Why Malus’ Law Uses the Angle Between Polariser Axes

A beam of light passes through two ideal polarising filters. Their transmission axes are labelled \(25^\circ\) and \(70^\circ\) above the horizontal. Which angle belongs in Malus’ law: \(70^\circ\), \(25^\circ\), or \(45^\circ\)?

It is \(45^\circ\). The second filter receives light that is already polarised along the first filter’s \(25^\circ\) axis. It responds to the separation between the two axes, not the angle either filter makes with the horizontal.

This is the reason Malus’ law uses the angle between the incoming polarisation direction and the analyser’s transmission axis. The filter transmits the component of the electric field parallel to its own axis. That projection depends on the relative orientation of the two directions, not their absolute positions in the laboratory.

There is one important qualification: if the incoming light is already plane-polarised, its original electric-field direction also matters when it reaches the first filter.

01The angle is a relationship, not a position

Malus’ law describes the intensity transmitted when plane-polarised light encounters another 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 electric-field direction and the filter’s transmission axis.

Why does this angle matter?

Imagine the incoming electric field as an oscillating arrow. The analyser only transmits the component parallel to its transmission axis. If the incoming field has amplitude \(E_{\text{in}}\), its transmitted component has amplitude

\[ E_{\parallel}=E_{\text{in}}\cos\theta \]

Light intensity is proportional to the square of electric-field amplitude. Therefore,

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

The cosine comes from the projection. The square comes from the relationship between electric-field amplitude and intensity.

Why absolute angles are misleading

Suppose P1 is at \(25^\circ\) and P2 is at \(70^\circ\), with both angles measured anticlockwise from the horizontal.

The incoming light at P2 is polarised along P1’s transmission axis. Therefore,

\[ \theta=70^\circ-25^\circ=45^\circ \]

Putting \(70^\circ\) into Malus’ law would compare P2 with the horizontal. But the horizontal is not the direction of the incoming electric field.

The distinction becomes especially important when angles are measured on opposite sides of the horizontal.

A polariser’s transmission axis is a line, not an arrow. An axis at \(0^\circ\) is physically identical to one at \(180^\circ\). Consequently, the angle between two transmission axes is normally expressed as their smaller separation, between \(0^\circ\) and \(90^\circ\).

For example, consider two axes at \(15^\circ\) and \(155^\circ\).

Their absolute difference is \(140^\circ\), but their smaller angular separation is

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

Using \(140^\circ\) in the cosine-squared expression would actually give the same intensity, because \(\cos^2(140^\circ)=\cos^2(40^\circ)\). However, \(40^\circ\) is the clearer physical description of the angle between the two unoriented axes.

The key idea: absolute angles describe where the filters point relative to a chosen reference. Malus’ law describes how far apart the incoming polarisation direction and the next filter’s axis are.

02Track the intensity before each filter

There is a second detail that goes hand in hand with relative angles: the intensity entering a filter is the intensity that must be used in its calculation.

Consider unpolarised light entering two ideal filters, P1 and P2.

The first filter creates plane-polarised light. The second filter then acts on that newly polarised light.

For unpolarised light, the first ideal polariser transmits half the incident intensity:

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

After P1, the transmitted light is polarised along P1’s transmission axis.

P2 therefore receives light polarised along P1, so its relative angle is the separation between the two filter axes:

\[ I_2=I_1\cos^2(\alpha_2-\alpha_1) \]

Here, \(\alpha_1\) and \(\alpha_2\) are the absolute orientations of P1 and P2, measured from the same reference direction. The angle difference must be interpreted using the smaller separation between the axes.

Notice that \(I_0\) is not used again in the second calculation. P2 does not receive the original beam. It receives the reduced intensity \(I_1\).

What if the incoming light is already polarised?

This changes the first calculation.

Suppose the incoming electric field is already polarised at an angle \(\phi\) to the horizontal. If P1 has a transmission axis at \(\alpha_1\), the intensity after P1 is

\[ I_1=I_0\cos^2(\alpha_1-\phi) \]

The first filter does not automatically halve the intensity. It applies Malus’ law to the polarised light reaching it.

After P1, the light is polarised along P1’s axis. The next filter then uses the angle between P1 and P2:

\[ I_2=I_1\cos^2(\alpha_2-\alpha_1) \]

The difference between these two situations is worth remembering:

This is also why you must update the polarisation direction after every filter. Once light passes through an ideal polariser, its transmitted electric field is aligned with that polariser’s transmission axis.

