How to Find Refractive Index from a Sine-Sine Graph

A ray of light enters a glass block from air and changes direction. You measure the angle of incidence and the angle of refraction, but how do you turn those measurements into the refractive index of the glass?

The answer is to plot the sine of the angle of incidence against the sine of the angle of refraction. The gradient of this graph gives the refractive index of the second medium relative to the first.

There is one detail that catches students out: the gradient depends on which quantity is on each axis and which medium the light enters. Get those two things right, and the calculation is straightforward.

01Finding refractive index from the gradient

Start with Snell’s law:

\[ n_1\sin i=n_2\sin r \]

where:

• \(n_1\) is the refractive index of the medium the light starts in (the incident medium).
• \(n_2\) is the refractive index of the medium the light enters (the refracting medium).
• \(i\) is the angle of incidence.
• \(r\) is the angle of refraction.

Rearrange the equation to make \(\sin i\) the subject:

\[ \sin i=\frac{n_2}{n_1}\sin r \]

Now compare this with the equation of a straight line:

\[ y=mx \]

If you plot:

• Vertical axis (\(y\)): \(\sin i\)
• Horizontal axis (\(x\)): \(\sin r\)

then:

\[ \boxed{\text{Gradient}=\frac{n_2}{n_1}} \]

This is the relative refractive index of the second medium compared with the first.

What does this mean for air and glass?

Suppose light travels from air into glass.

The refractive index of air is approximately \(1.00\), so:

\[ m=\frac{n_{\text{glass}}}{n_{\text{air}}} \]

\[ m=\frac{n_{\text{glass}}}{1.00} \]

Therefore:

\[ \boxed{n_{\text{glass}}\approx m} \]

For an air-to-glass experiment, with \(\sin i\) on the vertical axis and \(\sin r\) on the horizontal axis, the gradient gives the refractive index of the glass.

This is the result most commonly used in HSC-style sine-sine graph questions.

However, don’t memorise that the gradient always equals the refractive index. The axes and medium pair determine the relationship.

If the axes are swapped, the gradient becomes the reciprocal.

02Draw the normal before measuring angles

Before collecting any data, make sure you know exactly what the angles represent.

The normal is an imaginary line drawn at 90° to the boundary between two media, at the point where the light ray meets it.

The two angles are:

• Angle of incidence, \(i\): measured between the incoming ray and the normal.
• Angle of refraction, \(r\): measured between the refracted ray and the normal.

Both angles are measured from the normal, not from the surface.

Here’s a useful angle check.

Suppose a ray makes an angle of 55° with the surface of a glass block.

The angle of incidence is not 55°. It is:

\[ i=90^\circ-55^\circ=35^\circ \]

This is because the normal is perpendicular to the surface.

For HSC calculations, use positive angle magnitudes measured from the normal, usually in degrees. The sine values and graph gradient are dimensionless, so refractive index has no units.

03How the experiment produces a sine-sine graph

The purpose of the experiment is to investigate whether the sine of the angle of incidence is proportional to the sine of the angle of refraction.

A typical setup uses:

• A narrow light beam from a ray box or low-power laser.
• A semicircular glass block.
• A sheet of paper.
• A ruler and protractor.

The arrangement of the apparatus matters because you want to measure refraction at one boundary, without confusing it with what happens at another boundary.

Why a semicircular glass block is useful

Place a semicircular glass block on a sheet of paper, with its flat face along a straight line.

Mark the midpoint of the flat face. This point is also the centre of the semicircle.

Direct a narrow light ray from the air towards this midpoint.

The ray enters the glass through the flat face and bends towards the normal. It then travels through the glass towards the curved surface.

Here’s the useful part: because the ray starts at the centre of the semicircle, its path towards the curved surface is a radius.

The normal to a curved surface at any point is along the radius through that point. Therefore, the ray reaches the curved surface at normal incidence and exits without changing direction.

This isolates the refraction at the flat air-glass boundary.

