Why Polarisation Supports a Transverse Model of Light

Rotate a second polarising filter while keeping the light beam travelling in exactly the same direction. The intensity reaching a detector can fall from a maximum to almost zero.

Why should rotating a filter around the beam change how much light gets through?

The answer is that light carries an oscillation direction perpendicular to its direction of travel. The second filter selects how much of that oscillation lies along its own transmission axis. A purely longitudinal wave has no such transverse direction to select.

This is why polarisation supports a transverse model of light: it demonstrates that light has a directional oscillation in the plane perpendicular to its propagation.

The important evidence is not simply that a filter reduces brightness. It is that the intensity after a second filter depends on the relative orientation of the two filters.

01The experiment: follow the light through two filters

Consider a controlled experiment using a stable, unpolarised light source and two linear polarising filters.

The first filter, \(P_1\), stays fixed. The second filter, \(P_2\), can rotate around the direction of the light beam.

A light sensor measures the intensity at three positions:

• \(I_0\): immediately before the first polariser.
• \(I_1\): after the first polariser and immediately before the second.
• \(I_2\): after the second polariser.

The sensor can be moved between these positions while keeping the source output constant.

Before looking at the results, make a prediction.

If the first polariser is rotated, should the amount of light passing through it change?

For an ideal linear polariser receiving unpolarised light, the answer is no. The average transmitted intensity is half the incident intensity, regardless of the filter’s orientation.

The important change happens when we introduce the second polariser.

What happens at the first polariser?

Ordinary unpolarised light contains electric-field oscillations with no single preferred linear orientation.

An ideal linear polariser selects the component of the electric field aligned with its transmission axis.

After passing through \(P_1\), the light is linearly polarised. Its electric field now oscillates along the transmission axis of the first filter.

For unpolarised incident light:

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

Here:

• \(I_0\) is the intensity before the first polariser.
• \(I_1\) is the intensity after the first polariser.
• Both intensities are measured in \(\text{W m}^{-2}\).

For example, if the incident intensity is \(120\ \text{W m}^{-2}\), the ideal transmitted intensity is:

\[ I_1=\frac{120}{2} =60\ \text{W m}^{-2} \]

This means the second polariser receives \(60\ \text{W m}^{-2}\) of already polarised light.

Notice that the first filter has done two things: it has reduced the intensity, and it has established a particular polarisation direction.

That second effect is the crucial one.

What happens at the second polariser?

The second polariser is called an analyser because it tests the polarisation direction of the incoming light.

Its transmission axis can be rotated relative to the first polariser.

Define \(\theta\) as the smallest angle between the two transmission axes, measured in the plane perpendicular to the beam. We take \(0^\circ\) to mean parallel axes and \(90^\circ\) to mean crossed axes.

There is no need to measure either axis from the room’s vertical or horizontal direction.

The second filter transmits the component of the incoming electric field that lies along its transmission axis.

If the incoming electric-field amplitude is \(E\), the transmitted component is:

\[ E_{\text{transmitted}}=E\cos\theta \]

Intensity is proportional to the square of the electric-field amplitude.

Therefore:

\[ \boxed{I_2=I_1\cos^2\theta} \]

This is Malus’ law.

The important detail is that \(I_1\) is the intensity entering the second polariser, not the original intensity from the source.

We can now predict what happens as the analyser rotates.

• At \(0^\circ\), the axes are parallel. The second filter transmits the maximum intensity.
• At \(30^\circ\), some of the electric-field component is rejected.
• At \(60^\circ\), less intensity is transmitted.
• At \(90^\circ\), the axes are perpendicular. An ideal analyser transmits no light.

The incoming light has not changed direction. Only the relative orientation of the two filters has changed.

02Reading the experimental results

Suppose the source produces unpolarised light with an initial intensity of \(120\ \text{W m}^{-2}\).

Assume both polarisers are ideal and the source remains constant.

The following is an illustrative set of idealised results, not a record of an actual experiment.

Notice two important patterns.

First, the intensity entering the second polariser stays at \(60\ \text{W m}^{-2}\). The first polariser has not changed, and the source is stable.

Second, the intensity after the second polariser changes significantly as the relative angle changes.

This is the experimental pattern we need to explain.

The second filter is not simply reducing the intensity by a fixed amount. Its transmission depends on the orientation of the polarisation already established by the first filter.

Why the relative angle matters

Imagine rotating both polarisers together while keeping the angle between their transmission axes unchanged.

Would the intensity after the second filter change?

No. Under the ideal model, it stays the same.

For example, suppose the two axes are separated by \(60^\circ\). Rotating both filters together by \(35^\circ\) does not change their relative separation.

