Can Sound Waves in Air Be Polarised? A Physics Comparison

Place two polarising filters in front of a lamp and rotate the second one. At certain angles, the light becomes almost completely blocked. The filters are selecting a particular direction of oscillation in the light wave.

Could you do the same thing with sound? Could two filters block a sound wave when their axes are at right angles?

No. Ordinary sound waves travelling through air cannot be polarised like light. The reason comes down to how the two waves oscillate. Light is transverse, so its oscillations have a direction perpendicular to its travel. Sound in air is longitudinal, so air particles oscillate parallel to the direction the sound travels.

That difference determines whether polarisation is possible.

01What does polarisation actually mean?

Start by separating two directions:

• The direction of propagation: the direction the wave travels.
• The direction of oscillation: the direction in which the wave’s field or particles move.

For a transverse wave, these directions are perpendicular. For a longitudinal wave, they are parallel.

Imagine flicking one end of a rope sideways. The disturbance travels along the rope, but the rope itself moves up and down. You could also make the rope move sideways in another transverse direction if the arrangement allowed it.

That freedom to oscillate in different directions perpendicular to the wave’s travel is what makes polarisation possible.

Why light can be polarised

Light is an electromagnetic wave. Its electric field oscillates perpendicular to the direction in which the wave travels.

For linearly polarised light, the electric field oscillates in one consistent transverse direction.

A polarising filter has a transmission axis. It allows the component of the electric field aligned with that axis to pass through, while absorbing or rejecting the perpendicular component.

Rotate the filter, and you change which component is transmitted.

This is the physical basis of light polarisation.

Why sound in air cannot be polarised

Sound in air behaves differently.

As a sound wave travels, air particles oscillate backwards and forwards around their equilibrium positions. Their motion is parallel to the direction of propagation.

The wave consists of alternating compressions and rarefactions, which produce changes in air pressure.

There is no transverse oscillation direction for a polarising filter to select.

This is not because sound is too weak or because its wavelength is too large. It is a consequence of the type of wave.

If you need to revisit the particle motion behind this distinction, see Sound as a Longitudinal Pressure Wave: HSC Physics Guide.

02The polarising filter experiment: what happens to light?

A useful way to understand polarisation is to follow the intensity of light through two filters.

Consider an experimental arrangement with:

• A lamp producing unpolarised light.
• A first polarising filter, \(P_1\).
• A second polarising filter, \(P_2\), which can rotate.
• A light sensor measuring the transmitted intensity.

The light travels in a straight line through both filters. The first filter remains fixed, while the second rotates around the direction of propagation.

What should happen when the second filter rotates?

Before calculating anything, predict the two extreme cases:

• If both transmission axes are parallel, how much light should the second filter block?
• If the axes are perpendicular, should any light get through?

The first arrangement allows maximum transmission through the second filter. The second ideally blocks all the light.

To understand the intermediate angles, we need to track the intensity at each stage.

Step 1: Light enters the first polarising filter

Let the initial intensity of unpolarised light be \(I_0\).

An ideal polarising filter transmits half the intensity of unpolarised light.

Therefore,

\[
I_1=\frac{I_0}{2}
\]

The transmitted light is now linearly polarised along the transmission axis of the first filter.

Why does the intensity halve?

Unpolarised light contains electric field oscillations in randomly distributed transverse directions. The first polariser selects the component along its transmission axis.

When averaged over all incoming orientations, half the original intensity is transmitted.

Step 2: Light enters the second polarising filter

Now introduce the relative angle between the filters.

Call the transmission axis of the first filter A and the transmission axis of the second filter B.

Let \(\theta\) be the angle between A and B.

The second filter receives light that is already linearly polarised along A. It transmits the component aligned with B.

The transmitted intensity is given by Malus’s law:

\[
I_2=I_1\cos^2\theta
\]

Substituting the intensity after the first filter:

\[
\boxed{I_2=\frac{I_0}{2}\cos^2\theta}
\]

Notice something important: the intensity entering the second filter is \(I_1\), not the original intensity \(I_0\).

You must calculate the effect of each filter in sequence.

What happens at different relative angles?

Suppose the initial unpolarised light has an intensity of \(100\ \text{W m}^{-2}\).

The first filter always transmits:

\[
I_1=\frac{100}{2}=50\ \text{W m}^{-2}
\]

The following table uses illustrative theoretical values for ideal filters.

All intensities are in \(\text{W m}^{-2}\).

Three results are particularly important:

• At \(0^\circ\), the axes are parallel. The second filter transmits all the light reaching it.
• At \(45^\circ\), the second filter transmits half the light reaching it.
• At \(90^\circ\), the axes are perpendicular. An ideal second filter transmits no light.

Real filters are not perfect, so a small amount of light may still get through crossed filters.

