## Snell’s Law of Refraction

Snell’s law applies when the phenomenon of **refraction** occurs.

**Refraction is **the process of how waves change direction when they travel between different mediums in which light propagates at different speeds.

These mediums have different refractive indices. The refractive index of a medium, \( n \), is given by

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

where \( c \) is the speed of light in vacuum and \( v \) is the speed of light in the medium, both measured in meters per second \( (\mathrm{m/s}) \). As the units of the numerator and denominator in the refraction index equation are the same, the value for \( n \) must be unitless —it is a ratio. For example, the refraction index for air is approximately 1, and for water, is 1.33. The refractive index of a medium tells you how fast light travels in it: the higher the value of \( n \), the slower light travels.

## Snell’s Law Derivation

Fig. 1. A diagram showing a light ray refracting at a boundary between two mediums.

The diagram above shows the path of a light ray that moves from a medium with refractive index \( n_1 \) to one with refractive index \( n_2 \), where \( n_2 > n_1 \), and so the ray bends towards the normal. You might have already come across the equation that relates the speed of light to the wavelength and frequency of a wave:

$$c=v\lambda,$$

where \( c \) and \( v \) represent the same quantities as in the first equation, and \( \lambda \) is the wavelength of the light measured in meters.

When a light ray moves from one material to another, its frequency does not change, but its speed does, so this means that its wavelength must change. In the situation shown above, as light moves from the first medium to the next, its speed decreases, and so the wavelength must also have decreased. The diagram shows the wavefronts for the incoming and outgoing light rays. You can see that they are more widely spaced after crossing the boundary. We can use trigonometry along with the equations previously stated to prove Snell's Law.

Some of the useful distances and angles are labeled on the diagram. \( A \) and \( B \) represent distances that the incoming and outgoing rays travel in a certain time period (they could represent the wavelengths, and this will lead to the same result). They are related to the speeds of light in the different mediums by:

\[ A = v_1 T \]

\[ B = v_2 T\]

In these equations, \( v_1 \) and \( v_2 \) represent the speeds of light in mediums \( 1 \) and \( 2 \), respectively, and \( T \) is an arbitrary time period measured in seconds.

Fig. 2. Two triangles can be constructed from the wavefronts on either side of the boundary.

We can identify two right triangles in the boundary. Then, using trigonometry, we can relate the angles \( \theta_1 \) and \( \theta_2 \) to the known sides.

$$\sin(\theta_1)=\frac AC$$

$$\sin(\theta_2)=\frac BC$$

Hence:

$$\frac{\sin(\theta_1)}{\sin(\theta_2)}=\frac AB$$

$$\frac{\sin(\theta_1)}{\sin(\theta_2)}=\frac{v_1\mathrm T}{v_2\mathrm T}$$

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

Then, we can express the velocities in each medium in terms of their refractive index by using the definition of the refractive index.

$$ v=\frac cn \Rightarrow \frac{v_1}{v_2}=\frac{\displaystyle\frac c{n_1}}{\displaystyle\frac c{n_2}}=\frac{n_2}{n_1}$$

This leads to the expression:

$$\frac{\sin(\theta_1)}{\sin(\theta_2)}=\frac{n_2}{n_1},$$

which can be rearranged to give Snell's Law:

$$n_1\sin(\theta_1)=n_2\sin(\theta_2).$$

## Snell's Law Equation

As derived above, Snell's Law is

$$n_1\sin(\theta_1)=n_2\sin(\theta_2),$$

where \( n_1 \) is the refractive index of the material on the side of the boundary where the light ray starts and \( n_2 \) is the refractive index of the second material. The diagram shown below illustrates what the angles in the equation represent. Snell's Law can be used to find these quantities when light is incident on different boundaries. The procedure for doing this is explained in the next section.

Fig. 3. Snell's law describes how light refracts at a boundary.

The diagram above is called a **ray diagram,**** **and it has several key features to remember:

The line perpendicular to the surface of the new medium is called the

**normal**. If the incident ray is along the normal, no refraction will occur, and the refracted ray will also be along the normal.The angle between the incoming light ray and the normal is known as the

**angle of incidence**.The angle between the outgoing ray and the normal on the opposite side of the boundary is known as the

**angle of refraction**.

## Snell’s Law Procedure

Snell’s Law can be used in many problems to work out how a light ray will behave upon striking a boundary between two mediums of different refractive indices. There is a useful way of checking your calculations when working with Snell’s Law. Consider a car driving from a smooth road across a boundary into a muddy field. The car will move slower in the field than on the road. If the car were to move into the field at an angle, the side of the car entering the field first would move slower than the side still on the road, which would cause the car to turn in the direction of the slower-moving side until both sides move at the same speed. We can think of a light ray in the same way:

If a light ray moves from a medium with a lower refractive index to a higher one, then the path of the ray will bend

**towards**the normal (the light ray moves slower in the higher refractive index medium, just like how the car moves slower in the field).If a light ray moves from a medium with a higher refractive index to a lower one, then the path of the ray will bend

**away**from the normal.

A light ray is incident on a glass block. The angle of incidence is \( 45^\circ \). What is the angle of refraction? The refractive indices of air and glass are \( 1 \) and \( 1.5 \), respectively.

Before you make your calculation, think about if the light ray will bend towards or away from the normal.