Collecting the measurements

For each trial:

1. Trace the outline of the glass block onto the paper.
2. Direct the light beam towards the midpoint of the flat face.
3. Mark the incident and refracted ray paths.
4. Draw the normal at the point where the beam enters the glass.
5. Measure \(i\) and \(r\) from the normal using a protractor.
6. Repeat with different angles of incidence.

Keep the same glass block and light wavelength throughout the experiment. Refractive index can vary with wavelength, so changing the colour of the light introduces another variable.

Record the measurements in a table.

A common experimental mistake

A rectangular glass block can also be used to investigate refraction. However, it introduces a problem if you measure the wrong ray.

When light enters a rectangular block, it refracts at the first surface. It then refracts again when it exits through the second, parallel surface.

The emerging ray is parallel to the original incident ray, although it is laterally displaced.

A student who measures the angle of the emerging ray and calls it \(r\) is using the wrong angle. That angle is the angle of emergence, not the angle of refraction inside the glass.

The fix is to trace the ray inside the block and measure its angle at the first boundary, or use the semicircular arrangement described above.

04Converting your measurements into a graph

Once you have collected your angles, calculate their sine values.

Consider the following illustrative data for light travelling from air into glass with a refractive index of approximately 1.50.

These are idealised example readings, not measurements from a recorded experiment.

Now plot:

• \(\sin r\) on the horizontal axis.
• \(\sin i\) on the vertical axis.

You should obtain an approximately straight line passing through the origin.

Why should the points form a straight line?

Snell’s law tells us:

\[ \sin i=\frac{n_2}{n_1}\sin r \]

If the two media remain unchanged, their refractive indices remain constant for the chosen wavelength.

Therefore, the ratio between the two sine values is constant. This produces a linear relationship.

In an ideal experiment, the line passes exactly through the origin. In a real experiment, small measurement errors may produce some scatter or a slight intercept.

Calculating the gradient

The gradient of a straight-line graph is:

\[ m=\frac{\text{change in vertical value}} {\text{change in horizontal value}} \]

For this graph:

\[ m=\frac{\Delta(\sin i)}{\Delta(\sin r)} \]

Choose two widely separated points on your best-fit line.

Using the approximate endpoints from the illustrative data:

\[ \begin{aligned} m&=\frac{0.866-0.174}{0.577-0.116}\\ &=\frac{0.692}{0.461}\\ &=1.50 \end{aligned} \]

Since the light travels from air into glass:

\[ m=\frac{n_{\text{glass}}}{n_{\text{air}}} \]

Taking \(n_{\text{air}}=1.00\):

\[ \boxed{n_{\text{glass}}=1.50} \]

The refractive index of the glass is 1.50, with no units.

In an actual experiment, calculate the gradient using two widely separated points on the best-fit line, rather than simply choosing two adjacent experimental readings. This reduces the effect of measurement uncertainty.

05Worked examples

06The most common sine-sine graph misconception

Consider the same glass, with a refractive index of 1.50, and air, with a refractive index of 1.00.

For light travelling from air into glass, with \(\sin i\) vertically and \(\sin r\) horizontally:

\[ m=\frac{n_{\text{glass}}}{n_{\text{air}}} =1.50 \]

Now reverse the direction of the light. It travels from glass into air.

The same graph convention now gives:

\[ m=\frac{n_{\text{air}}}{n_{\text{glass}}} \]

\[ m=\frac{1.00}{1.50} \]

\[ \boxed{m=0.667} \]

The glass has not changed its refractive index. The light is simply travelling through the media in the opposite order.

This is why you must identify the incident medium and refracting medium before interpreting the gradient.

A gradient of 0.667 does not automatically mean that the glass has a refractive index of 0.667.

A reliable method for HSC questions

Whenever you encounter a sine-sine graph, follow this sequence:

1. Identify the incident medium and the refracting medium.
2. Check that both angles are measured from the normal.
3. Read the horizontal and vertical axis labels carefully.
4. Calculate the gradient using two widely separated points on the best-fit line.
5. Write Snell’s law and rearrange it to match the graph’s axes.
6. Use the gradient to calculate the relative refractive index.
7. If necessary, use the known refractive index of the first medium to calculate the absolute refractive index of the second medium.

