How to Trace a Ray Through a Parallel-Sided Glass Block

A narrow beam of light enters a rectangular glass block at an angle. It bends as it enters the glass, travels through the block, and bends again as it leaves. But where should the emerging ray go?

You might predict that it continues in the new direction after leaving the glass. In fact, when the block is surrounded by air on both sides, the emerging ray is parallel to the original incident ray. It is simply shifted sideways.

To trace the ray correctly, draw a normal at each boundary, measure every angle from the relevant normal, apply Snell’s law at the first surface, and then repeat the process at the second surface.

01The correct method for tracing a ray

The easiest way to avoid mistakes is to treat the two surfaces as separate boundaries.

At the first surface, light travels from air into glass. At the second, it travels from glass back into air. The direction of refraction changes because the ray is moving between different pairs of media.

Step 1: Identify the two boundaries

A parallel-sided glass block has two opposite faces that are parallel.

The incoming ray meets the first face at the point of incidence. After entering the glass, it travels towards the opposite face, where it exits.

Call these points A and B.

The glass between A and B is assumed to be homogeneous. This means the refractive index is constant throughout the material, so the ray travels in a straight line inside the block.

Step 2: Draw the normals

A normal is an imaginary line drawn perpendicular to a surface at the point where the ray meets it.

For a ray passing through a parallel-sided block, you need two normals:

• The first normal passes through the entry point A.
• The second normal passes through the exit point B.

Both normals must be at 90° to their respective surfaces.

Because the two glass faces are parallel, the two normals are also parallel.

This is important. It allows us to compare the angle of the ray inside the glass at the two boundaries.

Step 3: Measure angles from the normal, not the surface

This is one of the most common mistakes in HSC refraction questions.

There are four angles to recognise:

All angles are measured from the normal, not from the glass surface.

Suppose a ray makes an angle of \(35^\circ\) with the surface.

The angle of incidence is:

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

\[ i=55^\circ \]

You must use \(55^\circ\) in Snell’s law, not \(35^\circ\).

A useful rule is to draw the normal before measuring anything. It makes the correct angle much easier to identify.

Step 4: Trace the ray as it enters the glass

Now consider what happens at the first boundary.

Light travels from air into glass. Glass has a higher refractive index than air, so light travels more slowly in the glass.

The ray bends towards the normal.

This means:

\[ r<i \]

The relationship between the two angles is described by Snell’s law:

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

where:

• \(n_1\) is the refractive index of the incident medium.
• \(n_2\) is the refractive index of the transmitted medium.
• \(i\) is the angle of incidence.
• \(r\) is the angle of refraction.

For ordinary HSC calculations, air is approximated as having a refractive index of \(1.00\).

For glass with a refractive index of \(1.50\), light entering from air gives:

\[ 1.00\sin i=1.50\sin r \]

The resulting refracted angle is smaller than the incident angle.

Step 5: Draw the ray through the glass

Once you have found the refracted angle, draw the internal ray from A towards B.

Inside a uniform glass block, the ray travels in a straight line.

Do not draw another bend halfway through the glass. Refraction occurs at the boundary, not continuously throughout a uniform material.

Step 6: Trace the ray as it leaves the glass

The ray now reaches the second face.

This time, it travels from glass into air.

Air has a lower refractive index than glass, so the ray bends away from the normal.

At this boundary, Snell’s law becomes:

\[ n_{\text{glass}}\sin i_2 = n_{\text{air}}\sin e \]

Notice that the media have changed places.

The incident medium is now glass, and the transmitted medium is air.

Because the two faces are parallel, their normals are parallel. The internal ray therefore makes the same acute angle with both normals:

\[ i_2=r \]

This gives us a useful shortcut when tracing a ray through a parallel-sided block.

You still need to apply Snell’s law at the second boundary, but you do not need to measure the internal angle again if you already know \(r\).

