Why an Ideal Polariser Transmits Half of Unpolarised Light

Place an ideal polariser in front of a light meter. The reading drops from \(200\ \mathrm{W\,m^{-2}}\) to \(100\ \mathrm{W\,m^{-2}}\). Why exactly half? The filter hasn’t blocked half the beam’s area. It has selected one direction of electric-field oscillation, and unpolarised light contains equal average intensity in any two perpendicular directions. The selected direction therefore carries half the original intensity.

That’s the reason an ideal polariser transmits exactly 50% of unpolarised light.

Before going further, make a prediction. If you rotate the first polariser by \(60^\circ\), will the transmitted intensity change? No. Unpolarised light has no preferred direction, so every orientation of an ideal first polariser transmits the same average intensity.

The interesting part is understanding why.

01What does a polariser actually select?

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

A linear polariser has a particular direction called its transmission axis. It allows the electric-field component parallel to this axis to pass and blocks the perpendicular component.

For example, imagine a polariser with a horizontal transmission axis.

• A horizontally oscillating electric field passes through completely.
• A vertically oscillating electric field is blocked completely.
• An electric field at \(45^\circ\) has only part of its amplitude aligned with the transmission axis.

The key is that a polariser selects an electric-field direction, not a direction of travel.

Why the transmitted fraction depends on angle

Start with light that is already linearly polarised.

Let:

• \(E_0\) be its incident electric-field amplitude.
• \(\theta\) be the angle between its polarisation direction and the polariser’s transmission axis.
• \(E_{\parallel}\) be the transmitted electric-field amplitude.

The component parallel to the transmission axis is

\[ E_{\parallel}=E_0\cos\theta \]

Intensity is proportional to the square of electric-field amplitude. Therefore, the fraction of incident intensity transmitted is

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

This is the relationship behind Malus’ law.

Consider what happens at different angles:

Notice something important. A polariser doesn’t automatically transmit half the intensity. A linearly polarised beam aligned with its axis loses none of its intensity in the ideal model.

The half-intensity result requires something different about the incoming light.

02Why unpolarised light gives an average of one-half

Unpolarised light has no fixed preferred direction of electric-field oscillation.

Its polarisation directions are randomly distributed over the measurement interval. You can picture a large collection of electric-field vectors pointing in different directions across the plane perpendicular to the beam.

Some are aligned with the polariser. Some are perpendicular. Most are somewhere between.

The polariser transmits a different fraction from each orientation.

For example:

• Light polarised at \(0^\circ\) transmits \(100\%\).
• Light polarised at \(30^\circ\) transmits \(75\%\).
• Light polarised at \(60^\circ\) transmits \(25\%\).
• Light polarised at \(90^\circ\) transmits \(0\%\).

The total transmitted intensity is the average over all these orientations, weighted by how much incident intensity each orientation contributes.

Because unpolarised light has no preferred orientation, every direction is represented equally in the average.

Mathematically, the average transmission fraction is

\[ \left\langle\cos^2\theta\right\rangle =\frac{1}{2} \]

You can understand why using the trigonometric identity

\[ \cos^2\theta=\frac{1+\cos 2\theta}{2} \]

Across a complete distribution of polarisation directions, the positive and negative contributions from \(\cos 2\theta\) cancel.

That leaves

\[ \left\langle\cos^2\theta\right\rangle =\frac{1+0}{2} =\frac12 \]

Therefore,

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

where:

• \(I_0\) is the original intensity of the unpolarised light, measured in \(\mathrm{W\,m^{-2}}\).
• \(I_1\) is the intensity after the first ideal polariser, measured in \(\mathrm{W\,m^{-2}}\).

The factor of one-half is an average over many randomly oriented contributions. It is not a claim that every individual contribution loses exactly half its intensity.

Another way to see the same result

Choose two perpendicular directions:

• \(x\), parallel to the polariser’s transmission axis.
• \(y\), perpendicular to the transmission axis.

For unpolarised light, the average intensity is distributed equally between these two directions.

So,

\[ I_x=I_y=\frac{I_0}{2} \]

The ideal polariser transmits the \(x\)-component and blocks the \(y\)-component.

Therefore,

\[ I_{\text{out}}=I_x=\frac{I_0}{2} \]

This is often the quickest explanation to use in an HSC response.

The first approach explains the result through randomly distributed angles. The second explains it through two perpendicular components. Both describe the same physical situation.

One distinction is worth remembering: unpolarised light does not mean light has no electric field. It means there is no fixed preferred electric-field direction over the measurement interval.