The important step is number five. It prevents you from accidentally using the reciprocal of the gradient.

07Questions and solutions

Question 1

A student investigates light travelling from air into a transparent plastic.

They plot \(\sin i\) on the vertical axis and \(\sin r\) on the horizontal axis. The best-fit line has a gradient of 1.48.

Given that the refractive index of air is 1.00, calculate the refractive index of the plastic.

Solution 1

The refractive index of the plastic is 1.48.

Since the light travels from air into plastic, Snell’s law gives:

\[ n_{\text{air}}\sin i =n_{\text{plastic}}\sin r \]

Rearranging:

\[ \sin i= \frac{n_{\text{plastic}}}{n_{\text{air}}} \sin r \]

The graph has \(\sin i\) vertically and \(\sin r\) horizontally, so:

\[ m=\frac{n_{\text{plastic}}}{n_{\text{air}}} \]

Substitute the gradient:

\[ 1.48=\frac{n_{\text{plastic}}}{1.00} \]

Therefore:

\[ \boxed{n_{\text{plastic}}=1.48} \]

The refractive index is dimensionless.

Question 2

Light travels from water into an unknown transparent solid.

A graph of \(\sin i\) against \(\sin r\) has a gradient of 1.31.

One experimental reading records the incident ray at 55° to the surface of the water-solid boundary. The refracted ray makes an angle of 26° with the normal.

Given that the refractive index of water is 1.33:

(a) Calculate the refractive index of the solid using the graph.

(b) Check whether the measured angles are consistent with the graph gradient.

(c) State whether the light bends towards or away from the normal.

Solution 2

The refractive index of the solid is approximately 1.74. The measured angles are consistent with the graph, and the ray bends towards the normal.

Part (a): Calculate the refractive index.

The graph has:

• \(\sin i\) on the vertical axis.
• \(\sin r\) on the horizontal axis.

Therefore:

\[ m=\frac{n_2}{n_1} \]

Here, \(n_1=1.33\) for water.

Rearrange:

\[ n_2=mn_1 \]

Substitute:

\[ n_2=1.31\times1.33 \]

\[ \boxed{n_2=1.74} \]

Part (b): Check the angle measurements.

The incident angle is measured from the surface, so convert it to an angle from the normal:

\[ i=90^\circ-55^\circ \]

\[ i=35^\circ \]

The refracted angle is already measured from the normal:

\[ r=26^\circ \]

Calculate the ratio of the sine values:

\[ \begin{aligned} \frac{\sin i}{\sin r} &=\frac{\sin35^\circ}{\sin26^\circ}\\ &=\frac{0.574}{0.438}\\ &\approx1.31 \end{aligned} \]

This agrees with the graph gradient of 1.31.

Part (c): Determine the direction of bending.

The incident angle is 35°, while the refracted angle is 26°.

The refracted angle is smaller, so the ray bends towards the normal.

This also makes physical sense because the solid has a higher refractive index than water.

Question 3

A light ray travels from glass into air.

A graph is plotted with \(\sin i\) on the vertical axis and \(\sin r\) on the horizontal axis. The gradient is 0.668.

A student claims that the refractive index of the glass is 0.668.

(a) Explain why this conclusion is incorrect and calculate the actual refractive index of the glass.

(b) If the angle of incidence in the glass is 40°, calculate the angle of refraction in air.

Take the refractive index of air as 1.00.

Solution 3

The refractive index of the glass is 1.50, not 0.668. The refracted angle is approximately 74.6°.

Part (a): Calculate the refractive index.

The incident medium is glass, and the refracting medium is air.

Therefore:

\[ n_1=n_{\text{glass}} \]

\[ n_2=n_{\text{air}} \]

The graph has \(\sin i\) vertically and \(\sin r\) horizontally.

Therefore:

\[ m=\frac{n_2}{n_1} \]

Substitute:

\[ 0.668=\frac{1.00}{n_{\text{glass}}} \]

Rearrange:

\[ n_{\text{glass}}=\frac{1.00}{0.668} \]

\[ \boxed{n_{\text{glass}}=1.50} \]

The student’s mistake was treating the gradient as the absolute refractive index of the glass. It actually represents the ratio of the refractive indices of the two media.

