turns out to be $\sqrt{h_2 ^2 + \left ( w-x \right ) }$. From all the given information we can rewrite all the expression using sine and $\theta_2$. We get:

$\frac{w-x} {\sqrt{h_2 ^2 + \left ( w-x \right ) ^2}} = \sin \theta_2$

Rewrite the expression involving $t ' \left ( x \right )$ using sine.

Remember that the expression was:

$t ' \left ( x \right ) = \frac{x} {v_1 \sqrt{x^2 + h_1 ^2}} - \frac{w-x} {v_2 \sqrt{h_2 ^2 + \left ( w-x \right ) ^2}} = 0$

When we rewrite it using sine the expression is:

$t ' \left ( x \right ) = \frac{\sin \theta_1} {v_1} - \frac{\sin \theta_2} {v_2} = 0$

This gives the desired:

$\frac{\sin \theta_1} {v_1} = \frac{\sin \theta_2} {v_2}$

Now note that we already established what the velocity is in terms of the speed of light in a vacuum in a medium.

Substitute $v_1 = \frac{c} {n_1}$ and $v_2 = \frac{c} {n_2}$.

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

Rewrite the expression.

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

Multiply both sides by $\frac {1} {c}$.

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

And, we have Snell's Law derived from Fermat's Principle.

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

## Practice Problems

Here are some exercises to develop and practice your understanding of the concepts involved in Snell's Law. The explanations above should provide enough information to answer all the questions. The answers are available through a link at the end.

Problem 1: You walk into your hotel lobby and see two mirrors that are perpendicular to each other. Your friend pulls out his cellphone that has a bright light and point it in the direction of one of the mirrors. The ingoing light from the cellphone and the outgoing light from the other mirror are parallel. Will this always be the case for any ray of light, why or why not? Look at Image 8 for a hint.

Image 8

Problem 2: Rank the media according to their indices of refraction in Image 9.

Image 9

Problem 3: Light traveling through glass (n=1.5) has an angle of incidence of 30o. What is the angle of refraction?

Below is a sketch of what happens in problem 1. We want to show that ingoing and outgoing rays are parallel or not. The rays are going in the opposite direction so if they are parallel, one will be turned 180 degrees from the other. The ingoing ray reflects-off the edge of the mirror. It turns at an angle that is $180^o - 2 \alpha$. When the ray hits the second mirror a second reflection occurs. The second reflection turns through an angle that is $180^o - 2 \theta$.

So the ray turns through an angle that totals to: $\left ( 180 ^o - 2 \alpha \right ) + \left ( 180 ^o - 2 \theta \right )$.

We can rewrite it as such:

$360^o - 2 \left ( \alpha + \theta \right )$

In the image there is a triangle. We can now relate $\alpha$ and $\theta$.

$\alpha + \theta = 90^o$

If we substitute the relationship we found with $\alpha$ and $\theta$ into

$360^o - 2 \left ( \alpha + \theta \right )$

Then, we get:

$360^o - 2 \left ( 90^o \right )$
$360^o - 180^o$
$180^o$

This is what we wanted to prove. The light turns a total of 180o. So it will always be true that any ray of light in this situation. The ingoing and outgoing light are parallel.

Image 10

Answer to Problem 2: $n_3 > n_1 > n_2$

In this problem, you want to see which of the rays bend toward the normal the most. You can think of it as which of the lines is closest to a vertical position. In this case, the ray in medium three is the closest, then it's the ray in medium 1 and finally the ray in medium 2.

This was in fact a plug and chug problem. So, anyone who likes to crunch numbers really would like this problem. The answer is approximately 19.5o.

Snell's Law states that:

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

Light leaves the air. Air has an index of refraction of 1. So,

$n_1 = 1$

We are given the angle of incidence which is 30o. Light travels through glass which has an index of refraction of $1.5$. With all this information we solve for $\theta_2$.

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

We plug everything in the equation above.

$1 * \sin 30 = 1.5 * \sin \theta_2$

Recall, $\sin 30 = .5$

$.5 = 1.5 * \sin \theta_2$

Divide both side by $1.5$.

$\frac{.5} {1.5} = \sin \theta_2$

Take the inverse of sine to solve for theta.

