Now that we know that the degree of bending, or the angle of refraction, is dependent on the
refractive index of a medium, how do we calculate the angle of refraction? The answer to this
question was discovered by a Dutch physicist called Willebrord Snell in 1621 and is now called
Snell's Law or the Law of Refraction.

Snell's law

\[\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}\]
where
\[\begin{array}{rl}
n_1 &= \text{Refractive index of material 1} \\
n_2 &= \text{Refractive index of material 2} \\
\theta_1 &= \text{Angle of incidence} \\
\theta_2 &= \text{Angle of refraction} \\
\end{array}\]

Remember that angles of incidence and refraction are measured from the normal, which is an imaginary
line perpendicular to the surface.

Suppose we have two media with refractive indices \(n_1\) and \(n_2\). A light ray is incident on the
surface between these materials with an angle of incidence \(\theta_1\). The
refracted ray that passes through the second medium will have an angle of refraction
\(\theta_2\).

Snell never published his discovery of the Law of Refraction. His work was actually published
by another prominent physicist of the time, Christiaan Huygens, who gave credit to
Snell.

A recommended project for formal assessment on verifying Snell's laws and determining the
refractive index of an unknown transparent solid should be done. Learners will need a
glass block, ray box, 0-360 protractor, a glass block, a transparent block of an unknown
material and A4 paper. This project is given as two formal experiments, which learners
can complete and then write up as a project. The project should include some background
information on Snell's law, all the steps taken in the experiments, key experimental
results and a conclusion about the experiment (including identification of the unknown
solid from the refractive index).

Verifying Snell's Law

Aim

To verify Snell's law

Apparatus

glass block, ray box, \(\text{360}\)\(\text{°}\) protractor, 5 pieces of A4
paper, pencil, ruler

Method

This experiment will require you to follow the steps below 5 times (once for each
piece of A4 paper).

Place the glass block in the middle of the A4 piece of paper so that
its sides are parallel to each of the sides of the paper and
draw around the block with a pencil to make its outline on the
piece of paper.

Turn on the ray box and aim the light ray towards the glass block so
that it makes an angle with the nearest surface of the block as
shown in the picture. For each piece of paper, change the angle
of the incoming ray.

You will now need to mark on the paper, the path of the incoming and
outgoing light rays. Do this by first drawing a dot on the paper
somewhere along the incoming light ray. Now draw a second dot on
the paper at the point where the incoming light ray hits the
surface of the block. Do the same thing for the outgoing light
ray; mark the point where it leaves the block and some other
point along its path.

Now switch off the ray box and remove the glass block from the paper.
Use a ruler to join the dots of the incoming ray. Now join the
dots of the outgoing ray. Lastly, draw a line which joins the
point where the incoming ray hits the block and where the
outgoing ray leaves the block. This is the path of the light ray
through the glass.

The aim of this experiment is to verify Snell's law. i.e. \(n_1 \sin
\theta_1 = n_2 \sin \theta_2\). We know the refractive index of
our two media:

For air, \(n_1 =\text{1,0}\)

For glass, \(n_2 =\text{1,5}\)

Now we need to measure the two angles, \(\theta_1\) and \(\theta_2\).
To do this, we need to draw the normal to the surface where the
light ray enters the block. Use the protractor to measure an
angle of \(\text{90}\)\(\text{°}\) to the entry surface and
draw the normal. At this point, the drawing on your piece of
paper should look something like the picture:

Now measure \(\theta_1\) and \(\theta_2\) using the protractor. Enter
the values you measured into a table which looks like:

\(\theta_1\)

\(\theta_2\)

\(n_1 \sin \theta_1\)

\(n_2 \sin \theta_2\)

Discussion

Have a look at your completed table. You should have 5 rows filled in, one for each
of your pieces of A4 paper. For each row, what do you notice about the values in
the last two columns? Do your values agree with what Snell's law predicts?

Using Snell's law to determine the refractive index of an unknown material

Aim

To determine the refractive index of an unknown material

Apparatus

ray box, \(\text{360}\)\(\text{°}\) protractor, 5 pieces of A4 paper, a block of
unknown transparent material, pencil, ruler

Method

This experiment will require you to follow the steps below at least 5 times. The
steps are the same as you followed in the previous experiment.

Place the block in the middle of the A4 piece of paper so that its
sides are parallel to each of the sides of the paper and draw
around the block with a pencil to make its outline on the piece
of paper.