03How an experiment tests the half-intensity prediction

Suppose you want to check whether the intensity really halves when unpolarised light passes through an ideal polariser.

You need a stable light source, a polariser that can rotate, and a light detector.

Arrange them in this order:

1. An approximately unpolarised light source produces a beam.
2. The beam passes through a rotatable polariser.
3. A photodiode behind the polariser measures the transmitted light.

The photodiode stays in the same position throughout the investigation.

What should the experiment show?

First, remove the polariser and record the incident intensity, \(I_0\).

Then insert the first polariser and record the transmitted intensity, \(I_1\).

For an ideal polariser and unpolarised input, the prediction is

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

Next, rotate the first polariser while keeping the source and detector fixed.

The reading should remain unchanged. Rotating the polariser changes the direction it selects, but the incoming unpolarised light has no preferred direction.

Now introduce a second polariser behind the first.

This gives a further test. The first polariser has already made the transmitted light linearly polarised, with its electric field oscillating along the first polariser’s transmission axis.

The second polariser should therefore behave differently depending on its orientation.

For ideal filters, the expected readings are:

The final result is particularly useful. The first polariser has not merely reduced the intensity. It has changed the polarisation state of the transmitted light.

What could make the experimental result different?

A real polariser is not perfectly ideal.

It can absorb some light that is aligned with its transmission axis. It can also allow a small amount of the perpendicular component through.

Other experimental issues include:

• Fluctuations in the light source.
• Background light reaching the detector.
• Changes in detector position or beam alignment.
• Partial polarisation in the original source.

For this reason, a real experiment might measure \(48\%\) or \(47\%\) transmission rather than exactly \(50\%\).

That does not automatically disprove the model. The ideal prediction assumes complete transmission of the parallel component and complete rejection of the perpendicular component.

A better investigation keeps the detector fixed, measures and subtracts the background reading, checks that the source is stable, and rotates the first polariser through several orientations.

This helps distinguish the predicted half-intensity effect from losses caused by the equipment.

There is another important experimental limitation: one reading of half the original intensity does not, by itself, prove that the incoming light was unpolarised. We’ll return to this when comparing two different possible sources.

05Why the second polariser does not automatically halve the intensity

This is where students often make an incorrect prediction.

They remember that an ideal polariser halves unpolarised light, so they apply another factor of one-half whenever a second polariser appears.

That is not correct.

After the first polariser, the light is already linearly polarised. Its electric field oscillates in one fixed direction: the transmission axis of the first polariser.

The second polariser now receives polarised light, not unpolarised light.

Its transmitted intensity depends on the angle between the two transmission axes.

Let \(\Delta\theta\) represent this relative angle.

Then,

\[ I_2=I_1\cos^2(\Delta\theta) \]

Here:

• \(I_1\) is the intensity reaching the second polariser.
• \(I_2\) is the intensity leaving the second polariser.
• \(\Delta\theta\) is the angle between the first and second transmission axes.

The relative angle is essential. It is not automatically the angle of the second polariser measured from horizontal.

Consider three possibilities:

Parallel axes:

\[ \Delta\theta=0^\circ \]

\[ I_2=I_1\cos^2(0^\circ)=I_1 \]

No additional intensity is lost in the ideal model.

Axes at \(45^\circ\):

\[ \Delta\theta=45^\circ \]

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

The second polariser halves the intensity.

Perpendicular axes:

\[ \Delta\theta=90^\circ \]

\[ I_2=I_1\cos^2(90^\circ)=0 \]

No light is transmitted.

The important distinction is that the first polariser receives unpolarised light, while the second receives light that has already been polarised.

For a full explanation of the projection calculation and its applications, see Malus’ Law for HSC Physics: Intensity and Polariser Angles.

06A half-intensity reading does not prove that light is unpolarised

There is a subtle problem with using a single polariser to identify unpolarised light.

Imagine a beam that is already linearly polarised at \(45^\circ\) to horizontal.

Place a horizontal ideal polariser in its path.

The transmitted intensity is

\[ I=I_0\cos^2(45^\circ) \]

Since

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

the polariser transmits exactly half the incident intensity.

That is the same result you would obtain from unpolarised light.

So the following two situations can produce identical detector readings:

Does that mean the two beams are physically identical? No.

The unpolarised beam has no preferred electric-field direction. The other beam has a fixed polarisation direction at \(45^\circ\).

To distinguish them, rotate the polariser while the original beam is still entering it.

• For unpolarised light, the transmitted intensity remains constant.
• For light polarised at \(45^\circ\), the transmitted intensity changes as the polariser rotates.