The direction of refraction at each boundary

Keep this table in mind:

At normal incidence, the ray travels along the normal. Its direction does not change at either boundary, even though its speed changes when it enters or leaves the glass.

02Why does the emerging ray remain parallel?

This is the feature that makes parallel-sided glass blocks particularly interesting.

The first surface bends the ray towards the normal. The second surface bends it away from the normal.

For air on both sides, these changes in direction cancel.

We can prove this using Snell’s law.

At the first boundary:

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

At the second boundary:

\[ n_{\text{glass}}\sin i_2 = n_{\text{air}}\sin e \]

Since the faces are parallel:

\[ i_2=r \]

Substituting this into the second equation gives:

\[ n_{\text{glass}}\sin r = n_{\text{air}}\sin e \]

Comparing the two equations:

\[ \sin i=\sin e \]

For the acute angles in this ray-tracing situation:

\[ e=i \]

Therefore, the emerging ray is parallel to the incident ray.

This result assumes:

• The two glass faces are parallel.
• The ray successfully exits the block.
• The surrounding medium is the same on both sides.
• The block is homogeneous.

There is an important distinction here: parallel does not mean that the two rays follow exactly the same line.

The emerging ray is displaced sideways.

Understanding lateral displacement

The sideways separation between the original incident ray, extended through the block, and the emerging ray is called lateral displacement.

The ray travels diagonally through the glass, so it exits at a different position from where it would have travelled if the glass were not there.

You can calculate lateral displacement using:

\[ d=\frac{t\sin(i-r)}{\cos r} \]

where:

• \(d\) is the lateral displacement, measured in metres or centimetres.
• \(t\) is the perpendicular thickness of the glass block.
• \(i\) is the angle of incidence.
• \(r\) is the angle of refraction.

The formula comes from the geometry of the ray inside the block.

The internal ray travels a distance \(L\), where:

\[ L=\frac{t}{\cos r} \]

The angle between the internal ray and the original incident direction is \(i-r\).

Therefore:

\[ d=L\sin(i-r) \]

Substituting for \(L\) gives the lateral displacement formula.

There are two useful predictions:

• A thicker block produces a greater lateral displacement for the same incident angle and material.
• At normal incidence, \(i=r=0^\circ\), so lateral displacement is zero.

03How to trace the ray experimentally

You can investigate this using a ray box and a rectangular glass block.

The challenge is that the glass block hides part of the ray’s path from the paper underneath it. You need to record enough information to reconstruct the internal ray.

Apparatus

You need:

• A ray box with a narrow slit.
• A rectangular glass block.
• A sheet of plain paper.
• A pencil.
• A ruler.
• A protractor.

A narrow beam is important. A wide beam makes it harder to identify the exact direction of the ray.

Setting up the experiment

Step 1: Trace the glass block.

Place the glass block on the paper.

Carefully draw around its edges with a pencil. Remove the block and label the two opposite faces that the beam will cross.

Replace the block in exactly the same position.

Step 2: Direct the incident beam towards the block.

Position the ray box so its narrow beam approaches the first face at an angle.

Make sure the beam enters one face and exits through the opposite face, rather than striking one of the side edges.

Step 3: Mark the incident and emerging rays.

Mark two points along the incident beam before it enters the glass.

Then mark two points along the emerging beam after it leaves the glass.

Using two points for each ray allows you to reconstruct its direction accurately.

Step 4: Reconstruct the internal ray.

Switch off the ray box and remove the glass block.

The incident ray intersects the first face at point A. The emerging ray intersects the opposite face at point B.

Join A and B with a ruler.

This gives you the internal ray path. The straight line is justified because the glass is assumed to be homogeneous.

Step 5: Draw the normals and measure the angles.

At A and B, draw lines perpendicular to the glass surfaces.

Measure:

• \(i\), between the incident ray and the first normal.
• \(r\), between the internal ray and the first normal.
• \(i_2\), between the internal ray and the second normal.
• \(e\), between the emerging ray and the second normal.