Turn on the ray box and aim the light ray towards the block so that
it makes an angle with the nearest surface of the block. For
each piece of paper, change the angle of the incoming ray.

You will now need to mark on the paper, the path of the incoming and
outgoing light rays. Do this by first drawing a dot on the paper
somewhere along the incoming light ray. Now draw a second dot on
the paper at the point where the incoming light ray hits the
surface of the block. Do the same thing for the outgoing light
ray; mark the point where it leaves the block and some other
point along its path.

Now switch off the ray box and remove the block from the paper. Use a
ruler to join the dots of the incoming ray. Now join the dots of
the outgoing ray. Lastly, draw a line which joins the point
where the incoming ray hits the block and where the outgoing ray
leaves the block. This is the path of the light ray through the
block.

The aim of this experiment is to determine the refractive index
\(n_2\) of the unknown material using Snell's law which states
\(n_1 \sin \theta_1 = n_2 \sin \theta_2\). We know that for air
\(n_1\) = \(\text{1,0}\) and we can measure the angle of
incidence, \(\theta_1\) and the angle of refraction,
\(\theta_2\). Then we can solve the equation for the unknown,
\(n_2\).

To measure \(\theta_1\) and \(\theta_2\), we need to draw the normal
to the surface where the light ray enters the block. Use the
protractor to measure an angle of \(\text{90}\)\(\text{°}\)
to the entry surface and draw the normal. Now use the protractor
to measure \(\theta_1\) and \(\theta_2\).

Enter the values you measured into a table which looks like the one
below and calculate the resulting value for \(n_2\).

\(\theta_1\)

\(\theta_2\)

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

Question

What do you notice about all your values in the last column of the table?

Discussion

You should have noticed that the values in the last column of the table are similar
but not identical. This is due to measurement errors when you measured the
angles of incidence and angles of refraction. These sorts of errors are common
in all physics experiments and lead to a measure of uncertainty in the final
extracted value. However, since we did the same experiment 5 times, we can
average the 5 independent measurements of \(n_2\) to get a good approximation to
the real value for our unknown material.

If
\[n_2 > n_1\]
then from Snell's Law,
\[\sin \theta_1 > \sin \theta_2\]

For angles smaller than \(\text{90}\)\(\text{°}\), \(\sin{\theta}\) increases as \(\theta\)
increases. Therefore,
\[\theta_1 > \theta_2.\]
This means that the angle of incidence is greater than the angle of refraction and the light ray
is bent toward the normal.

Similarly, if
\[n_2 < n_1\]
then from Snell's Law,
\[\sin \theta_1 < \sin \theta_2.\]
For angles smaller than \(\text{90}\)\(\text{°}\), \(\sin{\theta}\) increases as \(\theta\)
increases. Therefore,
\[\theta_1 < \theta_2.\]
This means that the angle of incidence is less than the angle of refraction and the light ray is
bent away from the normal.

What happens to a ray that lies along the normal line? In this case, the angle of incidence is
\(\text{0}\)\(\text{°}\) and
\begin{align*}
\sin \theta_2 &= \frac{n_1}{n_2} \sin \theta_1 \\
&= \text{0} \\
\therefore \theta_2 &= \text{0}.
\end{align*}

This shows that if the light ray is incident at \(\text{0}\)\(\text{°}\), then the angle of
refraction is also \(\text{0}\)\(\text{°}\).
The direction of the light ray is unchanged, however, the speed of the light will change as it
moves into the new medium. Therefore refraction still occurs although it will not be easily
observed.

Let's use an everyday example that you can more easily imagine. Imagine you are pushing a lawnmower
or a cart through short grass. As long as the grass is pretty much the same length everywhere,
it is easy to keep the mower or cart going in a straight line. However, if the wheels on one
side of the mower or cart enter an area of grass that is longer than the grass on the other
side, that side of the mower will move more slowly since it is harder to push the wheels through
the longer grass. The result is that the mower will start to turn inwards, towards the longer
grass side as you can see in the picture. This is similar to light which passes through a medium
and then enters a new medium with a higher refractive index or higher optical density. The light
will change direction towards the normal, just like the mower in the longer grass. The opposite
happens when the mower moves from an area of longer grass into an area with shorter grass. The
side of the mower in the shorter grass will move faster and
the mower will turn outwards, just like a light ray moving from a medium with high refractive
index into a medium with low refractive index moves away from the normal.

