How to Apply Malus’ Law to Three Polarising Filters

An already plane-polarised beam enters three filters. Their transmission axes are \(25^\circ\), \(65^\circ\), and \(140^\circ\) above the horizontal. Which angle should you use for the third application of Malus’ law: \(140^\circ\), \(65^\circ\), or something else?

The correct angle is \(75^\circ\). After passing through the second filter, the surviving light is polarised along its \(65^\circ\) transmission axis. The third filter must therefore be compared with \(65^\circ\), not with the original beam direction or the horizontal.

The essential method is to update two things after every filter: the intensity and the direction of polarisation.

01The rule: apply Malus’ law one filter at a time

Malus’ law describes the intensity transmitted when plane-polarised light passes through a polarising filter:

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

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 angle between the incoming polarisation direction and the filter’s transmission axis.

For three filters, call them P1, P2, and P3, the calculation becomes:

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

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

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

The important detail is that the intensity entering each filter becomes the starting intensity for the next calculation.

The angles also change. Once light passes through a polariser, its surviving electric field is aligned with that polariser’s transmission axis. The next filter therefore compares its axis with the previous filter’s axis.

This is why you cannot simply use all three absolute angles in separate applications of Malus’ law.

Keep track of the intensity before each filter

Use the following notation:

• \(I_0\): intensity entering P1.
• \(I_1\): intensity leaving P1 and entering P2.
• \(I_2\): intensity leaving P2 and entering P3.
• \(I_3\): final intensity leaving P3.

The calculation should follow this sequence:

Notice that \(I_0\) is not used again after the first filter. Each new calculation starts with the intensity that actually reaches that filter.

Finding the correct angle between filters

HSC questions often give transmission axes as absolute angles measured from the horizontal.

For example:

• P1 is at \(20^\circ\).
• P2 is at \(55^\circ\).
• P3 is at \(125^\circ\).

The relative angles are:

\[
\Delta\theta_2=55^\circ-20^\circ=35^\circ
\]

\[
\Delta\theta_3=125^\circ-55^\circ=70^\circ
\]

There is one additional detail to remember: a polariser’s transmission axis is a line, not an arrow.

A filter at \(0^\circ\) has exactly the same transmission axis as a filter at \(180^\circ\). Consequently, the smallest angle between two axes is the one you normally use.

If the absolute difference is greater than \(90^\circ\), subtract it from \(180^\circ\).

For example, the difference between \(160^\circ\) and \(20^\circ\) is \(140^\circ\), but the smallest angle between the axes is:

\[
\Delta\theta=180^\circ-140^\circ=40^\circ
\]

What happens if the incoming light is unpolarised?

Before applying Malus’ law, identify the type of light entering the first filter.

If the incoming light is already plane-polarised, apply Malus’ law directly.

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

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

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

You can then apply Malus’ law normally to P2 and P3.

Do not halve the intensity again at each filter. The half-intensity rule applies when unpolarised light first encounters an ideal polariser.

02Worked examples

03What a three-filter experiment should show

Malus’ law is not just a calculation rule. It makes a prediction about what you should measure when light passes through a sequence of polarising filters.

Consider a simple experimental arrangement.

The apparatus contains:

• A light source producing approximately unpolarised light.
• Three polarising filters mounted in sequence.
• A light detector that measures the transmitted intensity.

The light travels from the source through P1, P2, and P3 before reaching the detector.

Each filter can be rotated around the direction of light travel to change its transmission axis.

The prediction

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

The filters are arranged as follows:

• P1: \(0^\circ\).
• P2: \(30^\circ\).
• P3: \(75^\circ\).

The predicted intensities are:

All values are in \(\mathrm{W\,m^{-2}}\).

The detector should therefore measure approximately \(18.75\ \mathrm{W\,m^{-2}}\) after the third filter, assuming ideal filters and a stable source.

Measuring each stage matters

Suppose you only place the detector after P3.

You obtain a final intensity, but you cannot tell exactly where an unexpected reduction occurred. Was P2 responsible, or was P3?

A better experimental design records the intensity after every stage.

One way to do this is to use a collimated beam and keep the detector in a fixed position. Remove the filters downstream of the measurement stage to record the intensity after P1, after P2, and finally after P3.

Keep the source output, detector position, and alignment unchanged.