You should find that:

\[ i_2\approx r \]

and:

\[ e\approx i \]

The approximation accounts for practical measurement errors.

What should the experiment show?

The original prediction is that the ray bends towards the normal when entering glass, then away from the normal when leaving.

The observations should support three conclusions:

1. The internal ray makes a smaller angle with the normal than the incident ray.
2. The emerging ray makes approximately the same angle with its normal as the original incident ray.
3. The emerging ray is parallel to the incident ray but laterally displaced.

The experiment has some limitations.

The glass block might move slightly when replaced. The beam might be too wide to mark accurately, or the pencil marks might be too close together to establish the ray’s direction.

To improve the method, use a narrower beam, keep the block aligned with its traced outline, and space the marking points further apart. Repeat the experiment at several incident angles.

A small difference between \(i\) and \(e\) does not automatically mean the block is not parallel-sided. Consider the uncertainty in the measurements first.

04Worked examples

05A contrasting case: What if the faces are not parallel?

The result \(e=i\) depends on the faces being parallel.

Consider a slightly wedge-shaped piece of glass. Its two surfaces are not parallel, so their normals are not parallel either.

The internal ray therefore makes different angles with the two normals.

You must calculate the incidence angle at the second boundary using the second normal, rather than assuming it equals the original refraction angle.

This is also why you should never assume that a ray emerging from any piece of glass must be parallel to its incident ray.

There is another condition to watch: the ray must actually be able to leave the second boundary. If it travels from a higher refractive index to a lower one at an angle greater than the critical angle, total internal reflection occurs instead.

For a separate method of deciding whether a ray refracts or undergoes total internal reflection at the second boundary, see How to Decide if a Ray Refracts or Reflects at a Second Boundary.

06Questions and solutions

The following questions move from straightforward ray tracing to situations where you need to reconsider the assumptions behind the parallel-sided block model.

Unless stated otherwise, treat air as having a refractive index of \(1.00\), and assume the glass is homogeneous. All angles are positive acute angles measured from the relevant normal.

Question 1

A ray travels from air into a parallel-sided glass block with a refractive index of \(1.50\).

The angle of incidence is \(30.0^\circ\).

Calculate:

a) The angle of refraction inside the glass.

b) The angle of emergence.

c) Whether the emerging ray is parallel to the incident ray or travels along the same line.

Solution 1

The angle of refraction is \(19.5^\circ\), and the angle of emergence is \(30.0^\circ\). The emerging ray is parallel to the incident ray, but it is laterally displaced.

Step 1: Apply Snell’s law at the first boundary.

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

\[ 1.00\sin30^\circ=1.50\sin r \]

\[ \sin r=\frac{0.500}{1.50} \]

\[ r=\sin^{-1}(0.3333) \]

\[ r=19.5^\circ \]

Step 2: Apply Snell’s law at the second boundary.

Because the surfaces are parallel:

\[ i_2=r=19.5^\circ \]

Therefore:

\[ 1.50\sin19.5^\circ=1.00\sin e \]

\[ e=30.0^\circ \]

Step 3: Compare the incident and emerging angles.

The incident and emerging angles are equal:

\[ e=i=30.0^\circ \]

The ray therefore emerges parallel to its original direction.

However, because it travelled diagonally through the block, its final position is shifted sideways.

Question 2

A ray approaches a parallel-sided glass block from air.

A student measures the angle between the incoming ray and the front surface as \(28^\circ\).

The refractive index of the glass is \(1.60\), and air surrounds both sides of the block.

Calculate:

a) The correct angle of incidence.

b) The angle of refraction.

c) The angle of emergence.

d) The incorrect angle of refraction that would be obtained if the student mistakenly used \(28^\circ\) as the angle of incidence.

Explain why the student’s original measurement must be converted before using Snell’s law.