Testing the relationship in an experiment

How could you test whether transmitted intensity depends on the separation between the filters rather than their absolute orientations?

Use a steady light source, two polarising filters, and a light sensor.

Arrange them in the order that light travels:

1. An initially unpolarised lamp produces a beam.
2. The beam passes through a fixed polariser, P.
3. It then passes through a rotatable analyser, A.
4. A light sensor behind A measures the transmitted intensity.

First, measure the intensity immediately after P. Call this \(I_P\). Then rotate A and measure the intensity after it.

Because the source is unpolarised, the first ideal polariser transmits half its original intensity. Suppose the measured intensity immediately before A is \(100\ \mathrm{W\,m^{-2}}\).

What should the sensor read as A rotates?

The ideal model predicts:

These are calculated ideal predictions, not measurements from an actual experiment.

Two results are particularly useful:

• At \(0^\circ\), the axes are parallel. The analyser transmits all the incoming electric-field component.
• At \(90^\circ\), the axes are crossed. The incoming electric field has no component parallel to the analyser’s transmission axis, so an ideal analyser transmits zero intensity.

The half-intensity result at \(45^\circ\) is another useful check. The electric-field amplitude is reduced by \(\cos45^\circ\), but the intensity is reduced by \(\cos^245^\circ=0.5\).

A useful test: rotate both filters together

Now imagine rotating P and A anticlockwise by \(30^\circ\), without changing their separation.

Their new orientations are \(30^\circ\) and \(75^\circ\), respectively. The separation is still \(45^\circ\).

Would the detector reading change?

For an initially unpolarised source, it would not. The first ideal polariser still transmits half the original intensity, and the analyser still transmits half the intensity reaching it.

The entire arrangement has changed its orientation relative to the laboratory, but none of the relevant relationships has changed.

This is a practical demonstration that there is no special horizontal direction in Malus’ law.

However, repeat the experiment with a source that is already plane-polarised, and keep the source’s electric-field direction fixed while rotating both filters. The result can change.

Why? The first filter now has a changing angle relative to the incoming electric field. Its transmitted intensity changes, even though the angle between P and A stays constant.

This is the distinction to keep:

• For the first filter, compare its axis with the incoming light’s original polarisation direction.
• For the next filter, compare its axis with the previous filter’s transmission axis.

What could make the experimental readings imperfect?

The simple experiment assumes ideal filters and a constant light source.

Real filters absorb some light, and they do not necessarily block every component perfectly. A crossed pair can therefore produce a small non-zero detector reading.

There is another possible problem. If the lamp becomes brighter or dimmer while you rotate the analyser, the detector readings may change even when the relative angle is unchanged.

A better measurement records the intensity immediately before the analyser as well as the intensity after it. Calculate the transmission ratio:

\[ T=\frac{I_{\text{after}}}{I_{\text{before}}} \]

Comparing this ratio at different angles helps separate changes caused by the analyser from changes in the incoming beam.

A reference detector monitoring the beam before A can improve the experiment further by tracking source fluctuations continuously.

The experiment tests the relative-angle model, while the non-zero crossed-filter reading reminds us that real apparatus does not behave exactly like an ideal mathematical model.

For a fuller derivation of the cosine-squared relationship, see Malus’ Law for HSC Physics: Intensity and Polariser Angles.

03Worked examples

04Questions and solutions

Question 1

An initially unpolarised beam has an intensity of \(240\ \mathrm{W\,m^{-2}}\). It passes through two ideal polarising filters:

• P1 has its transmission axis at \(15^\circ\).
• P2 has its transmission axis at \(75^\circ\).

Both angles are measured anticlockwise from the horizontal.

Calculate the intensity after each filter.

Solution 1

The intensity after P1 is \(120\ \mathrm{W\,m^{-2}}\), and the final intensity after P2 is \(30.0\ \mathrm{W\,m^{-2}}\).

Step 1: Apply the half-intensity rule to the first filter.

The incoming light is unpolarised.

\[ I_1=\frac{I_0}{2} =\frac{240}{2} =120\ \mathrm{W\,m^{-2}} \]

The light is now polarised at \(15^\circ\).

Step 2: Find the angle between the filters.

\[ \theta=75^\circ-15^\circ=60^\circ \]

Step 3: Apply Malus’ law to P2.

\[ \begin{aligned} I_2&=I_1\cos^2\theta\\ &=120\cos^2(60^\circ)\\ &=120(0.25)\\ &=30.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

The final intensity is \(30.0\ \mathrm{W\,m^{-2}}\).