This is a useful diagnostic. A single half-intensity reading is not enough to establish the polarisation state of an unknown beam.

07Questions and solutions

Question 1

An unpolarised light beam has an intensity of \(180\ \mathrm{W\,m^{-2}}\). It passes through one ideal linear polariser.

Calculate the transmitted intensity. Explain whether rotating the polariser by \(55^\circ\) changes your answer.

Solution 1

The transmitted intensity is \(90\ \mathrm{W\,m^{-2}}\), and rotating the polariser does not change this result.

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

Using

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

Substitute:

\[ \begin{aligned} I_1&=\frac{180}{2}\\ &=90\ \mathrm{W\,m^{-2}} \end{aligned} \]

The transmitted intensity is therefore

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

Rotating the polariser changes its transmission axis, but unpolarised light has no preferred electric-field direction. Its average intensity is distributed equally across perpendicular directions.

The tempting mistake is to apply \(\cos^2 55^\circ\) to the original intensity. That would be appropriate if the incoming beam were already linearly polarised at a known angle. Here, it is unpolarised, so the correct first-filter rule is simply to halve the intensity.

Question 2

An unpolarised beam has an intensity of \(400\ \mathrm{W\,m^{-2}}\). It passes through two ideal polarisers.

Initially:

• P1 has its transmission axis \(20^\circ\) above horizontal.
• P2 has its transmission axis \(80^\circ\) above horizontal.

Calculate the intensity after each polariser.

P1 is then rotated to \(35^\circ\) above horizontal, while P2 remains at \(80^\circ\).

Calculate the new final intensity and explain why it changes even though the original light is still unpolarised.

Solution 2

Initially, the final intensity is \(50.0\ \mathrm{W\,m^{-2}}\). After rotating P1, the final intensity increases to approximately \(100\ \mathrm{W\,m^{-2}}\).

The first polariser receives unpolarised light, so its transmission is independent of its orientation.

Initial arrangement

Step 1: Calculate the intensity after P1.

\[ \begin{aligned} I_1&=\frac{I_0}{2}\\ &=\frac{400}{2}\\ &=200\ \mathrm{W\,m^{-2}} \end{aligned} \]

The light is now linearly polarised at \(20^\circ\) above horizontal.

Step 2: Calculate the relative angle between P1 and P2.

\[ \Delta\theta=80^\circ-20^\circ \]

\[ \Delta\theta=60^\circ \]

Step 3: Apply Malus’ law.

\[ \begin{aligned} I_2&=I_1\cos^2(60^\circ)\\ &=200(0.25)\\ &=50.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

After rotating P1

Step 4: Recalculate the relative angle.

P1 is now at \(35^\circ\), while P2 remains at \(80^\circ\).

\[ \Delta\theta=80^\circ-35^\circ \]

\[ \Delta\theta=45^\circ \]

Step 5: Calculate the new final intensity.

The first polariser still receives unpolarised light, so the intensity after P1 remains \(200\ \mathrm{W\,m^{-2}}\).

Therefore,

\[ \begin{aligned} I_2'&=200\cos^2(45^\circ)\\ &=200(0.5)\\ &=100\ \mathrm{W\,m^{-2}} \end{aligned} \]

The new final intensity is

\[ \boxed{100\ \mathrm{W\,m^{-2}}} \]

The important point is that rotating P1 does not change the amount of unpolarised light it transmits. However, it changes the direction of the polarised light leaving it.

That changes the relative angle at P2, which changes the second polariser’s transmission.

The tempting mistake is to assume that because the original light is unpolarised, rotating P1 cannot affect the final intensity. It cannot affect the intensity immediately after P1, but it can affect everything that happens afterwards.

Question 3

A physics student investigates the intensity of light passing through two real polarising filters.

The source is stable, the detector remains fixed, and the readings have been corrected for background light.

The following illustrative data are obtained:

Answer the following:

(a) What does the first set of readings suggest about the incoming light?

(b) Explain why rotating P1 does not change the detector reading.

(c) Explain why adding a parallel P2 produces only a small additional reduction.

(d) Why does the crossed arrangement produce a reading close to, but not exactly, zero?

Solution 3

The results are consistent with an approximately unpolarised input beam. P1 produces linearly polarised light, and P2 reduces the intensity according to its orientation.

(a) Incoming light

The detector initially reads \(8.0\ \mathrm{mW}\).

After P1, it reads \(4.0\ \mathrm{mW}\).

The transmission fraction is

\[ \frac{I_1}{I_0} =\frac{4.0}{8.0} =0.50 \]

P1 therefore transmits \(50\%\) of the incident light.