The measured ratios can then be compared with the predictions:

\[
\frac{I_2}{I_1}=\cos^2\Delta\theta_2
\]

\[
\frac{I_3}{I_2}=\cos^2\Delta\theta_3
\]

If the measurements approximately agree, they support the sequential Malus’ law model.

Real filters are not perfectly ideal. They absorb some light, may transmit a small amount of the unwanted polarisation, and can introduce reflections. Source fluctuations, detector calibration, and misalignment can also affect the readings.

These effects explain why experimental results may differ slightly from theoretical predictions. HSC calculation questions normally remove these complications by specifying ideal polarising filters.

04Common mistakes to avoid

A useful final check is to ask yourself two questions after every filter:

1. What is the intensity of the light leaving this filter?
2. What is the new polarisation direction?

If you can answer both, you have the information needed for the next application of Malus’ law.

05Questions and solutions

The following questions move from standard three-filter calculations to more challenging situations involving axis orientation, experimental measurements, and optimisation.

Question 1

Unpolarised light has an intensity of \(180\ \mathrm{W\,m^{-2}}\). It passes through three ideal polarising filters.

Their transmission axes are:

• P1: \(40^\circ\) above horizontal.
• P2: \(65^\circ\) above horizontal.
• P3: \(110^\circ\) above horizontal.

Calculate the intensity after each filter and determine the final transmitted intensity.

Solution 1

The intensities after P1, P2, and P3 are \(90.0\), \(74.0\), and \(37.0\ \mathrm{W\,m^{-2}}\), respectively.

Start by recognising that the incoming light is unpolarised.

The first ideal filter therefore transmits half the initial intensity:

\[
\begin{aligned}
I_1&=\frac{180}{2}\\
&=90.0\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

After P1, the light is polarised at \(40^\circ\).

The relative angle between P1 and P2 is:

\[
\Delta\theta_2=65^\circ-40^\circ=25^\circ
\]

Apply Malus’ law:

\[
\begin{aligned}
I_2&=I_1\cos^2(25^\circ)\\
&=90.0(0.8214)\\
&=73.9\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

The surviving light is now polarised at \(65^\circ\).

The relative angle between P2 and P3 is:

\[
\Delta\theta_3=110^\circ-65^\circ=45^\circ
\]

Therefore:

\[
\begin{aligned}
I_3&=I_2\cos^2(45^\circ)\\
&=73.9(0.5)\\
&=37.0\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

The final intensity is:

\[
\boxed{I_3=37.0\ \mathrm{W\,m^{-2}}}
\]

The common mistake would be to use \(65^\circ\) and \(110^\circ\) directly as the Malus’ law angles. Those are absolute orientations, not the angles between consecutive filters.

Question 2

A plane-polarised beam has an intensity of \(150\ \mathrm{W\,m^{-2}}\). Its polarisation direction is \(165^\circ\) anticlockwise from horizontal.

The beam passes through three ideal polarising filters:

• P1: \(5^\circ\).
• P2: \(45^\circ\).
• P3: \(110^\circ\).

Calculate the intensity after each filter.

Explain why the angle used for the first application of Malus’ law is not \(160^\circ\).

Solution 2

The intensities after P1, P2, and P3 are approximately \(132.5\), \(77.7\), and \(13.9\ \mathrm{W\,m^{-2}}\).

The first relative angle is \(20^\circ\), not \(160^\circ\), because polariser axes are equivalent when rotated through \(180^\circ\).

Step 1: Calculate the intensity after P1.

The absolute difference between the incoming polarisation and P1 is:

\[
|165^\circ-5^\circ|=160^\circ
\]

The smallest angle between the two axes is:

\[
\Delta\theta_1=180^\circ-160^\circ=20^\circ
\]

Therefore:

\[
\begin{aligned}
I_1&=150\cos^2(20^\circ)\\
&=150(0.8830)\\
&=132.5\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

The surviving light is polarised at \(5^\circ\).

Step 2: Calculate the intensity after P2.

The relative angle is:

\[
\Delta\theta_2=45^\circ-5^\circ=40^\circ
\]

Therefore:

\[
\begin{aligned}
I_2&=132.5\cos^2(40^\circ)\\
&=132.5(0.5868)\\
&=77.7\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

The surviving light is now polarised at \(45^\circ\).

Step 3: Calculate the intensity after P3.