Solution 2

The correct angle of incidence is \(62^\circ\), the angle of refraction is approximately \(33.5^\circ\), and the angle of emergence is \(62^\circ\).

Using \(28^\circ\) directly would produce an incorrect internal angle of approximately \(17.1^\circ\).

Step 1: Convert the surface angle into an angle measured from the normal.

The normal is perpendicular to the surface.

Therefore:

\[ i=90^\circ-28^\circ \]

\[ i=62^\circ \]

Step 2: Calculate the correct angle of refraction.

Apply Snell’s law:

\[ 1.00\sin62^\circ = 1.60\sin r \]

Rearrange:

\[ r=\sin^{-1} \left(\frac{\sin62^\circ}{1.60}\right) \]

\[ r=33.5^\circ \]

Step 3: Calculate the angle of emergence.

The faces are parallel, and the surrounding medium is air on both sides.

Therefore:

\[ e=i=62^\circ \]

Step 4: Investigate the student’s incorrect method.

If the student uses \(28^\circ\) directly:

\[ 1.00\sin28^\circ=1.60\sin r \]

\[ r=\sin^{-1} \left(\frac{\sin28^\circ}{1.60}\right) \]

\[ r=17.1^\circ \]

This is incorrect because Snell’s law uses angles measured from the normal.

The tempting mistake is to use the angle that is easiest to measure on the diagram. However, the angle between the ray and the surface is not the angle required by Snell’s law.

Question 3

A ray enters a piece of glass from air at an angle of \(45^\circ\) to the normal of the first face.

The glass has a refractive index of \(1.50\).

Unlike a parallel-sided block, the second face is slightly tilted. Its normal is rotated by \(4.0^\circ\) towards the internal ray relative to the first normal.

Calculate:

a) The angle of refraction at the first boundary.

b) The angle of incidence at the second boundary.

c) The angle of emergence into air.

d) The direction of the emerging ray relative to the first normal.

Is the emerging ray parallel to the original incident ray?

Solution 3

The emerging ray makes approximately \(41.8^\circ\) with the first normal. It is therefore not parallel to the incident ray, which makes \(45^\circ\) with that normal.

The difference occurs because the two normals are no longer parallel.

Step 1: Calculate the first angle of refraction.

At the first boundary, light travels from air into glass.

\[ 1.00\sin45^\circ = 1.50\sin r \]

\[ r=\sin^{-1} \left(\frac{\sin45^\circ}{1.50}\right) \]

\[ r=28.1^\circ \]

Step 2: Calculate the incidence angle at the second face.

The second normal has rotated \(4.0^\circ\) towards the internal ray.

Therefore, the angle between the internal ray and the second normal is:

\[ i_2=28.1^\circ-4.0^\circ \]

\[ i_2=24.1^\circ \]

This is the important difference from a parallel-sided block.

You cannot simply use \(i_2=r\), because the normals are not parallel.

Step 3: Calculate the angle of emergence.

The ray travels from glass into air.

Apply Snell’s law:

\[ 1.50\sin24.1^\circ = 1.00\sin e \]

\[ e=\sin^{-1}(1.50\sin24.1^\circ) \]

\[ e\approx37.8^\circ \]

Step 4: Find the emerging direction relative to the first normal.

The second normal is rotated \(4.0^\circ\) towards the internal ray.

The emerging ray is therefore:

\[ 37.8^\circ+4.0^\circ \]

\[ =41.8^\circ \]

The original incident ray made an angle of \(45^\circ\) with the first normal.

The emerging ray makes an angle of \(41.8^\circ\).

They are not parallel.

The key lesson is that the parallel-emergence result requires parallel faces. Changing the orientation of the second face changes the normal, which changes the angle at which the ray exits.

Question 4

A student performs a ray-tracing experiment using a rectangular glass block surrounded by air.

They obtain the following illustrative measurements:

Assume each measured angle has an estimated uncertainty of approximately \(1^\circ\).

The student calculates the refractive index using:

\[ n=\frac{\sin40^\circ}{\sin41^\circ} \]

They obtain a value below 1 and conclude that the glass has a lower refractive index than air.

a) Calculate the correct refractive index of the glass.

b) Explain the student’s mistake.

c) Does the measured emergence angle provide convincing evidence that the glass block’s faces are not parallel?

d) Suggest one improvement to the experimental method.

Solution 4

The correct refractive index is approximately \(1.52\). The student’s mistake is using the angle of emergence instead of the internal angle of refraction.

The measured emergence angle of \(41^\circ\) is consistent with the predicted \(40^\circ\), given the stated measurement uncertainty.

Step 1: Identify the correct pair of angles.

To calculate the refractive index of glass from air-to-glass refraction, use:

• The incident angle in air.
• The refracted angle inside the glass.

These are \(40^\circ\) and \(25^\circ\), respectively.

Step 2: Apply Snell’s law.

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

Rearrange:

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

Substitute:

\[ n_{\text{glass}} = \frac{1.00\sin40^\circ}{\sin25^\circ} \]

\[ n_{\text{glass}} = \frac{0.6428}{0.4226} \]

\[ n_{\text{glass}}=1.52 \]

This is the correct result.

Step 3: Explain the incorrect calculation.

The student used:

\[ n=\frac{\sin40^\circ}{\sin41^\circ} \]

This compares the angle of incidence in air with the angle of emergence in air.

It does not compare the angles on opposite sides of a single boundary, so it cannot be used to calculate the refractive index of the glass.

The relevant angle inside the glass is \(25^\circ\), not \(41^\circ\).

Step 4: Evaluate the experimental result.

For parallel faces with air on both sides, the predicted relationship is:

\[ e=i \]

Therefore, the expected emergence angle is:

\[ e=40^\circ \]

The measured value is \(41^\circ\).

The difference is only \(1^\circ\), which is consistent with the stated measurement uncertainty.

There is no convincing evidence that the faces are not parallel.

Step 5: Improve the experiment.

One useful improvement is to use a narrower beam from the ray box.

A narrow beam makes it easier to identify the ray’s centreline and measure its direction accurately.

The student should also ensure that the block remains aligned with its traced outline throughout the experiment.

Question 5

A glass block has two parallel faces, but the surrounding media are different.

Water is on the incident side of the block, while air is on the exit side.

The refractive indices are:

• Water: \(n_{\text{water}}=1.33\)
• Glass: \(n_{\text{glass}}=1.50\)
• Air: \(n_{\text{air}}=1.00\)

A ray enters the glass from water at an angle of incidence of \(40^\circ\).

a) Calculate the angle of refraction inside the glass.

b) Calculate the critical angle for the glass-to-air boundary.

c) Determine whether the ray exits into air or undergoes total internal reflection.

d) If the ray exits, calculate its angle of emergence.

e) Find the limiting angle of incidence in water at the first boundary. Above this angle, the ray cannot emerge into air through the second face.

Explain why the usual rule that a parallel-sided block produces an emerging ray parallel to the incident ray does not apply here.

Solution 5

The ray refracts inside the glass at approximately \(34.8^\circ\), then exits into air at approximately \(58.7^\circ\). It does not undergo total internal reflection at the second boundary.

The limiting angle of incidence in water is approximately \(48.8^\circ\). Above this angle, total internal reflection occurs at the glass-air boundary.

This question exposes an important limitation of the parallel-sided block rule: the surrounding media must be the same on both sides for the emerging ray to be parallel to the incident ray.

Step 1: Calculate the angle of refraction at the water-glass boundary.

Apply Snell’s law:

\[ n_{\text{water}}\sin i = n_{\text{glass}}\sin r \]

Substitute:

\[ 1.33\sin40^\circ = 1.50\sin r \]

Rearrange:

\[ r=\sin^{-1} \left( \frac{1.33\sin40^\circ}{1.50} \right) \]

\[ r=34.8^\circ \]

The ray bends towards the normal because it enters a material with a higher refractive index.

Step 2: Calculate the critical angle at the glass-air boundary.

The ray travels from glass into air.

Since glass has a higher refractive index than air, total internal reflection is possible.

The critical angle is:

\[ \theta_c= \sin^{-1} \left( \frac{n_{\text{air}}}{n_{\text{glass}}} \right) \]

Substitute:

\[ \theta_c= \sin^{-1} \left( \frac{1.00}{1.50} \right) \]

\[ \theta_c=41.8^\circ \]

Step 3: Determine whether the ray exits.

Because the two faces are parallel, their normals are parallel.

Therefore, the internal incidence angle at the second boundary is:

\[ i_2=r=34.8^\circ \]

Compare this with the critical angle:

\[ 34.8^\circ<41.8^\circ \]

The incidence angle is smaller than the critical angle.

Therefore, the ray refracts into the air rather than undergoing total internal reflection.

Step 4: Calculate the angle of emergence.

Apply Snell’s law at the glass-air boundary:

\[ 1.50\sin34.8^\circ = 1.00\sin e \]

Rearrange:

\[ e=\sin^{-1} (1.50\sin34.8^\circ) \]

\[ e=58.7^\circ \]

The ray emerges at a larger angle than its original \(40^\circ\) incidence angle.

This makes sense because the initial and final media are different.

Step 5: Find the limiting angle of incidence in water.

At the limiting condition, the ray reaches the second boundary at the critical angle.

Therefore:

\[ i_2=\theta_c \]

Since the faces are parallel:

\[ r=\theta_c \]

Apply Snell’s law at the first boundary:

\[ n_{\text{water}}\sin i_{\text{lim}} = n_{\text{glass}}\sin\theta_c \]

At the critical angle:

\[ n_{\text{glass}}\sin\theta_c = n_{\text{air}} \]

Therefore:

\[ 1.33\sin i_{\text{lim}}=1.00 \]

Rearrange:

\[ i_{\text{lim}} = \sin^{-1} \left( \frac{1.00}{1.33} \right) \]

\[ i_{\text{lim}}=48.8^\circ \]

The outcomes are:

• If \(i<48.8^\circ\), the ray can emerge into air.
• If \(i=48.8^\circ\), the refracted ray travels along the second boundary.
• If \(i>48.8^\circ\), total internal reflection occurs.

Why the usual parallel-ray rule fails

A common prediction is that parallel faces always produce an emerging ray parallel to the incident ray.

That is incomplete.

For parallel faces, provided the ray transmits through both boundaries, Snell’s law gives:

\[ n_{\text{initial}}\sin i = n_{\text{final}}\sin e \]

The internal glass angle cancels because the two normals are parallel.

For the emerging ray to be parallel to the incident ray, we need:

\[ n_{\text{initial}}=n_{\text{final}} \]

In an ordinary air-glass-air arrangement, this condition is satisfied.

In this question, the initial medium is water and the final medium is air. Their refractive indices are different.

The emerging ray therefore changes direction overall, even though the glass faces remain parallel.

This is the more general rule to remember: parallel surfaces help determine the geometry of the internal ray, but the initial and final media determine whether the emerging ray returns to its original direction.

07Where to go next

Once you can trace a ray through both faces, the next useful skill is deciding what happens when the second boundary produces total internal reflection rather than refraction.

You can build on this method using How to Decide if a Ray Refracts or Reflects at a Second Boundary.

If you also need to explain why light’s speed and wavelength change when entering glass while its frequency remains constant, see Why Light’s Frequency Stays Constant When It Enters Glass.

The key skill is to treat every boundary separately: identify the two media, draw the normal, measure the correct angle, and only then apply Snell’s law.