Worked example 2: Using Snell's Law

Light is refracted at the boundary between water and an unknown medium. If the angle
of incidence is \(\text{25}\)\(\text{°}\), and the angle of refraction is
\(\text{20,6}\)\(\text{°}\), calculate the refractive index of the unknown
medium and use Table 5.1 to
identify the material.

Determine what is given and what is being asked

The angle of incidence \(\theta_{1}=\)\(\text{25}\)\(\text{°}\)

The angle of refraction \(\theta_{2}=\)\(\text{20,6}\)\(\text{°}\)

We can look up the refractive index for water in Table 5.1: \(n_{1} =\text{1,333}\)

We need to calculate the refractive index for the unknown medium and identify it.

Determine how to approach the problem

We can use Snell's law to calculate the unknown refractive index, \(n_{2}\)

According to Table 5.1, typical glass
has a refractive index between \(\text{1,5}\) to \(\text{1,9}\). Therefore the
unknown medium is typical glass.

Worked example 3: Using Snell's law

A light ray with an angle of incidence of \(\text{35}\)\(\text{°}\) passes from
water to air. Find the angle of refraction using Snell's Law and Table 5.1. Discuss the
meaning of your answer.

Determine the refractive indices of water and air

From Table 5.1, the refractive index
is \(\text{1,333}\) for water and about \(\text{1}\) for air. We know the angle
of incidence, so we are ready to use Snell's Law.

Since \(\text{130,3}\)\(\text{°}\) is larger than
\(\text{90}\)\(\text{°}\), the solution is:
\[\theta_2 = \text{49,7}\text{°}\]

Discuss the answer

The light ray passes from a medium of high refractive index to one of low refractive
index. Therefore, the light ray is bent away from the normal.

Worked example 4: Using Snell's Law

A light ray passes from water to diamond with an angle of incidence of
\(\text{75}\)\(\text{°}\). Calculate the angle of refraction. Discuss the
meaning of your answer.

Determine the refractive indices of water and air

From Table 5.1, the refractive index
is \(\text{1,333}\) for water and \(\text{2,42}\) for diamond. We know the angle
of incidence, so we are ready to use Snell's Law.

The light ray passes from a medium of low refractive index to one of high refractive
index. Therefore, the light ray is bent towards the normal.

temp text

Snell's Law

Textbook Exercise 5.4

State Snell's Law.

Snell's law states that the angle of incidence times the refractive
index of medium 1 is equal to the angle of refraction times the
refractive index of medium 2. Mathematically this is: \(n_{1}
\sin \theta_{1} = n_{2} \sin \theta_{2}\)

Light travels from a region of glass into a region of glycerine,
making an angle of incidence of \(\text{40}\)\(\text{°}\).

Draw the incident and refracted light rays on the diagram and
label the angles of incidence and refraction.

Calculate the angle of refraction.

The refractive index of air is \(\text{1}\) and the
refractive index of glycerine is \(\text{1,4729}\).

A ray of light travels from silicon to water. If the ray of light in
the water makes an angle of \(\text{69}\)\(\text{°}\) to
the normal to the surface, what is the angle of incidence in the
silicon?

The refractive index of silicon is \(\text{4,01}\) and the refractive
index of water is \(\text{1,333}\).

Light travels from a medium with \(n = \text{1,25}\) into a medium of
\(n = \text{1,34}\), at an angle of
\(\text{27}\)\(\text{°}\) from the normal.

What happens to the speed of the light? Does it increase,
decrease, or remain the same?

The speed of light decreases as it enters the second medium.

What happens to the wavelength of the light? Does it
increase, decrease, or remain the same?

The wavelength of the light remains the same. Wavelength is
related to frequency and the frequency of light does not
change as it moves from one medium to another.

Does the light bend towards the normal, away from the normal,
or not at all?

Towards the normal.

Light travels from a medium with \(n = \text{1,63}\) into a medium of
\(n = \text{1,42}\).

What happens to the speed of the light? Does it increase,
decrease, or remain the same?

The speed of light increases as it enters the second medium.

What happens to the wavelength of the light? Does it
increase, decrease, or remain the same?

The wavelength of the light remains the same. Wavelength is
related to frequency and the frequency of light does not
change as it moves from one medium to another.

Does the light bend towards the normal, away from the normal,
or not at all?