Malus’ law still uses:

\[ I_2=I_1\cos^2(60^\circ) \]

The transmitted intensity is unchanged.

This rules out an important misconception: there is no universally special vertical or horizontal orientation that makes a polariser transmit more light.

The experiment depends on the relationship between the two axes, not their absolute orientation in the laboratory.

Why a longitudinal model struggles to explain this

Now consider two different models of a wave.

A transverse wave oscillates perpendicular to its direction of propagation. A longitudinal wave oscillates parallel to its direction of propagation.

For a wave travelling horizontally, these models make very different predictions about polarisation.

A transverse wave has different possible oscillation directions across the plane perpendicular to its motion.

A longitudinal wave does not. Its oscillation is along the direction of travel.

Think about what this means for the experiment.

If light were purely longitudinal, its oscillations would point forwards and backwards along the beam. Rotating a polariser around the beam would not reveal different transverse oscillation directions because none would exist.

Yet the experiment shows that the transmitted intensity depends on the relative orientation of the filters.

The transverse model explains this naturally:

1. The first polariser selects a particular transverse electric-field direction.
2. The second polariser receives light with that established direction.
3. Its transmission depends on the component of the electric field aligned with its own axis.
4. Rotating the second polariser changes that component and therefore changes the transmitted intensity.

A purely longitudinal model cannot account for this directional behaviour without introducing an additional transverse orientation property.

That is the key reason polarisation supports a transverse model of light.

03A useful contrasting case: one polariser is not enough

Suppose you place just one ideal polariser in front of an unpolarised light source.

You then rotate it through different orientations.

The transmitted intensity remains approximately half the original intensity.

Why doesn’t this observation establish the transverse nature of light as clearly?

Because a simple reduction in brightness could be explained by absorption. Many different physical models could account for a filter reducing the amount of transmitted energy.

There is no second filter testing an orientation established by the first.

The evidence becomes much stronger when a second polariser is introduced and its transmission changes as its axis rotates relative to the first.

This distinction is worth remembering in an HSC explanation:

• One polariser demonstrates intensity reduction.
• Two polarisers demonstrate orientation-dependent transmission.
• The dependence on relative orientation supports the existence of a transverse oscillation direction.

04Worked examples

05What the experiment establishes, and what it doesn’t

Polarisation provides strong evidence that light has transverse properties.

The experiment shows that light can carry an orientation perpendicular to its direction of propagation, and that a polariser can select a component of that orientation.

A purely longitudinal wave cannot explain this behaviour using longitudinal oscillations alone.

For light, the transverse oscillations are described by the electromagnetic wave model. The electric field and magnetic field both oscillate perpendicular to the direction of propagation, and they are perpendicular to each other.

For polarisation, we normally describe the orientation using the electric field.

There is one important limitation to the conclusion.

Polarisation does not prove that light must be described exclusively as a classical wave. Modern physics also describes light in terms of photons, which have polarisation states.

The experiment specifically supports the transverse character of light. It does not, by itself, establish every aspect of the nature of light.

For a broader explanation of the underlying wave behaviour, see Polarisation of Light and Transverse Waves for HSC Physics.

06Questions and solutions

Question 1

Unpolarised light with an initial intensity of \(96\ \text{W m}^{-2}\) passes through two ideal polarisers.

The transmission axes are separated by \(60^\circ\).

Calculate:

a) The intensity after the first polariser.

b) The intensity after the second polariser.

Solution 1

The intensity after the first polariser is \(48\ \text{W m}^{-2}\), and the final intensity is \(12\ \text{W m}^{-2}\).

Part a: Intensity after the first polariser

The incident light is unpolarised, so an ideal first polariser transmits half its intensity.

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

\[ I_1=\frac{96}{2} \]

\[ \boxed{I_1=48\ \text{W m}^{-2}} \]

Part b: Intensity after the second polariser

The light entering the second polariser is already linearly polarised.

Use Malus’ law:

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

\[ I_2=48\cos^2(60^\circ) \]

\[ I_2=48(0.25) \]

\[ \boxed{I_2=12\ \text{W m}^{-2}} \]

The second polariser transmits one-quarter of the intensity entering it.

A common mistake is to halve the intensity again. The one-half rule applies to unpolarised light entering the first ideal polariser, not to already polarised light entering the second.

Question 2

A stable, unpolarised light source produces an intensity of \(240\ \text{W m}^{-2}\).

After passing through the first polariser, the measured intensity is \(108\ \text{W m}^{-2}\).

An ideal analyser is placed after the first polariser.

The relative angle between the two transmission axes is initially \(30^\circ\). It is then increased to \(60^\circ\).

Calculate:

a) The transmitted intensity at each relative angle.

b) The factor by which the transmitted intensity changes.

Explain why the original source intensity should not be halved again when calculating the second filter’s output.

Solution 2

The transmitted intensities are \(81\ \text{W m}^{-2}\) at \(30^\circ\) and \(27\ \text{W m}^{-2}\) at \(60^\circ\). The second reading is one-third of the first.

Part a: Calculate both transmitted intensities

The intensity entering the analyser is the measured value:

\[ I_1=108\ \text{W m}^{-2} \]

At \(30^\circ\):

\[ I_2=108\cos^2(30^\circ) \]

\[ I_2=108(0.75) \]

\[ \boxed{I_2=81\ \text{W m}^{-2}} \]

At \(60^\circ\):

\[ I_2=108\cos^2(60^\circ) \]

\[ I_2=108(0.25) \]

\[ \boxed{I_2=27\ \text{W m}^{-2}} \]

Part b: Calculate the change

Compare the two transmitted intensities:

\[ \frac{81}{27}=3 \]

Therefore, the intensity at \(30^\circ\) is three times the intensity at \(60^\circ\).

The intensity decreases by:

\[ \frac{81-27}{81}\times100 =66.7\% \]

The tempting approach is to start with the original \(240\ \text{W m}^{-2}\) and apply the one-half factor again.

However, the first polariser’s output has already been measured as \(108\ \text{W m}^{-2}\). That is the intensity entering the analyser.

Using the measured intensity ensures that losses in a real first polariser are properly accounted for.

Question 3

A student conducts two experiments using the same stable, unpolarised light source.

In both experiments, the intensity entering the first polariser is unchanged, and the relative angle between the two transmission axes is \(45^\circ\).

In the second experiment, both polarisers are rotated together around the beam direction. Their relative angle remains \(45^\circ\).

The student observes that the transmitted intensities are approximately equal.

The student concludes:

The vertical orientation of the polariser must determine how much light passes through it.

Evaluate this conclusion using the experimental observations. Explain why the relative angle is more important than the absolute orientation of the filters.

Solution 3

The student’s conclusion is incorrect. The transmitted intensity depends on the relative angle between the transmission axes, not their absolute orientation in the laboratory.

Malus’ law gives:

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

In both experiments:

\[ \theta=45^\circ \]

Therefore:

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

Since:

\[ \cos^2(45^\circ)=0.5 \]

the transmitted intensity is:

\[ I_2=0.5I_1 \]

Rotating both filters together does not change this result because their relative angle stays the same.

The tempting explanation is that a particular direction, such as vertical, must be better at transmitting light.

However, if this were true, rotating both filters together would change the transmitted intensity even though their relative separation remained unchanged.

The observations instead support a model in which the first polariser establishes an electric-field orientation and the second responds to that orientation.

There is no universally preferred vertical or horizontal direction.

Question 4

A student investigates the transmission of light through two polarisers.

The intensity after the first, real polariser is measured as \(90\ \text{W m}^{-2}\).

The second polariser is ideal, and the detector has a constant background reading of \(5\ \text{W m}^{-2}\), even when no light reaches it.

The following readings are recorded:

a) Correct the readings for the detector background.

b) Show whether the corrected results agree with Malus’ law.

c) Explain why the non-zero detector reading at \(90^\circ\) does not disprove the prediction that crossed ideal polarisers block light.

Solution 4

The corrected transmitted intensities are \(90\), \(45\), and \(0\ \text{W m}^{-2}\), respectively. They agree exactly with the ideal Malus’ law predictions.

Part a: Correct for detector background

The detector always adds \(5\ \text{W m}^{-2}\) to its reading.

Therefore:

\[ I_{\text{corrected}} =I_{\text{measured}}-5 \]

The corrected readings are:

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

Part b: Compare with Malus’ law

The intensity entering the second polariser is:

\[ I_1=90\ \text{W m}^{-2} \]

At \(0^\circ\):

\[ I_2=90\cos^2(0^\circ) =90\ \text{W m}^{-2} \]

At \(45^\circ\):

\[ I_2=90\cos^2(45^\circ) =90(0.5) =45\ \text{W m}^{-2} \]

At \(90^\circ\):

\[ I_2=90\cos^2(90^\circ) =0\ \text{W m}^{-2} \]

The corrected readings match all three predictions.

Part c: Interpret the non-zero reading

The reading of \(5\ \text{W m}^{-2}\) is caused entirely by the detector background, not by transmitted light.

The tempting conclusion is that crossed polarisers must transmit some light because the detector does not display zero.

However, measurements must account for background signals before drawing conclusions.

After correcting the background, the intensity is zero, exactly as predicted for ideal crossed polarisers.

In a real experiment, additional factors such as imperfect polariser extinction and scattered light may also prevent a measured reading from reaching zero. These limitations do not invalidate the transverse-wave explanation.

Question 5

An investigator proposes testing two competing models of light.

• Model A: Light carries a transverse electric-field direction. The first polariser establishes this direction, and the second polariser transmits the component aligned with its own transmission axis.
• Model B: Light is purely longitudinal and carries only an intensity value after passing through the first filter. The second filter can respond to its own orientation, but the light carries no information about the orientation of the first filter.

A stable, unpolarised source produces an intensity of \(160\ \text{W m}^{-2}\).

Two ideal polarisers are placed in the beam. In every experiment, the intensity measured between the two polarisers is \(80\ \text{W m}^{-2}\).

The second polariser remains fixed while the first polariser is rotated.

The following results are obtained:

The investigator then rotates both polarisers together, keeping their relative angle at \(45^\circ\). The intensity after the second polariser remains \(40\ \text{W m}^{-2}\).

a) Explain why Model A predicts the first three results.

b) Explain why Model B cannot account for the changing intensity if it assumes that the light carries only a scalar intensity and no orientation information.

c) Explain what the final joint-rotation experiment establishes.

d) State why these observations support a transverse model without proving that light can only be described as a classical wave.

Solution 5

Model A correctly predicts all the observations because the first polariser establishes a transverse orientation that determines how much light the second polariser transmits. Model B cannot explain the changing output using intensity alone.

Part a: Explain the predictions of Model A

The source produces unpolarised light with:

\[ I_0=160\ \text{W m}^{-2} \]

After the first ideal polariser:

\[ I_1=\frac{160}{2} =80\ \text{W m}^{-2} \]

The light leaving the first polariser is now linearly polarised.

When the two axes are parallel:

\[ I_2=80\cos^2(0^\circ) \]

\[ \boxed{I_2=80\ \text{W m}^{-2}} \]

When the axes are separated by \(45^\circ\):

\[ I_2=80\cos^2(45^\circ) \]

\[ \boxed{I_2=40\ \text{W m}^{-2}} \]

When the axes are crossed:

\[ I_2=80\cos^2(90^\circ) \]

\[ \boxed{I_2=0\ \text{W m}^{-2}} \]

These are exactly the observed results.

Part b: Explain the problem with Model B

Model B assumes that the first polariser only reduces the intensity of the incoming light.

However, the intensity entering the second polariser is identical in all three experiments:

\[ I_1=80\ \text{W m}^{-2} \]

The second polariser also remains fixed.

If the incoming light carries only a scalar intensity and no information about the first polariser’s orientation, Model B predicts that the second polariser should produce the same output in all three experiments.

But the measured output changes from \(80\) to \(40\) to \(0\ \text{W m}^{-2}\).

The missing ingredient is an orientation carried by the light between the filters.

Model A provides this through the transverse electric-field direction.

A model that tries to rescue Model B by allowing the first filter to imprint a directional property on the light must explain how that property is oriented in the plane perpendicular to the beam.

In doing so, it has introduced the transverse directional feature that the original model lacked.

Part c: Explain the joint-rotation experiment

Rotating both polarisers together does not change their relative angle.

The second polariser still receives light whose electric field is oriented at \(45^\circ\) to its transmission axis.

Therefore:

\[ I_2=80\cos^2(45^\circ) \]

\[ \boxed{I_2=40\ \text{W m}^{-2}} \]

The unchanged reading shows that the result does not depend on a special absolute orientation in the laboratory.

It depends on the relative orientation of the incoming polarisation and the analyser.

Part d: State the overall conclusion

The experiments strongly support the transverse-wave model because the light carries a direction of oscillation perpendicular to its propagation.

A purely longitudinal wave cannot provide the different transverse orientations selected by the two filters.

However, this does not prove that light is exclusively a classical wave. Modern photon descriptions also include polarisation.

The experiment establishes the transverse character of light, rather than every detail of its physical description.

The key reasoning to carry into other HSC Physics questions is simple: a change in intensity alone is not the strongest evidence. A change in intensity caused by changing the relative orientation of two polarisers reveals that light carries a transverse directional property.

The next useful connection is the distinction between transverse and longitudinal mechanical waves. Comparing light with sound makes it clearer why sound waves travelling through air cannot be polarised in the same way.