The relative angle is:

\[
\Delta\theta_3=110^\circ-45^\circ=65^\circ
\]

Therefore:

\[
\begin{aligned}
I_3&=77.7\cos^2(65^\circ)\\
&=77.7(0.1786)\\
&=13.9\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

The final result is:

\[
\boxed{I_3=13.9\ \mathrm{W\,m^{-2}}}
\]

The tempting approach is to use \(160^\circ\) directly in Malus’ law. However, the transmission axis is an unoriented line. An axis at \(5^\circ\) is equivalent to one at \(185^\circ\), making its smallest separation from \(165^\circ\) just \(20^\circ\).

This distinction becomes especially important when filter orientations cross the \(180^\circ\) boundary.

Question 3

A plane-polarised beam with an initial intensity of \(100\ \mathrm{W\,m^{-2}}\) travels horizontally.

Three ideal polarising filters have transmission axes at \(20^\circ\), \(50^\circ\), and \(100^\circ\).

Two different arrangements are tested:

• Arrangement A: \(20^\circ \rightarrow 50^\circ \rightarrow 100^\circ\).
• Arrangement B: \(50^\circ \rightarrow 20^\circ \rightarrow 100^\circ\).

Calculate the final transmitted intensity for both arrangements.

Determine which arrangement transmits more light and calculate how many times greater its final intensity is.

Explain why changing the order of the filters changes the result, even though the same three filters are used.

Solution 3

Arrangement A transmits approximately \(27.4\ \mathrm{W\,m^{-2}}\), while Arrangement B transmits approximately \(0.934\ \mathrm{W\,m^{-2}}\).

Arrangement A therefore transmits approximately 29.3 times as much light.

The reason is that the order of the filters changes the relative angle encountered at each stage.

Step 1: Calculate the intensity for Arrangement A.

The incoming light is polarised horizontally, at \(0^\circ\).

The successive relative angles are:

\[
\begin{aligned}
\Delta\theta_1&=20^\circ\\
\Delta\theta_2&=50^\circ-20^\circ=30^\circ\\
\Delta\theta_3&=100^\circ-50^\circ=50^\circ
\end{aligned}
\]

Apply Malus’ law at each stage:

\[
\begin{aligned}
I_A&=100\cos^2(20^\circ)\\
&\quad\times\cos^2(30^\circ)\\
&\quad\times\cos^2(50^\circ)\\
&=100(0.8830)(0.7500)(0.4132)\\
&=27.4\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

Step 2: Calculate the intensity for Arrangement B.

The first filter is now at \(50^\circ\).

The successive relative angles become:

\[
\begin{aligned}
\Delta\theta_1&=50^\circ\\
\Delta\theta_2&=50^\circ-20^\circ=30^\circ\\
\Delta\theta_3&=100^\circ-20^\circ=80^\circ
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
I_B&=100\cos^2(50^\circ)\\
&\quad\times\cos^2(30^\circ)\\
&\quad\times\cos^2(80^\circ)\\
&=100(0.4132)(0.7500)(0.03015)\\
&=0.934\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

Step 3: Compare the two arrangements.

The ratio of the final intensities is:

\[
\begin{aligned}
\frac{I_A}{I_B}
&=\frac{27.4}{0.934}\\
&\approx29.3
\end{aligned}
\]

Arrangement A transmits approximately 29.3 times more intensity than Arrangement B.

The tempting prediction is that rearranging the same filters should not matter. However, each filter changes the polarisation direction of the surviving light.

In Arrangement A, the final filter is only \(50^\circ\) away from the incoming polarisation at that stage.

In Arrangement B, the final filter is \(80^\circ\) away from the incoming polarisation.

That much larger final angle causes a substantial reduction in intensity.

The key constraint is the sequence of relative angles, not simply the set of filter orientations.

Question 4

An experiment uses an initially unpolarised light source with an intensity of \(200\ \mathrm{W\,m^{-2}}\).

Three ideal polarising filters are arranged in sequence:

• P1 has its transmission axis at \(0^\circ\).
• P2 has its transmission axis at \(30^\circ\).
• P3 has an unknown transmission axis.

The following illustrative detector measurements are recorded:

Determine the possible orientations of P3 between \(0^\circ\) and \(180^\circ\).

Explain why the measurements do not identify a unique orientation.