The fact that this reading remains approximately constant at \(0^\circ\), \(45^\circ\), and \(90^\circ\) supports the conclusion that the incoming beam is unpolarised.

(b) Rotating P1

The detector reading remains unchanged because the incoming unpolarised light has no preferred polarisation direction.

An ideal first polariser transmits half the average intensity, regardless of its orientation.

The real filter also has approximately the same transmission efficiency at each tested orientation.

(c) Parallel second polariser

The second polariser is aligned with the first.

The light reaching P2 is already polarised along P1’s transmission axis. Therefore, P2 allows almost all of that light through.

The small reduction from \(4.0\ \mathrm{mW}\) to \(3.9\ \mathrm{mW}\) is consistent with additional absorption or reflection in the real filter.

It would be incorrect to halve the \(4.0\ \mathrm{mW}\) reading simply because another polariser has been introduced.

(d) Perpendicular second polariser

For ideal crossed polarisers,

\[ I_2=I_1\cos^2(90^\circ)=0 \]

However, real polarisers do not block the perpendicular component perfectly.

A small amount of light can leak through because of the filter’s finite extinction ratio. This explains the residual \(0.12\ \mathrm{mW}\) reading.

The experiment supports the ideal model while also showing why actual laboratory measurements may differ slightly from theoretical predictions.

Question 4

A beam consists of two independent light components:

• An unpolarised component with an intensity of \(160\ \mathrm{W\,m^{-2}}\).
• A linearly polarised component with an intensity of \(80\ \mathrm{W\,m^{-2}}\), polarised \(30^\circ\) above horizontal.

The two components overlap to form one beam. Their intensities add.

The beam passes through an ideal polariser.

Calculate the total transmitted intensity when the polariser is:

(a) Horizontal.

(b) Vertical.

Explain which component causes the transmitted intensity to change when the polariser is rotated.

Solution 4

The horizontal polariser transmits \(140\ \mathrm{W\,m^{-2}}\), while the vertical polariser transmits \(100\ \mathrm{W\,m^{-2}}\).

The reason is that the unpolarised component always contributes half its intensity, but the linearly polarised component depends on the relative angle.

(a) Horizontal polariser

Step 1: Calculate the transmitted unpolarised component.

An ideal first polariser transmits half the unpolarised intensity.

\[ I_{\text{unpol}} =\frac{160}{2} =80\ \mathrm{W\,m^{-2}} \]

Step 2: Calculate the transmitted linearly polarised component.

The incoming polarisation direction is \(30^\circ\) above horizontal.

The horizontal transmission axis is \(0^\circ\), so the relative angle is \(30^\circ\).

Using Malus’ law,

\[ \begin{aligned} I_{\text{pol}} &=80\cos^2(30^\circ)\\ &=80(0.75)\\ &=60\ \mathrm{W\,m^{-2}} \end{aligned} \]

Step 3: Add the independent contributions.

\[ \begin{aligned} I_{\text{total}} &=80+60\\ &=140\ \mathrm{W\,m^{-2}} \end{aligned} \]

Therefore,

\[ \boxed{I_{\text{horizontal}}=140\ \mathrm{W\,m^{-2}}} \]

(b) Vertical polariser

Step 4: Calculate the unpolarised contribution.

It is still unpolarised, so the transmitted intensity remains

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

Step 5: Calculate the new relative angle.

The vertical transmission axis is at \(90^\circ\).

The incoming linearly polarised component is at \(30^\circ\).

Therefore,

\[ \Delta\theta=90^\circ-30^\circ =60^\circ \]

Step 6: Calculate the transmitted polarised component.

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

Step 7: Add the two contributions.

\[ \begin{aligned} I_{\text{total}} &=80+20\\ &=100\ \mathrm{W\,m^{-2}} \end{aligned} \]

Therefore,

\[ \boxed{I_{\text{vertical}}=100\ \mathrm{W\,m^{-2}}} \]

Only the linearly polarised component changes its transmitted intensity as the polariser rotates.

The unpolarised component always contributes half its original intensity, regardless of the transmission axis.

The tempting mistake is to halve the entire \(240\ \mathrm{W\,m^{-2}}\) beam for both orientations. That would ignore the fact that the beam contains a component with a definite polarisation direction.

Question 5

Two light sources produce beams with the same initial intensity:

\[ I_0=200\ \mathrm{W\,m^{-2}} \]

However, their polarisation states are different.

• Source A produces genuinely unpolarised light.
• Source B produces fully linearly polarised light at \(45^\circ\) above horizontal.