Part (b): Calculate the refracted angle.

Use Snell’s law:

\[ n_1\sin i=n_2\sin r \]

Substitute:

\[ 1.50\sin40^\circ=1.00\sin r \]

Rearrange:

\[ \sin r=1.50\sin40^\circ \]

\[ \sin r=1.50\times0.6428 \]

\[ \sin r=0.9642 \]

Take the inverse sine:

\[ r=\sin^{-1}(0.9642) \]

\[ \boxed{r=74.6^\circ} \]

The ray bends away from the normal because it travels from a higher refractive index to a lower refractive index.

The result is also physically possible: the critical angle for this glass-air boundary is approximately 41.8°, and the incident angle of 40° is below it.

Question 4

A student uses a rectangular glass block to investigate Snell’s law.

They measure:

• \(i\): the angle of incidence in air at the first surface.
• \(e\): the angle of emergence in air at the second surface.

The two surfaces are parallel, and air surrounds the block.

The student plots \(\sin i\) vertically against \(\sin e\) horizontally. Their graph has a gradient of 1.02.

They conclude that the refractive index of the glass is 1.02.

Explain why this conclusion is incorrect.

Describe how the geometry of the experiment affects the measurements, and suggest one way to modify the experiment to obtain a valid refractive-index measurement.

Solution 4

The gradient of 1.02 does not represent the refractive index of the glass. The experiment measures the relationship between the incident and emerging angles, rather than the incident and refracted angles at one boundary.

The key is that the light refracts twice.

First boundary: air into glass

At the first surface, light enters the glass and bends towards the normal.

The angle inside the glass is the actual angle of refraction, \(r\).

Second boundary: glass into air

At the second surface, light exits the glass and bends away from the normal.

Because the two surfaces are parallel, their normals are also parallel.

For a ray that emerges into the same external medium, the emerging ray is parallel to the original incident ray.

Therefore:

\[ e=i \]

This means:

\[ \sin e=\sin i \]

A graph of \(\sin i\) against \(\sin e\) should ideally have a gradient of:

\[ m=1 \]

The measured gradient of 1.02 is reasonably close to one and can be explained by experimental uncertainty.

It does not give the refractive index of the glass.

How to improve the experiment

There are two suitable options:

• Trace the ray inside the rectangular glass block and measure the actual angle of refraction at the first surface.
• Use a semicircular glass block, directing the ray through the midpoint of its flat face so that it exits the curved surface normally.

Then plot \(\sin i\) against \(\sin r\).

For an air-to-glass experiment, the gradient of this corrected graph gives the refractive index of the glass relative to air.

Question 5

An unknown transparent material, X, is investigated using three separate refraction experiments.

All experiments use the same wavelength of light and the same graph convention:

• Vertical axis: \(\sin i\).
• Horizontal axis: \(\sin r\).

The following gradients are obtained:

Take the refractive index of air as 1.00 and the refractive index of water as 1.33.

Assume the material is homogeneous and isotropic, and that the quoted gradients are accurate to the stated precision.

(a) Calculate the refractive index of X independently using experiments A and B. Explain whether the results contradict each other.

(b) Calculate the expected gradient for experiment C. Explain why it should be approximately the reciprocal of the gradient in experiment B.

(c) A ray travels from X into water at an angle of incidence of 60°. Calculate its angle of refraction.

(d) The incident angle is increased to 70°, but no refracted ray is observed. Explain why this does not contradict the straight-line relationship predicted by Snell’s law. Calculate the critical angle and identify what happens to the light.

Solution 5

The refractive index of X is approximately 1.50. All three graph gradients are consistent with this value. At 60°, the refracted angle is approximately 77.6°. At 70°, total internal reflection occurs because the incident angle exceeds the critical angle of approximately 62.5°.

This question tests an important distinction: a graph gradient gives a ratio between two refractive indices, not necessarily the absolute refractive index of either material.