Suggest an additional measurement that would distinguish between the possibilities.

Solution 4

P3 can have its transmission axis at either \(75^\circ\) or \(165^\circ\).

The detector measurements determine the magnitude of the relative angle, but they do not reveal which side of P2 the final filter occupies.

Step 1: Check the first filter.

The source is unpolarised, so an ideal first polariser transmits half the intensity:

\[
I_1=\frac{200}{2}=100\ \mathrm{W\,m^{-2}}
\]

This agrees with the measured value.

Step 2: Determine the angle between P1 and P2.

The intensity changes from \(100\) to \(75.0\ \mathrm{W\,m^{-2}}\).

Using Malus’ law:

\[
75.0=100\cos^2\Delta\theta_2
\]

Rearranging:

\[
\cos^2\Delta\theta_2=0.75
\]

Since \(\cos^2 30^\circ=0.75\):

\[
\Delta\theta_2=30^\circ
\]

This agrees with the given orientation of P2.

Step 3: Determine the angle between P2 and P3.

The intensity changes from \(75.0\) to \(37.5\ \mathrm{W\,m^{-2}}\).

Therefore:

\[
37.5=75.0\cos^2\Delta\theta_3
\]

Rearranging:

\[
\cos^2\Delta\theta_3=\frac{37.5}{75.0}
\]

\[
\cos^2\Delta\theta_3=0.5
\]

This gives:

\[
\Delta\theta_3=45^\circ
\]

P3 must therefore be \(45^\circ\) away from P2.

Step 4: Find both possible orientations.

P2 is at \(30^\circ\).

P3 could be \(45^\circ\) anticlockwise from P2:

\[
\alpha_3=30^\circ+45^\circ=75^\circ
\]

Alternatively, P3 could be \(45^\circ\) clockwise from P2:

\[
\alpha_3=30^\circ-45^\circ=-15^\circ
\]

Expressing the second orientation between \(0^\circ\) and \(180^\circ\):

\[
\alpha_3=165^\circ
\]

Therefore:

\[
\boxed{\alpha_3=75^\circ\text{ or }165^\circ}
\]

Step 5: Propose an additional measurement.

Place another ideal horizontal analyser after P3.

If P3 is at \(75^\circ\), the relative angle to the new horizontal analyser is \(75^\circ\).

The predicted intensity becomes:

\[
\begin{aligned}
I_4&=37.5\cos^2(75^\circ)\\
&=2.51\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

If P3 is at \(165^\circ\), the smallest angle to the horizontal analyser is \(15^\circ\).

The predicted intensity becomes:

\[
\begin{aligned}
I_4&=37.5\cos^2(15^\circ)\\
&=35.0\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

The two predictions are very different, so this additional measurement would distinguish the orientations.

The important lesson is that intensity measurements involving \(\cos^2\theta\) can hide directional information. A squared cosine does not distinguish between positive and negative angle differences.

Question 5

A plane-polarised beam has an intensity of \(80.0\ \mathrm{W\,m^{-2}}\) and is polarised at \(20^\circ\) above horizontal.

It passes through three ideal polarising filters:

• P1 is fixed at \(20^\circ\).
• P2 can be rotated to any transmission axis between \(0^\circ\) and \(180^\circ\).
• P3 is fixed at \(140^\circ\).

Determine the orientation of P2 that maximises the final transmitted intensity.

A student argues that P2 should be set at \(80^\circ\), halfway between \(20^\circ\) and \(140^\circ\).

Is this correct?

Find the actual maximum transmitted intensity and explain why the arithmetic midpoint gives a different result.

Solution 5

The maximum transmitted intensity is \(45.0\ \mathrm{W\,m^{-2}}\), achieved when P2 is set at \(170^\circ\).

The student’s \(80^\circ\) setting is not optimal because polariser axes are lines, not directed arrows. The smallest angular separation between the outer filters is \(60^\circ\), not \(120^\circ\).

Step 1: Identify the intensity entering P1.

The incoming light is already polarised at \(20^\circ\), and P1 is also at \(20^\circ\).

The first filter therefore transmits all the incoming intensity:

\[
I_1=80.0\ \mathrm{W\,m^{-2}}
\]

The problem is now to choose P2 so that the product of the next two Malus’ law factors is as large as possible.

Step 2: Examine the separation between the outer filters.