This is a problem in which we utilize Snell’s Law:

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

We can plug in the values as labeled in the figure:

$$\sin(45)=1.5\sin(x)$$

This expression can be rearranged to find \( x\):

$$x=\sin^{-1}(\frac{\sin(45)}{1.5})=28^\circ.$$

## Snell's Law Example

A ray of light is incident on a triangular-shaped prism, as shown in the diagram below. The angle of incidence on the first face is \( 45^\circ \), and the angle at the top of the prism is \( 40^\circ \). What is the angle of refraction from the second face of the prism (when the ray emerges from the prism into the air)? Consider that the prism is made from the same glass as in the previous example.

This problem can seem complicated at first, but all that is needed is to use trigonometry and two applications of Snell’s Law. It is helpful to label the angles that you think might be needed, as in the figure above. The angle that we want to find is the angle of refraction from the second face – this is labeled with the letter \( y \). The angle \( x \) is in fact, the same as in the first example. It can be found through the use of Snell’s Law for the light ray incident on the first boundary between the glass and the prism:

$$\sin(45)=1.5\sin(x)$$

$$x=\sin^{-1}(\frac{\sin(45)}{1.5})=28^\circ$$

The angles \( x \) and \( a \) make up the right angle between the normal and the boundary, so they add to 90 degrees, and angle \( a \) can be found:

$$a=90-x=62^\circ.$$

The angle at the top vertex of the triangle, A, is \( 40^\circ \). It can be seen that this vertex also forms a triangle with the points where the light rays meet the two boundaries. We know that the angles in a triangle add to \( 180^\circ \), and angle \( a \) has just been found, so now angle \( b \) can be computed:

$$40+a+b=180$$

$$b=140-a=78^\circ$$

Once again, angles \( b \) and \( c \) add together to form the right angle between the normal and the boundary on the right-hand side of the prism, so \( c \) can be found:

$$90-b=c=12^\circ.$$

Finally, Snell’s Law needs to be applied again in order to find the angle \( y \), which is the angle of refraction when the light ray leaves the prism:

$$1.5\sin(12)=\sin(y)$$

$$y=\sin^{-1}(\frac{\sin(12)}{1.5})=8^\circ$$

There is a very particular and interesting case of the refracting angle that we can explore using Snell's Law. If we increase the incident angle of a light ray striking a boundary, the angle of refraction will increase as well, until it is greater than \( 90^\circ \). Then, all of the light is reflected instead of moving out of the first medium —this is called **total internal reflection**. This can only occur when the light ray bends away from the normal —when it moves from a lower refractive index material to a higher refractive index material.

**Total internal reflection** is the phenomenon of when light is completely reflected from a boundary when its angle of incidence is large enough.

Snell's Law is

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

It can be rearranged to

$$\frac{\sin(\theta_1)}{\sin(\theta_2)}=\frac{n_2}{n_1}$$

We want to find the angle that the light is refracted along the boundary, called the **critical angle**.

The **critical angle** for a boundary is the incident angle for a light ray at which it is refracted along the boundary line.

When light is reflected along the boundary, the refracted angle is \( 90^\circ \), so

$$\sin(\theta_2)=\sin(90)=1$$

and our equation becomes

$$\sin(\theta_c)=\frac{n_2}{n_1},$$

where \( \theta_c \) is the critical angle. This can be rearranged to find the critical angle:

$$\theta_c=\sin^{-1}(\frac{n_2}{n_1}).$$

Total internal reflection occurs when the angle of incidence is greater than the critical angle. The phenomenon of total internal reflection is used in many different applications, such as sending information down optical fibers and observing minuscule objects through microscopes!

Sending information using optical fibers is more efficient than traditional copper wires, as optical fibers can carry more information with less loss in the signal. These fibers allow transferring around 1 Gigabyte of data per second! It is amazing what we can achieve by understanding Snell's Law!

## Snell's Law - Key takeaways

Refraction is

The refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium.

Snell’s Law relates the refractive indices of two mediums on either side of their boundary and also the angles of incidence and refraction.

Snell’s Law can be derived by considering the distance between wavefronts for either side of a boundary that a light ray has crossed.

If a light ray strikes a boundary parallel to the normal, no refraction will occur and the ray will continue along the normal.

A light ray moving from a higher refractive index medium to a lower one will bend away from the normal.

A light ray moving from a lower refractive index medium to a higher one will bend towards the normal.

Total internal reflection is the phenomenon of when light is completely reflected from a boundary when its angle of incidence is large enough.

The critical angle for a boundary is the incident angle for a light ray at which it is refracted along the boundary line.

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##### Frequently Asked Questions about Snell's law

What is Snell's Law?

Snell's Law is the relation between the refractive indices of the two mediums and the angles of incidence and refraction.

How can you prove Snell's law?

You can prove Snell's Law by considering the wavefronts before and after the boundary. The distance that the wavelengths travel in a certain time can be used to arrive at the correct equation.

What does Snell's Law state?

Snell's Law states that a light ray moves through a boundary and gets refracted in such a way that the product of the refractive index of the first medium times the sine of the incident angle equals the product of the refractive index of the second medium times the sine of the refracted angle. We can also express this as n1sin(incident angle)=n2sin(refracted angle).

What are n1 and n2 in Snell's Law?

In Snell's Law, n1 and n2 are the refractive indices of the materials on either side of a boundary that the light ray is passing through.

When is Snell's Law not valid?

Snell's Law is not valid when the angle of incidence is zero degrees as the angle of refraction will be zero as well.

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