$\arcsin \left ( \frac{.5} {1.5} \right ) = \theta_2$

Simplify.

$\arcsin \left ( .3333 \right ) = \theta_2$

Solve for theta.

$19.471 = \theta_2$

The angle of refraction is about 19.5o if we round to the tenths place.

# Why It's Interesting

The concept of Snell's Law is very important in the real world. Snell's Law has a lot to do with you making long distance telephone calls, watching television and surfing the web.

Above it was mentioned that light bends away from the normal if it travels from a higher medium to a lower medium. Therefore, the angle of refraction is greater than the angle of incidence.

As the angle of incidence increases, so does the angle of refraction. At some angle of incidence, known as the critical angle, light traveling from a higher medium to a lower medium (or vice versa) will be refracted at 90°. What happens when the angle of refraction is 90 degrees or more? More importantly, what does this mean visually?

At this stage, total internal reflection occurs. It is an optical phenomenon that happens when a ray of light strikes a medium and the angle of refraction is 90 degrees or larger. The angle of incidence is at its critical angle. There is no set value for the critical angles, becasue there are so many objects with specific indices of refraction. In the case of air and water the critical angle would be 48.6 degrees. During total internal reflection, a light that leaves the first medium can pass through the other medium. The light is then reflected within the first medium.

Let's first test this out the concept of a critical angle and total internal reflection by using the equation of Snell's Law and then interpreting what exactly will happen. The angle of refraction $\theta_2$ is 90°. It's important to remember that $\sin 90 = 1$ also, remember that sine of any number greater than 90° does not exist. The critical angle (denoted as such $\theta_c$) is then $\theta_2$. Thus, we have the following:

State Snell's Law.

$n_1 \sin \theta_1 = n2 \sin \theta_2$

The angle of refraction $\theta_2$ is 90°

$n_1 \sin \theta_1 = n2 \sin 90$

Recall: $\sin 90 = 1$

$n_1 \sin \theta_1 = n2$

Solve for the critical angle.

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

The resulting value of $\theta_1$ is equal to the critical angle $\theta_c$.

So:

$\sin \theta_c = \frac{n_1}{n_2}$
The expression above means that a ray of light traveling at a critical angle from the first medium will not penetrated the second medium. No light is transmitted. The light will be reflected back into the first medium. The process occurring is called total internal reflection.

Total internal reflection is used in binoculars. Glass, if oriented appropriately, can be a perfect reflector. Binoculars have glass prisms that reflect internally so light can provide a longer path and the observer can see at a farther distance.

But more importantly, total internal reflection is used in optic fiber cables. Optic fiber cables are thin glass structures that carry digital information (this is made out of light) from a long distance to your television, cellphone etc. The light travels along the cables by reflecting off the walls. The light hit an angle greater than the critical angle so that all light is reflected back into the cable and is never lost.

Besides, total internal reflection, we see the concepts of Snell's law in mirages. Mirages are optical phenomena that result from refracted light. Usually people think of deserts, when mirages are mentioned. In the desert, a mirage gives the appearance that a large body of water is distant. In reality, an image of the sky is being refracted from the warm air above the sand.

On a hot day, the surface of a road may appear wet, oily, or shimmery. This is an example of a mirage that most people see. What you are actually seeing is refracted light from the sky. Light passes through through many media and refracts at each medium. This is the wonder of the property of light and the connection with Snell's Law.

## Refraction and Images

Snell's law is actually the underlying principal behind anything that uses lenses (your eyes, glasses, cameras, microscopes, etc). We can determine the nature (meaning the location and size of an object) of an image just by following light rays starting at the object that either strikes a lens.

Snell' Law explains the formation of images and the relationship they have with mirrors. In general, there are two types of images. First, there is a real image, where light converges at a point. The other type of image is a virtual image, where light appears to converge to a point.

A mathematical equation that relates the image distance, the image height, the object's height and the distance from the object to mirror. This can help explain why we see things the way we do. For instance, if you wanted to buy a long mirror, you might wonder if the mirror has to touch the floor in order for you to see yourself form head to toes. A variation of Snell's Law would tell you that the mirror only has to be half the height of your eyes above the floor. Snell's Law is important in many things that you would never have noticed.

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