The absolute difference between their axis labels is:

\[
140^\circ-20^\circ=120^\circ
\]

However, the smallest angle between their transmission axes is:

\[
180^\circ-120^\circ=60^\circ
\]

This means the shorter route between the outer axes crosses the \(180^\circ\) boundary.

An intermediate axis at \(170^\circ\) lies halfway along this shorter route:

\[
20^\circ\rightarrow170^\circ\rightarrow140^\circ
\]

The relative angles are:

\[
\Delta\theta_2=30^\circ
\]

\[
\Delta\theta_3=30^\circ
\]

Step 3: Calculate the intensity for this arrangement.

Applying Malus’ law twice:

\[
\begin{aligned}
I_3&=80.0\cos^2(30^\circ)\cos^2(30^\circ)\\
&=80.0(0.75)(0.75)\\
&=45.0\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

So the \(170^\circ\) setting produces \(45.0\ \mathrm{W\,m^{-2}}\).

Step 4: Compare this with the student’s \(80^\circ\) setting.

The student has calculated the ordinary arithmetic midpoint:

\[
\frac{20^\circ+140^\circ}{2}=80^\circ
\]

However, this places P2 halfway along the longer route between the two outer axes.

The relative angles are now:

\[
\Delta\theta_2=80^\circ-20^\circ=60^\circ
\]

\[
\Delta\theta_3=140^\circ-80^\circ=60^\circ
\]

The resulting intensity is:

\[
\begin{aligned}
I_3&=80.0\cos^2(60^\circ)\cos^2(60^\circ)\\
&=80.0(0.25)(0.25)\\
&=5.00\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

The student’s arrangement transmits only \(5.00\ \mathrm{W\,m^{-2}}\), compared with \(45.0\ \mathrm{W\,m^{-2}}\) for the \(170^\circ\) arrangement.

The better arrangement transmits nine times as much intensity.

Step 5: Confirm that \(170^\circ\) is the global maximum.

It is not enough to find a setting that gives a larger intensity. We need to establish that no other orientation of P2 can produce a better result.

Let the adjustable orientation of P2 be \(\theta\).

The transmitted intensity is:

\[
I_3=80.0\cos^2(\theta-20^\circ)
\cos^2(140^\circ-\theta)
\]

The product-to-sum trigonometric identity gives:

\[
\cos A\cos B=
\frac{1}{2}[\cos(A+B)+\cos(A-B)]
\]

Choose:

\[
A=\theta-20^\circ
\]

\[
B=140^\circ-\theta
\]

Then:

\[
A+B=120^\circ
\]

and:

\[
A-B=2\theta-160^\circ
\]

Substituting:

\[
\begin{aligned}
\cos A\cos B
&=\frac{1}{2}[
\cos120^\circ\\
&\qquad+\cos(2\theta-160^\circ)]
\end{aligned}
\]

Since \(\cos120^\circ=-0.5\):

\[
\cos A\cos B=
\frac{1}{2}[
\cos(2\theta-160^\circ)-0.5]
\]

Cosine cannot be smaller than \(-1\) or greater than \(1\).

Therefore, the largest possible magnitude of this product is:

\[
|\cos A\cos B|\leq0.75
\]

Squaring both sides:

\[
\cos^2 A\cos^2 B\leq0.5625
\]

The maximum intensity is consequently:

\[
\begin{aligned}
I_{3,\max}&=80.0(0.5625)\\
&=45.0\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

This maximum occurs when:

\[
\cos(2\theta-160^\circ)=-1
\]

Within the allowed range of \(0^\circ\) to \(180^\circ\):

\[
2\theta-160^\circ=180^\circ
\]

\[
2\theta=340^\circ
\]

\[
\boxed{\theta=170^\circ}
\]

The result is therefore the global maximum, not just a favourable trial setting.

The conceptual trap is treating polariser orientations as directed angles. A polariser at \(20^\circ\) is the same axis as one at \(200^\circ\). Once you account for this \(180^\circ\) symmetry, the correct halfway position is \(170^\circ\), not \(80^\circ\).

The central habit in every three-filter problem is the same: carry forward the intensity and reset the polarisation direction after each filter. Once that is automatic, you can tackle more complicated polarisation arrangements and apply the same amplitude-first reasoning to interference, where electric fields combine before intensity is calculated.