Both beams are passed through an ideal horizontal polariser. Both detectors record

\[ 100\ \mathrm{W\,m^{-2}} \]

A student concludes that both sources must therefore produce unpolarised light.

You are asked to investigate this claim.

(a) Explain why the student’s conclusion is not justified.

(b) Design a measurement using one rotatable ideal polariser that distinguishes the two original sources. Predict the transmitted intensities when its axis is at \(0^\circ\), \(45^\circ\), and \(90^\circ\) above horizontal.

(c) Suppose the student instead keeps the original horizontal polariser in place and puts a second rotatable analyser behind it. Even if the second analyser is rotated through every possible angle, can the student distinguish the sources? Explain carefully.

Solution 5

The student’s conclusion is incorrect. Both sources produce the same reading through one horizontal polariser, but their original polarisation states are different.

(a) Why the first measurement is insufficient

For Source A, the incoming light is unpolarised.

An ideal first polariser transmits half its intensity:

\[ I_A=\frac{200}{2} =100\ \mathrm{W\,m^{-2}} \]

For Source B, the incoming light is already linearly polarised at \(45^\circ\) to horizontal.

The horizontal polariser transmits

\[ \begin{aligned} I_B&=I_0\cos^2(45^\circ)\\ &=200(0.5)\\ &=100\ \mathrm{W\,m^{-2}} \end{aligned} \]

Both sources therefore produce the same detector reading.

However, the mechanisms are different:

• Source A loses half its intensity through averaging over random polarisation directions.
• Source B loses half because the horizontal polariser transmits only the horizontal component of a field oriented at \(45^\circ\).

A half-intensity reading alone cannot distinguish these possibilities.

(b) Distinguishing the original sources

Place one rotatable ideal polariser directly in the path of each original beam, before any other polariser.

Measure the transmitted intensity at three orientations.

For Source A, the input is unpolarised. Its transmitted intensity remains half the original intensity at every orientation.

Therefore,

\[ I_A=100\ \mathrm{W\,m^{-2}} \]

for all three angles.

For Source B, use the relative angle between its fixed \(45^\circ\) polarisation direction and the analyser.

At \(0^\circ\):

\[ \begin{aligned} I_B&=200\cos^2(45^\circ)\\ &=100\ \mathrm{W\,m^{-2}} \end{aligned} \]

At \(45^\circ\):

\[ \begin{aligned} I_B&=200\cos^2(0^\circ)\\ &=200\ \mathrm{W\,m^{-2}} \end{aligned} \]

At \(90^\circ\):

\[ \begin{aligned} I_B&=200\cos^2(45^\circ)\\ &=100\ \mathrm{W\,m^{-2}} \end{aligned} \]

The predicted results are:

Source A gives a constant reading, while Source B reaches its maximum when the analyser aligns with its polarisation direction.

The two sources can therefore be distinguished by measuring the original light at more than one analyser orientation.

(c) Why a second analyser cannot distinguish the sources

No. Once the original horizontal polariser has been passed, both beams have the same polarisation direction and the same intensity.

This is the crucial part of the problem.

For Source A, the first polariser selects the horizontal component of the unpolarised light. The transmitted beam is now horizontally polarised, with an intensity of \(100\ \mathrm{W\,m^{-2}}\).

For Source B, the first polariser selects the horizontal component of the \(45^\circ\) polarised light. Its transmitted beam is also horizontally polarised, with an intensity of \(100\ \mathrm{W\,m^{-2}}\).

The two beams leaving the first polariser are now physically identical in the ideal model.

If a second analyser is placed behind the horizontal polariser at an angle \(\phi\), both beams produce

\[ I_2=100\cos^2\phi \]

Their readings will remain identical at every possible angle.

The tempting approach is to keep adding measurements downstream, hoping to recover information about the original beams. But the first polariser has already removed the difference between them.

The correct test must happen before that information is lost.

This is the deeper lesson: a polariser doesn’t simply reduce intensity. It also changes the polarisation state of the transmitted light.

08Where to take this concept next

The half-intensity rule is the starting point for understanding more complicated polariser arrangements.

Once the first polariser has converted unpolarised light into linearly polarised light, every subsequent ideal polariser is handled using the relative angle between consecutive transmission axes.

The next useful step is Malus’ Law for HSC Physics: Intensity and Polariser Angles.

When working with actual detector measurements, also keep intensity changes caused by polarisation separate from changes caused by distance. Moving a detector farther from a source can reduce the measured intensity through the inverse-square law, even when no polarising filter has been introduced.