Away from the normal

Light is incident on a rectangular glass prism. The prism is
surrounded by air. The angle of incidence is
\(\text{23}\)\(\text{°}\). Calculate the angle of
reflection and the angle of refraction.

The angle of reflection is the same as the angle of incidence:
\(\text{23}\)\(\text{°}\).

To find the angle of refraction we need the indices of refraction for
glass and air. The index of refraction for air is \(\text{1}\)
and for the prism it is \(\text{1,5}\) - \(\text{1,9}\) (we
assume that the prism is made of typical glass). Since typical
glass has a range of refractive indices we will calculate the
maximum and minimum possible angles of refraction.

Light is refracted at the interface between air and an unknown
medium. If the angle of incidence is
\(\text{53}\)\(\text{°}\) and the angle of refraction is
\(\text{37}\)\(\text{°}\), calculate the refractive index
of the unknown, second medium.

The index of refraction for air is \(\text{1}\). The refractive index
of the unknown medium is:

Light is refracted at the interface between a medium of refractive
index \(\text{1,5}\) and a second medium of refractive index
\(\text{2,1}\). If the angle of incidence is
\(\text{45}\)\(\text{°}\), calculate the angle of
refraction.

A ray of light strikes the interface between air and diamond. If the
incident ray makes an angle of \(\text{30}\)\(\text{°}\)
with the interface, calculate the angle made by the refracted
ray with the interface.

The refractive index of air is 1 and the refractive index of diamond
is \(\text{2,419}\). The angle of refraction is:

This is the angle between the normal and the refracted ray. The
question actually wants the angle between the refracted ray and
the interface. This angle is:

The angles of incidence and refraction were measured in five unknown
media and recorded in the table below. Use your knowledge about
Snell's Law to identify each of the unknown media A–E. Use
Table 5.1 to
help you.

Zingi and Tumi performed an investigation to identify an unknown
liquid. They shone a ray of light into the unknown liquid,
varying the angle of incidence and recording the angle of
refraction. Their results are recorded in the following table:

Angle of incidence

Angle of refraction

\(\text{0,0}\)\(\text{°}\)

\(\text{0,00}\)\(\text{°}\)

\(\text{5,0}\)\(\text{°}\)

\(\text{3,76}\)\(\text{°}\)

\(\text{10,0}\)\(\text{°}\)

\(\text{7,50}\)\(\text{°}\)

\(\text{15,0}\)\(\text{°}\)

\(\text{11,2}\)\(\text{°}\)

\(\text{20,0}\)\(\text{°}\)

\(\text{14,9}\)\(\text{°}\)

\(\text{25,0}\)\(\text{°}\)

\(\text{18,5}\)\(\text{°}\)

\(\text{30,0}\)\(\text{°}\)

\(\text{22,1}\)\(\text{°}\)

\(\text{35,0}\)\(\text{°}\)

\(\text{25,5}\)\(\text{°}\)

\(\text{40,0}\)\(\text{°}\)

\(\text{28,9}\)\(\text{°}\)

\(\text{45,0}\)\(\text{°}\)

\(\text{32,1}\)\(\text{°}\)

\(\text{50,0}\)\(\text{°}\)

\(\text{35,2}\)\(\text{°}\)

\(\text{55,0}\)\(\text{°}\)

\(\text{38,0}\)\(\text{°}\)

\(\text{60,0}\)\(\text{°}\)

\(\text{40,6}\)\(\text{°}\)

\(\text{65,0}\)\(\text{°}\)

\(\text{43,0}\)\(\text{°}\)

\(\text{70,0}\)\(\text{°}\)

?

\(\text{75,0}\)\(\text{°}\)

?

\(\text{80,0}\)\(\text{°}\)

?

\(\text{85,0}\)\(\text{°}\)

?

Write down an aim for the investigation.

To use refractive indices and Snell's law to determine an
unknown substance.

Make a list of all the apparatus they used.

Protractor, ruler, ray box, paper, pencils, beakers.

Identify the unknown liquid.

We can use any of the pairs of data to find the refractive
index of the unknown substance. We will use the second
pair of readings.

Predict what the angle of refraction will be for
\(\text{70}\)\(\text{°}\), \(\text{75}\)\(\text{°}\),
\(\text{80}\)\(\text{°}\) and
\(\text{85}\)\(\text{°}\).

The angle of refraction at \(\text{70}\)\(\text{°}\) is: