What Electric-Field Direction Means in Polarisation Diagrams

A light beam travels from left to right across a page. Beside it, a polarisation diagram shows a vertical electric-field arrow. Does that mean the light is travelling upwards?

No. The light travels horizontally, while its electric field oscillates vertically. The arrow represents the direction of the electric field’s oscillation, not the direction in which the light moves.

This distinction becomes important when light passes through polarising filters. The direction of the electric field determines how much light a filter transmits, so reading that arrow correctly is the first step towards applying Malus’ law.

01What does the electric-field direction actually represent?

Light is a transverse electromagnetic wave. This means its electric and magnetic fields oscillate perpendicular to the direction in which the wave travels.

For linearly polarised light, the electric field oscillates along one fixed direction.

Imagine a beam travelling horizontally. Its electric field oscillates vertically.

At one instant, the electric field might point upwards. Half a cycle later, it points downwards. It then reverses again, repeatedly.

The important detail is that the electric field continues oscillating along the same vertical line.

This is what we mean by the direction of polarisation.

Direction of oscillation versus direction of propagation

There are two different directions to keep separate:

• Direction of propagation: The direction in which the light wave travels.
• Electric-field direction: The direction along which the electric field oscillates.

For example, if light travels horizontally and its electric field oscillates vertically, the wave is vertically polarised.

The electric field does not make the light travel upwards and downwards. Instead, the field changes direction at each point as the electromagnetic wave propagates forwards.

A useful analogy is a swing moving backwards and forwards along a north-south line. Its direction of motion changes, but its line of oscillation remains north-south.

The analogy has a limit: an electric field is not a physical object swinging through space. It is a field whose value and direction change with time.

Why do some diagrams use double-headed arrows?

You might encounter two different ways of representing the electric field:

• A double-headed arrow represents the axis along which the electric field oscillates.
• A single-headed arrow can represent the instantaneous electric-field vector at a particular moment.

The distinction is important because an instantaneous electric-field vector reverses direction every half-cycle.

However, the polarisation direction of a linearly polarised wave remains unchanged.

This also explains why polarisation axes are treated as lines rather than arrows when calculating angles between filters.

An axis at \(0^\circ\) represents the same physical orientation as an axis at \(180^\circ\).

Is the electric-field direction also the direction of an electric force?

There is a useful connection here.

An electric field points in the direction of the electric force on a positive test charge placed at rest at that point.

Therefore, at a particular instant, the electric-field vector in a light wave has the same directional meaning.

However, there is an important difference between the situations:

• An electrostatic field around stationary charges can remain constant.
• The electric field in an electromagnetic wave oscillates and reverses direction as the wave passes.

If you want to explore how electric-field direction determines the force on positive and negative charges, see Force on a Charge in an Electric Field: HSC Physics Guide.

02How polarising filters use the electric-field direction

Now consider what happens when this light reaches a polarising filter.

A polarising filter has a property called its transmission axis. This is the direction along which the filter allows the electric-field component of the light to pass.

For example, a filter with a vertical transmission axis allows the vertical component of the incoming electric field through.

The perpendicular component is blocked in the ideal polariser model.

What happens when the axes are at an angle?

Suppose incoming light has an electric-field amplitude \(E\), and its polarisation direction makes an angle \(\theta\) with the filter’s transmission axis.

The filter transmits the component parallel to its transmission axis.

Using basic vector resolution:

\[ E_{\parallel}=E\cos\theta \]

The transmitted electric-field amplitude is therefore reduced by a factor of \(\cos\theta\).

However, light intensity is proportional to the square of the electric-field amplitude:

\[ I\propto E^2 \]

Squaring the amplitude relationship gives Malus’ law:

\[ I_{\text{out}}=I_{\text{in}}\cos^2\theta \]

Here:

• \(I_{\text{in}}\) is the intensity immediately before the filter, measured in \(\mathrm{W\,m^{-2}}\).
• \(I_{\text{out}}\) is the intensity immediately after the filter, measured in \(\mathrm{W\,m^{-2}}\).
• \(\theta\) is the smaller angle between the incoming electric-field oscillation axis and the filter’s transmission axis.

Three particular angles are worth remembering.

At \(0^\circ\), the electric field is already aligned with the transmission axis.

At \(45^\circ\), the filter transmits half the incoming intensity.

At \(90^\circ\), the electric field has no component parallel to the transmission axis, so an ideal filter transmits no light.

Notice that the \(45^\circ\) case transmits half the intensity, not half the electric-field amplitude. This is because intensity depends on the square of the amplitude.

03Why relative angles matter more than absolute angles

This is where reading a polarisation diagram correctly becomes especially useful.

Suppose light passes through two filters, P1 and P2.

P1 has a transmission axis at \(30^\circ\) to the horizontal.

P2 has a transmission axis at \(80^\circ\) to the horizontal.

Which angle should you use when applying Malus’ law to P2?

The answer is \(50^\circ\), not \(80^\circ\).

Why?

Once light passes through P1, its surviving electric field is polarised along P1’s transmission axis.

Therefore, the incoming electric field at P2 is oriented at \(30^\circ\), not at its original direction.

The relevant angle is the separation between the two filters:

\[ \Delta\theta=80^\circ-30^\circ \]

\[ \Delta\theta=50^\circ \]

This gives you an important rule:

After every ideal polarising filter, the transmitted light is polarised along that filter’s transmission axis.

The next filter must be compared with this new direction.

Remember that transmission axes are lines

Consider two filters with axes at \(165^\circ\) and \(20^\circ\), measured anticlockwise from the same reference direction.

Their absolute angle difference is:

\[ |165^\circ-20^\circ|=145^\circ \]

But the physical separation between the two axes is:

\[ \Delta\theta=180^\circ-145^\circ \]

\[ \Delta\theta=35^\circ \]

The axes are only \(35^\circ\) apart.

In general, when both axis angles are measured between \(0^\circ\) and \(180^\circ\), use:

\[ \Delta\theta= \min\left( |\alpha-\beta|, 180^\circ-|\alpha-\beta| \right) \]

Here, \(\alpha\) and \(\beta\) are the two transmission-axis angles.

The resulting angle lies between \(0^\circ\) and \(90^\circ\).

Keep track of the intensity before each filter

Finding the correct angle is only half the job. You must also use the intensity that actually reaches each filter.

For three filters, the sequence is:

The calculations follow:

\[ I_1=I_0\cos^2\Delta\theta_1 \]

\[ I_2=I_1\cos^2\Delta\theta_2 \]

\[ I_3=I_2\cos^2\Delta\theta_3 \]

Each new intensity becomes the input for the next filter.

There is one special case.

If the incoming light is initially unpolarised, the first ideal polariser transmits half its intensity:

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

The transmitted light is then plane-polarised along P1’s transmission axis.

You apply Malus’ law normally to subsequent filters. You do not halve the intensity again at every filter.

04The experiment that tests the meaning of the electric-field direction

How could you test whether the electric-field direction really determines the amount of light transmitted?

A simple experiment uses two polarising filters and a light detector.

Setting up the apparatus

Arrange the equipment in the order that the light travels:

1. Place a steady light source on the left.
2. Place a fixed polarising filter, P1, in front of the source.
3. Position a second polarising filter, P2, after P1. Mount P2 so that its transmission axis can rotate.
4. Place a light detector behind P2 to measure the transmitted intensity.

The beam travels horizontally through the apparatus. Both transmission axes lie in the plane perpendicular to the beam.

P1 remains fixed, while P2 rotates around the direction of light propagation.

What should the experiment show?

First, align the transmission axes of P1 and P2.

The axes are parallel, so the detector should record the maximum transmitted intensity.

Now rotate P2 gradually.

If Malus’ law is correct, the intensity should decrease as the relative angle increases.

For example, suppose the intensity immediately after P1 is \(100\) arbitrary intensity units.

The following table gives the predicted detector readings.

These are illustrative idealised predictions, not measurements from an actual experiment.

What does the observation tell us?

If the detector reading follows the predicted pattern, it supports the idea that a polariser transmits the electric-field component parallel to its transmission axis.

The detector does not directly measure the direction of the electric field. It measures light intensity, which depends on the square of the transmitted electric-field amplitude.

This is an important distinction.

The experiment supports the projection model, but it cannot tell you whether the electric field points upwards or downwards at a particular instant. The detector averages the rapidly oscillating light signal over time.

Why does the experimental design need controls?

Suppose the detector still records a small intensity when the two filters are crossed at \(90^\circ\).

Does that automatically mean Malus’ law is incorrect?

No.

The detector might be receiving background light, or the filters might not be perfectly aligned. Real polarising filters can also allow a small amount of unwanted light through.

A better experimental procedure includes several controls:

• Record the detector’s background reading with the beam blocked.
• Measure a reference intensity with the filters parallel.
• Subtract the background reading from subsequent measurements.
• Keep the light source, detector, and beam alignment unchanged.
• Compare the corrected readings with the theoretical predictions.

This makes it easier to distinguish genuine transmission from experimental error.

The HSC ideal-filter model assumes that the unwanted perpendicular component is completely blocked. Real equipment only approximates this behaviour.

05Worked examples

06Questions and solutions

Question 1

A linearly polarised light beam travels horizontally from left to right.

A polarisation diagram shows a single electric-field arrow pointing vertically upwards.

Explain:

(a) The direction in which the light propagates.

(b) What the electric-field arrow represents.

(c) What happens to the instantaneous electric-field vector half a cycle later. Does the polarisation direction change?

Solution 1

The light propagates horizontally, its electric field oscillates vertically, and the instantaneous electric-field vector reverses direction after half a cycle without changing the polarisation direction.

(a) The light travels horizontally from left to right. This is its direction of propagation.

(b) The vertical arrow represents the electric-field direction. For linearly polarised light, the electric field oscillates along a fixed vertical line, perpendicular to the direction of propagation.

(c) Half a cycle later, the instantaneous electric-field vector points downwards instead of upwards.

However, the electric field is still oscillating along the same vertical axis.

Therefore, the polarisation direction has not changed.

The key distinction is that the instantaneous vector can reverse while the oscillation axis remains fixed.

Question 2

A beam of vertically polarised light has an intensity of \(80\ \mathrm{W\,m^{-2}}\) immediately before an ideal analyser.

At one instant, the electric-field vector points upwards. Half a cycle later, it points downwards.

The analyser has a transmission axis at \(60^\circ\) to the vertical.

Calculate the transmitted intensity at both instants and explain why the downward-pointing electric field is not completely blocked.

Solution 2

The transmitted intensity is \(20\ \mathrm{W\,m^{-2}}\) at both instants because reversing the electric-field vector does not change its oscillation axis or the magnitude of its projection onto the analyser’s transmission axis.

Step 1: Identify the incoming intensity.

\[ I_{\text{in}}=80\ \mathrm{W\,m^{-2}} \]

Step 2: Find the angle between the polarisation axis and the analyser.

The incoming electric field oscillates vertically, and the analyser is at \(60^\circ\) to the vertical.

The smaller angle between their axes is:

\[ \theta=60^\circ \]

Step 3: Apply Malus’ law.

\[ \begin{aligned} I_{\text{out}} &=80\cos^2(60^\circ)\\ &=80(0.25)\\ &=20\ \mathrm{W\,m^{-2}} \end{aligned} \]

Step 4: Consider the second instant.

The electric-field vector now points downwards.

If you compare the directed vectors, their angle to the analyser can be described as \(120^\circ\). However:

\[ \cos^2(120^\circ)=0.25 \]

The transmitted intensity is therefore unchanged.

The filter responds to the electric-field component along its transmission axis, not to whether the instantaneous vector points upwards or downwards.

Question 3

A plane-polarised beam has an initial intensity of \(96\ \mathrm{W\,m^{-2}}\). Its electric-field oscillation axis is at \(10^\circ\) to the horizontal.

The beam passes through two ideal polarising filters:

• P1 has a transmission axis at \(40^\circ\).
• P2 has a transmission axis at \(130^\circ\).

Calculate the intensity after each filter.

Explain why comparing P2 directly with the original \(10^\circ\) polarisation direction gives an incorrect result.

Solution 3

The intensity after P1 is \(72\ \mathrm{W\,m^{-2}}\), and the intensity after P2 is zero for ideal filters.

The reason is that P2 is perpendicular to the polarisation direction produced by P1.

Step 1: Calculate the intensity after P1.

The incoming polarisation direction is \(10^\circ\), while P1 is at \(40^\circ\).

Therefore:

\[ \theta_1=40^\circ-10^\circ=30^\circ \]

Apply Malus’ law:

\[ \begin{aligned} I_1&=96\cos^2(30^\circ)\\ &=96(0.75)\\ &=72\ \mathrm{W\,m^{-2}} \end{aligned} \]

The light leaving P1 is now polarised at \(40^\circ\).

Step 2: Find the angle for P2.

This is where the important change occurs.

The incoming electric field at P2 is aligned with P1, at \(40^\circ\).

Therefore:

\[ \theta_2=130^\circ-40^\circ \]

\[ \theta_2=90^\circ \]

The two transmission axes are perpendicular.

Step 3: Calculate the intensity after P2.

\[ \begin{aligned} I_2&=72\cos^2(90^\circ)\\ &=72(0)\\ &=0\ \mathrm{W\,m^{-2}} \end{aligned} \]

No light is transmitted through the ideal second filter.

Why is comparing P2 with the original direction incorrect?

A student might compare P2’s \(130^\circ\) axis with the original \(10^\circ\) direction.

The smaller angle between those axes is \(60^\circ\), which might suggest some light should pass through.

But P2 does not receive light polarised at \(10^\circ\).

P1 has already changed the polarisation direction to \(40^\circ\). P2 must be compared with this new direction.

This is why tracking the electric-field direction after every filter is just as important as tracking the intensity.

Question 4

A beam of plane-polarised light passes through P1.

Immediately after P1:

• The intensity is \(64\ \mathrm{W\,m^{-2}}\).
• The electric-field oscillation axis is \(25^\circ\) anticlockwise from a reference direction.

The light then passes through a second ideal analyser, P2. A detector measures an intensity of \(16\ \mathrm{W\,m^{-2}}\) after P2.

All transmission-axis angles are measured between \(0^\circ\) and \(180^\circ\).

Determine the possible transmission-axis orientations of P2.

Can the detector reading tell you whether P2 is clockwise or anticlockwise from P1? Explain.

Solution 4

P2 can have a transmission axis at either \(85^\circ\) or \(145^\circ\). The detector reading cannot distinguish between these two orientations because Malus’ law depends on the square of the cosine.

Step 1: Identify the intensity before P2.

The intensity immediately before the second analyser is:

\[ I_{\text{in}}=64\ \mathrm{W\,m^{-2}} \]

The measured output is:

\[ I_{\text{out}}=16\ \mathrm{W\,m^{-2}} \]

Step 2: Apply Malus’ law.

\[ 16=64\cos^2\theta \]

Divide both sides by 64:

\[ \cos^2\theta=\frac{16}{64} \]

\[ \cos^2\theta=0.25 \]

Taking the square root:

\[ \cos\theta=0.5 \]

For the smaller angle between two transmission axes:

\[ \theta=60^\circ \]

Step 3: Find the possible absolute orientations of P2.

P1 is at \(25^\circ\).

P2 could be \(60^\circ\) anticlockwise from P1:

\[ 25^\circ+60^\circ=85^\circ \]

Alternatively, P2 could be \(60^\circ\) clockwise from P1:

\[ 25^\circ-60^\circ=-35^\circ \]

Because transmission axes repeat every \(180^\circ\):

\[ -35^\circ+180^\circ=145^\circ \]

Therefore, the two possible orientations are:

\[ \boxed{85^\circ\text{ or }145^\circ} \]

Step 4: Explain the experimental limitation.

Both orientations are \(60^\circ\) from P1’s transmission axis.

Malus’ law gives the same intensity for both because the cosine is squared.

A detector measuring intensity alone cannot distinguish between clockwise and anticlockwise orientations.

It tells you the magnitude of the angular separation, but not which side of P1 the second analyser occupies.

Question 5

A beam of light is already plane-polarised along the \(0^\circ\) transmission axis of P1.

Immediately after P1, its intensity is \(120\ \mathrm{W\,m^{-2}}\).

A second ideal polariser, M, is inserted after P1. Its transmission axis is set at an unknown angle \(\alpha\), where:

\[ 0^\circ\leq\alpha\leq90^\circ \]

A final ideal analyser, P3, has a fixed transmission axis at \(90^\circ\) to P1.

Two detectors record the following readings:

• A detector immediately after M reads \(90\ \mathrm{W\,m^{-2}}\).
• A detector after P3 reads \(22.5\ \mathrm{W\,m^{-2}}\).

A student argues that the final intensity must be zero because P1 and P3 are crossed (perpendicular).

Determine:

(a) The angle \(\alpha\).

(b) Why the final detector reading alone would not uniquely determine \(\alpha\).

(c) The maximum possible intensity after P3 if M is rotated to a more favourable orientation.

Explain physically why inserting M allows light through the crossed outer filters without creating energy.

Solution 5

The middle polariser is at \(30^\circ\), the final intensity reading alone also permits \(60^\circ\), and the maximum possible final intensity is \(30\ \mathrm{W\,m^{-2}}\), achieved when M is at \(45^\circ\).

The student’s claim is incorrect. Although P1 and P3 are crossed, M changes the polarisation direction of the light reaching P3.

Step 1: Use the detector reading immediately after M.

The intensity before M is:

\[ I_0=120\ \mathrm{W\,m^{-2}} \]

The measured intensity after M is:

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

Apply Malus’ law:

\[ 90=120\cos^2\alpha \]

Rearranging:

\[ \cos^2\alpha=\frac{90}{120} \]

\[ \cos^2\alpha=0.75 \]

Taking the square root:

\[ \cos\alpha=\sqrt{0.75} \]

Since \(0^\circ\leq\alpha\leq90^\circ\):

\[ \boxed{\alpha=30^\circ} \]

Step 2: Check the final detector reading.

After M, the light is polarised along M’s transmission axis, which is \(30^\circ\).

P3 is at \(90^\circ\).

Therefore, the relative angle at P3 is:

\[ \theta_3=90^\circ-30^\circ \]

\[ \theta_3=60^\circ \]

Apply Malus’ law again:

\[ \begin{aligned} I_2&=90\cos^2(60^\circ)\\ &=90(0.25)\\ &=22.5\ \mathrm{W\,m^{-2}} \end{aligned} \]

This agrees with the second detector reading.

Step 3: Explain why the final reading alone is insufficient.

Suppose we had only the original intensity and the final detector reading.

We would know:

\[ I_0=120\ \mathrm{W\,m^{-2}} \]

\[ I_2=22.5\ \mathrm{W\,m^{-2}} \]

Applying Malus’ law sequentially:

\[ I_2=120\cos^2\alpha \cos^2(90^\circ-\alpha) \]

Using the identity:

\[ \cos(90^\circ-\alpha)=\sin\alpha \]

we obtain:

\[ I_2=120\cos^2\alpha\sin^2\alpha \]

Using the double-angle identity:

\[ \sin(2\alpha)=2\sin\alpha\cos\alpha \]

gives:

\[ I_2=30\sin^2(2\alpha) \]

Substitute the final detector reading:

\[ 22.5=30\sin^2(2\alpha) \]

\[ \sin^2(2\alpha)=0.75 \]

For the given range of \(\alpha\), there are two possible solutions:

\[ 2\alpha=60^\circ \]

or

\[ 2\alpha=120^\circ \]

Therefore:

\[ \alpha=30^\circ \]

or

\[ \alpha=60^\circ \]

The final intensity alone cannot distinguish between these two arrangements.

The extra detector reading immediately after M resolves the ambiguity. At \(60^\circ\), the intensity after M would be only:

\[ 120\cos^2(60^\circ) =30\ \mathrm{W\,m^{-2}} \]

But the actual intermediate reading is \(90\ \mathrm{W\,m^{-2}}\), confirming that M is at \(30^\circ\).

Step 4: Find the maximum possible final intensity.

We have:

\[ I_2=30\sin^2(2\alpha) \]

The maximum possible value of \(\sin^2(2\alpha)\) is 1.

This occurs when:

\[ 2\alpha=90^\circ \]

Therefore:

\[ \alpha=45^\circ \]

The maximum final intensity is:

\[ \boxed{I_{\text{max}}=30\ \mathrm{W\,m^{-2}}} \]

At this orientation, the intensity after M is:

\[ I_1=120\cos^2(45^\circ) \]

\[ I_1=60\ \mathrm{W\,m^{-2}} \]

P3 is also \(45^\circ\) from M, so it transmits half of that intensity:

\[ I_2=60\cos^2(45^\circ) \]

\[ I_2=30\ \mathrm{W\,m^{-2}} \]

Why doesn’t this violate energy conservation?

The middle polariser does not create additional light.

It selects the component of the incoming electric field parallel to its own transmission axis. The surviving light is then polarised along that new axis.

P3 consequently receives light with a different polarisation direction from the original beam.

The intensity decreases at each stage:

\[ 120\rightarrow60\rightarrow30 \]

The final intensity is non-zero because M changes the polarisation state of the light reaching P3, not because it amplifies the beam.

This is the crucial physical idea: a filter changes the direction of the surviving electric-field oscillation, while reducing the intensity.

Once you can identify the electric-field direction and update it after each filter, you have the foundation for longer polarisation calculations. The next useful step is applying the same reasoning systematically to three filters in How to Apply Malus’ Law to Three Polarising Filters.