Part (a): Calculate the refractive index using experiments A and B.

For experiment A, light travels from air into X.

The gradient is:

\[ m=\frac{n_X}{n_{\text{air}}} \]

Therefore:

\[ 1.50=\frac{n_X}{1.00} \]

\[ \boxed{n_X=1.50} \]

Now consider experiment B.

Light travels from water into X, so:

\[ m=\frac{n_X}{n_{\text{water}}} \]

Substitute:

\[ 1.13=\frac{n_X}{1.33} \]

Rearrange:

\[ n_X=1.13\times1.33 \]

\[ \boxed{n_X=1.50} \]

More precisely, the second experiment gives approximately 1.503, which agrees with the first result to the stated precision.

The results do not contradict each other.

The gradients differ because the incident media are different. Experiment A compares X with air, while experiment B compares X with water.

The refractive index of X has not changed.

Part (b): Predict the gradient for experiment C.

Now the light travels from X into water.

The gradient is:

\[ m=\frac{n_{\text{water}}}{n_X} \]

Substitute:

\[ m=\frac{1.33}{1.50} \]

\[ \boxed{m=0.887} \]

The measured gradient is 0.89, which agrees with the predicted value after rounding.

There is another way to recognise this relationship.

Experiments B and C use the same two media, but the direction of light is reversed.

Therefore, their gradients should be reciprocals:

\[ m_C\approx\frac{1}{m_B} \]

\[ m_C=\frac{1}{1.13} \]

\[ m_C\approx0.885 \]

This is consistent with the measured gradient of 0.89.

Part (c): Calculate the refracted angle at 60°.

Light travels from X into water.

Therefore:

\[ n_1=1.50 \]

\[ n_2=1.33 \]

Use Snell’s law:

\[ n_1\sin i=n_2\sin r \]

Substitute:

\[ 1.50\sin60^\circ=1.33\sin r \]

Rearrange:

\[ \sin r=\frac{1.50}{1.33}\sin60^\circ \]

\[ \sin r=1.1278\times0.8660 \]

\[ \sin r=0.9767 \]

Take the inverse sine:

\[ r=\sin^{-1}(0.9767) \]

\[ \boxed{r\approx77.6^\circ} \]

The light refracts away from the normal, as expected when it travels from a higher refractive index to a lower refractive index.

Although the refracted angle is quite large, it is still less than 90°, so a refracted ray exists.

Part (d): Explain what happens at 70°.

The critical angle is the angle of incidence for which the refracted ray travels along the boundary, meaning:

\[ r=90^\circ \]

Using Snell’s law:

\[ n_1\sin c=n_2\sin90^\circ \]

Since \(\sin90^\circ=1\):

\[ \sin c=\frac{n_2}{n_1} \]

Substitute:

\[ \sin c=\frac{1.33}{1.50} \]

\[ \sin c=0.8867 \]

Therefore:

\[ \boxed{c=62.5^\circ} \]

The new incident angle is 70°, which is greater than the critical angle.

Therefore, total internal reflection occurs.

There is no refracted ray travelling into the water.

We can also see why this happens by trying to apply Snell’s law at 70°:

\[ \sin r=\frac{1.50}{1.33}\sin70^\circ \]

\[ \sin r\approx1.06 \]

But a real angle cannot have a sine greater than one.

The calculation is telling us that no transmitted refracted ray is physically possible.

This does not contradict Snell’s law. The sine-sine graph describes the relationship for angles at which a refracted ray exists. It cannot be extended beyond the critical angle as though every incident ray must produce a transmitted ray.

Instead, the light undergoes total internal reflection.

08Where this leads next

Finding refractive index from a sine-sine graph gives you the relationship between the optical properties of two materials.

You can connect that result to the speed of light using:

\[ n=\frac{c}{v} \]

This lets you calculate how fast light travels in the material. For a deeper explanation of that relationship, see Refractive Index: Using n = c/v in HSC Physics.

You can also use the refractive indices to predict whether a ray will refract or undergo total internal reflection at another boundary. That’s the next step when applying Snell’s law to more complicated optical systems.