Also, the half-intensity rule applies to the first filter only when the incoming light is unpolarised. If the incoming light is already polarised, you must use Malus’s law with the angle between its existing polarisation direction and the filter’s transmission axis.

03What happens if we add a third polarising filter?

Here is a result that initially seems strange.

Suppose you have two polarising filters with perpendicular transmission axes. No light should pass through both.

Now insert a third filter between them, with its transmission axis at an intermediate angle.

You might predict that the third filter cannot make any difference. After all, the first and last filters are still perpendicular.

That prediction is wrong.

The middle filter changes the polarisation direction of the light that passes through it. The final filter now receives light polarised along the middle filter’s axis, rather than the first filter’s axis.

Some light can therefore pass through the final filter.

04Why the same filter experiment does not work for sound in air

Now return to the original question.

Could you put two filters in front of a speaker and rotate the second one until the sound disappears?

Not through polarisation.

A sound wave travelling through air produces alternating compressions and rarefactions. Air particles oscillate parallel to the direction of propagation.

Imagine a speaker on the left producing sound that travels horizontally to the right.

The air particles move backwards and forwards horizontally. They do not oscillate vertically or sideways as part of an ordinary plane sound wave.

This means there is no equivalent of the electric field direction in a light wave for a polarising filter to select.

A sheet with vertical slots cannot select vertical sound oscillations because ordinary sound in air does not have those oscillations in the first place.

But can a physical barrier reduce sound intensity?

Absolutely.

A barrier can reduce the intensity of sound passing through it. It might absorb some acoustic energy, reflect sound back towards the source, or scatter the wave through diffraction.

For example, imagine placing a slotted panel between a speaker and a microphone. Rotating the panel could change the measured sound intensity because its geometry changes the way sound is transmitted or scattered.

That does not mean the sound has become polarised.

The distinction is important:

A reduction in intensity alone is not evidence of polarisation.

To demonstrate polarisation, you need a transverse oscillation direction that can be selectively transmitted or blocked.

A contrasting case: Can sound in a solid be polarised?

Here is an important qualification.

Not all mechanical waves are longitudinal.

Solids can support transverse shear waves. In these waves, particles oscillate perpendicular to the direction of propagation.

For example, imagine a shear wave travelling horizontally through a metal block while the material particles oscillate vertically.

A different shear wave could travel in the same direction but have particles oscillating in another perpendicular direction.

These transverse waves can be polarised.

Why can this happen in a solid but not in ordinary air?

A solid can resist shear deformation. When its particles are displaced sideways relative to neighbouring particles, restoring forces can support a transverse wave.

Air behaves as a fluid and has essentially no shear rigidity. It cannot support an ordinary freely propagating bulk transverse shear wave.

This is why ordinary sound travelling through air is longitudinal, whereas transverse mechanical waves can exist in solids.

Be careful with the wording in an exam:

• Ordinary sound waves in air cannot be polarised.
• Transverse mechanical waves, including shear waves in solids, can be polarised.

The statement that sound can never be polarised is too broad. The medium and type of wave matter.

05How to recognise whether a wave can be polarised

When an exam question asks whether a wave can be polarised, don’t start by memorising examples.

Start by identifying the direction of oscillation.

Use this decision process:

1. Identify the direction of propagation.

Which way does the wave travel?

2. Identify the direction of oscillation.

What is actually moving, or which field is oscillating?

3. Compare the two directions.

• Perpendicular: the wave is transverse and can be polarised.
• Parallel: the wave is longitudinal and cannot be polarised in the ordinary linear sense.

4. Identify what the filter selects.

A polarising filter must select a particular transverse oscillation direction. Simply absorbing or reflecting some wave energy is not polarisation.

There is one final trap worth remembering.

A sine-shaped graph does not automatically represent a transverse wave. A sound wave can be represented by a sinusoidal pressure-versus-position graph, but that graph shows how pressure varies. It does not show air particles physically moving up and down.

Always identify what the graph’s axes represent before deciding whether a wave is transverse or longitudinal.

06Questions and solutions

Question 1

Unpolarised light with an intensity of \(80\ \text{W m}^{-2}\) passes through two ideal polarising filters.

The transmission axis of the second filter is at \(45^\circ\) relative to the first.

Calculate:

a) The intensity immediately after the first filter.

b) The intensity immediately after the second filter.

c) The percentage of the original intensity that is transmitted.

Solution 1

The final intensity is \(20\ \text{W m}^{-2}\), which is \(25\%\) of the original intensity.

Part a: Intensity after the first filter

Since the incoming light is unpolarised, the first ideal filter transmits half its intensity.

\[
I_1=\frac{I_0}{2}
\]

\[
I_1=\frac{80}{2}
\]

\[
\boxed{I_1=40\ \text{W m}^{-2}}
\]

Part b: Intensity after the second filter

The second filter is at \(45^\circ\) relative to the first.

Apply Malus’s law:

\[
I_2=I_1\cos^2 45^\circ
\]

\[
I_2=40(0.7071)^2
\]

\[
\boxed{I_2=20\ \text{W m}^{-2}}
\]

Part c: Percentage transmitted

Compare the final intensity with the original intensity.

\[
\text{Percentage}
=\frac{I_2}{I_0}\times100
\]

\[
=\frac{20}{80}\times100
\]

\[
\boxed{25\%}
\]

The first filter transmits \(50\%\) of the original intensity, and the second transmits \(50\%\) of what reaches it.

The two filters therefore transmit \(25\%\) of the original intensity.

Question 2

A beam of already linearly polarised light has an intensity of \(64\ \text{W m}^{-2}\).

It enters a polarising filter P1. The incoming electric field direction is at \(20^\circ\) relative to the transmission axis A of P1.

A second filter P2 has transmission axis B at \(50^\circ\) relative to A, on the same side.

Calculate:

a) The intensity after P1.

b) The intensity after P2.

c) The intensity after P2 if P1 were removed.

Explain why the intensity in part (b) can be greater than the intensity in part (c).

Solution 2

The intensity after P2 is approximately \(23.4\ \text{W m}^{-2}\). Without P1, it would be approximately \(7.49\ \text{W m}^{-2}\).

The key difference from Question 1 is that the incoming light is already polarised. Therefore, the first filter does not automatically halve the intensity.

Part a: Intensity after P1

The incoming electric field is at \(20^\circ\) relative to transmission axis A.

Apply Malus’s law:

\[
I_1=64\cos^2 20^\circ
\]

\[
I_1=64(0.8830)
\]

\[
\boxed{I_1=56.5\ \text{W m}^{-2}}
\]

The light leaving P1 is now polarised along axis A.

Part b: Intensity after P2

The relative angle between axes A and B is \(50^\circ\).

The intensity entering P2 is \(56.5\ \text{W m}^{-2}\).

\[
I_2=56.5\cos^2 50^\circ
\]

\[
I_2=56.5(0.4132)
\]

\[
\boxed{I_2=23.4\ \text{W m}^{-2}}
\]

Part c: Remove P1

Without P1, the incoming polarisation direction remains at \(20^\circ\) relative to A.

Since B is a further \(50^\circ\) from A on the same side, the angle between the incoming electric field and B is:

\[
\theta=20^\circ+50^\circ
=70^\circ
\]

Apply Malus’s law directly:

\[
I=64\cos^2 70^\circ
\]

\[
I=64(0.1170)
\]

\[
\boxed{I=7.49\ \text{W m}^{-2}}
\]

The intensity is greater when P1 is present because P1 selects the component of the incoming electric field aligned with A. The light that leaves P1 is then polarised along A, making its relative angle to P2 smaller.

P1 absorbs some of the original light, but it changes the polarisation direction of the transmitted light.

This is why an intermediate polariser can increase the final transmitted intensity compared with using only the last filter.

Question 3

A student places a slotted panel between a loudspeaker and a microphone.

When the panel is rotated, the microphone reading decreases significantly. The student concludes that the sound wave must be polarised because its intensity depends on the panel’s orientation.

Explain why this conclusion is incorrect.

In your answer:

a) Describe the particle motion in the sound wave.

b) Identify two other physical effects that could explain the change in microphone reading.

c) Explain what feature would need to be demonstrated to establish polarisation.

Solution 3

The student’s conclusion is incorrect because a change in transmitted intensity does not establish polarisation.

Part a: Particle motion

Sound in air is longitudinal.

Air particles oscillate backwards and forwards parallel to the direction in which the sound wave travels.

The wave produces alternating compressions and rarefactions, rather than transverse particle oscillations.

There is therefore no transverse oscillation direction for the slotted panel to select.

Part b: Other possible explanations

Two possible explanations are:

• Reflection: the panel can reflect some of the incoming acoustic energy away from the microphone.
• Diffraction and scattering: the panel’s slots and edges can redirect sound waves, changing how much energy reaches the microphone.

Absorption and changes in the microphone’s directional response could also affect the reading.

Part c: What would establish polarisation?

The experiment would need to demonstrate that the wave contains a transverse oscillation with a preferred direction, and that an analyser selectively transmits the component aligned with its axis.

Ordinary sound in air has no such transverse oscillation direction.

The student has demonstrated that the panel changes sound transmission, not that sound has been polarised.

The distinction is between filtering energy and selecting a polarisation component.

Question 4

A mechanical wave travels horizontally through a solid metal block.

The particles in the block oscillate vertically, perpendicular to the direction of propagation.

A second wave could travel in the same direction while its particles oscillate in another perpendicular direction.

Explain:

a) Whether the first wave can be polarised.

b) Why this type of wave can exist in a solid.

c) Why an ordinary sound wave travelling through air cannot behave in the same way.

Solution 4

The wave in the metal block can be polarised because it is a transverse mechanical wave.

Part a: Polarisation

The wave travels horizontally, while the particles oscillate vertically.

The oscillation is perpendicular to the direction of propagation.

A transverse wave can have different oscillation directions within the plane perpendicular to propagation. A particular wave with a fixed oscillation direction is linearly polarised.

Therefore, the described wave is polarised.

Part b: Why it exists in a solid

Solids can resist shear deformation.

When particles are displaced sideways relative to their neighbours, restoring forces act to return them towards their equilibrium positions.

These forces allow transverse shear waves to propagate through the solid.

Part c: Why ordinary air sound is different

Air is a fluid and does not have the shear rigidity required to support an ordinary freely propagating bulk transverse shear wave.

Sound in air instead travels through compressions and rarefactions, with air particles oscillating parallel to propagation.

The important distinction is not simply whether the wave is mechanical. It is whether the medium and wave type allow transverse oscillations.

Question 5

A student attempts to demonstrate sound polarisation using a directional microphone.

A loudspeaker produces a steady sound wave travelling horizontally towards the microphone.

The microphone is mounted on a rotating stand. Its front-facing direction is rotated relative to the incoming sound.

For this question, assume an idealised directional microphone with a response proportional to \(\cos^2\phi\), where \(\phi\) is the angle between its front-facing direction and the incoming sound direction.

The microphone produces the following illustrative readings:

A second, calibrated omnidirectional pressure microphone at the same position records essentially unchanged pressure amplitude as the directional microphone rotates.

A particle-motion probe also indicates that the air particles oscillate parallel to the direction of propagation.

The student argues:

“The microphone response follows a cosine-squared relationship, just like Malus’s law. Therefore, the sound must be polarised.”

Evaluate this argument.

Your response should explain:

a) Why the directional microphone’s readings follow the observed pattern.

b) Why this is not evidence of sound polarisation.

c) The geometric difference between rotating a light polariser and rotating the microphone.

d) What the additional measurements tell us about the sound wave.

Solution 5

The student’s argument is incorrect. The cosine-squared relationship describes the directional microphone’s response, not the polarisation of the sound wave.

This is a particularly important distinction because the mathematical pattern resembles Malus’s law.

Part a: Why does the microphone reading change?

A directional microphone is more sensitive to sound arriving from certain directions than others.

When its front-facing direction points towards the speaker, it gives its maximum response.

As it rotates away, its sensitivity to the incoming sound decreases.

The question specifies an idealised response:

\[
R=\cos^2\phi
\]

At \(45^\circ\):

\[
R=\cos^2 45^\circ
=0.50
\]

At \(90^\circ\):

\[
R=\cos^2 90^\circ
=0
\]

The readings therefore follow the specified directional response.

Part b: Why is this not polarisation?

Polarisation requires a transverse oscillation direction that can be selected by an analyser.

The air particles are oscillating parallel to the direction of propagation.

The sound wave therefore does not have the transverse polarisation direction required for the light-style experiment.

The microphone is simply less sensitive when it is turned away from the incoming sound.

Part c: Compare the geometry

This is the most important part of the question.

When a light polariser rotates, the direction of light propagation remains fixed.

The filter’s transmission axis rotates within the plane perpendicular to the beam. It selects a different component of the light’s transverse electric field.

In the microphone experiment, the microphone’s front-facing direction changes relative to the direction from which the sound arrives.

The microphone is changing its directional sensitivity, not selecting a transverse oscillation component.

These are physically different situations, even though both can produce a cosine-squared mathematical relationship.

Part d: Interpret the additional measurements

The omnidirectional microphone records essentially unchanged pressure amplitude.

This indicates that the sound pressure at the measurement position has not changed significantly as the directional microphone rotates.

The particle-motion probe confirms that the air particles oscillate parallel to the direction of propagation.

Together, these measurements support the conclusion that the change in the first microphone’s readings is due to its directional response, rather than sound polarisation.

The deeper lesson is that observing a familiar mathematical relationship is not enough to establish that two physical systems behave in the same way. You must identify what the equation represents and what the apparatus is actually measuring.

07Where this concept leads next

Polarisation is a useful test of whether a wave is transverse, but it also introduces a broader way of thinking about wave behaviour.

When you study interference, diffraction, and standing waves, you will encounter situations where waves interact with barriers, filters, and one another.

The important habit is to distinguish the direction a wave travels from the direction in which its oscillations occur.

For light, that distinction makes polarisation possible. For ordinary sound in air, it explains why polarisation is not possible, regardless of how cleverly you arrange the filters.