The important detail is that P2 uses \(120\ \mathrm{W\,m^{-2}}\) as its incoming intensity and \(60^\circ\) as its relative angle. Neither the original intensity nor P2’s absolute angle should be reused.

Question 2

A beam of plane-polarised light has an intensity of \(72.0\ \mathrm{W\,m^{-2}}\). Its electric-field direction is \(160^\circ\) above the horizontal.

The beam passes through two ideal polarising filters:

• P1 has its transmission axis at \(25^\circ\).
• P2 has its transmission axis at \(100^\circ\).

Calculate the intensity after each filter. Explain why the first filter does not automatically halve the intensity.

Solution 2

The intensity after P1 is \(36.0\ \mathrm{W\,m^{-2}}\), and the final intensity after P2 is approximately \(2.41\ \mathrm{W\,m^{-2}}\).

Step 1: Find the angle between the incoming electric field and P1.

\[ 160^\circ-25^\circ=135^\circ \]

The smaller angle between the axes is

\[ \theta_1=180^\circ-135^\circ=45^\circ \]

Step 2: Calculate the intensity after P1.

The incoming light is already plane-polarised. Therefore, apply Malus’ law directly.

\[ \begin{aligned} I_1&=I_0\cos^2\theta_1\\ &=72.0\cos^2(45^\circ)\\ &=72.0(0.5)\\ &=36.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

Step 3: Find the angle between P1 and P2.

After P1, the light is polarised at \(25^\circ\).

Therefore,

\[ \theta_2=100^\circ-25^\circ=75^\circ \]

Step 4: Calculate the final intensity.

\[ \begin{aligned} I_2&=I_1\cos^2\theta_2\\ &=36.0\cos^2(75^\circ)\\ &=36.0(0.0670)\\ &=2.41\ \mathrm{W\,m^{-2}} \end{aligned} \]

The final intensity is approximately \(2.41\ \mathrm{W\,m^{-2}}\).

The first filter does not halve the intensity because the incoming beam is already plane-polarised. The amount transmitted depends on the angle between its existing electric-field direction and P1.

Question 3

An initially unpolarised beam has an intensity of \(200\ \mathrm{W\,m^{-2}}\). It passes through a fixed polariser P and a rotatable analyser A.

• P has its transmission axis at \(20^\circ\).
• Initially, A has its transmission axis at \(110^\circ\).
• A is then rotated \(30^\circ\) clockwise, towards P.

Calculate the intensity after P, the intensity after A before rotation, and the intensity after A following rotation.

Solution 3

The intensity after P is \(100\ \mathrm{W\,m^{-2}}\). The initial intensity after A is zero, while the final intensity after rotating A is \(25.0\ \mathrm{W\,m^{-2}}\).

Step 1: Calculate the intensity after P.

The incoming light is unpolarised, so

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

Step 2: Calculate the initial intensity after A.

Initially, the separation between the axes is

\[ \theta_{\text{initial}}=110^\circ-20^\circ=90^\circ \]

Therefore,

\[ \begin{aligned} I_{\text{initial}}&=100\cos^2(90^\circ)\\ &=0\ \mathrm{W\,m^{-2}} \end{aligned} \]

The filters are crossed, so the ideal analyser blocks the incoming polarised light.

Step 3: Find A’s new orientation.

A rotates \(30^\circ\) clockwise:

\[ \alpha_A=110^\circ-30^\circ=80^\circ \]

The new separation is

\[ \theta_{\text{final}}=80^\circ-20^\circ=60^\circ \]

Step 4: Calculate the final transmitted intensity.

\[ \begin{aligned} I_{\text{final}}&=100\cos^2(60^\circ)\\ &=100(0.25)\\ &=25.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

The final intensity is \(25.0\ \mathrm{W\,m^{-2}}\).

Notice the trap: the analyser rotated by \(30^\circ\), but its final separation from P is \(60^\circ\).

The rotation amount and the angle required in Malus’ law are not necessarily the same thing. You must determine the final orientations before calculating the intensity.

Question 4

A student investigates how transmitted intensity changes as an analyser rotates relative to a fixed polariser.

The student records the intensity immediately before and after the analyser at four different angles.

The following are illustrative sensor readings, not results from a reported experiment.

Answer the following:

a) Calculate the transmission ratio \(T=I_{\text{after}}/I_{\text{before}}\) at each angle.

b) Divide each transmission ratio by the ratio at \(0^\circ\). Compare the results with the predictions of Malus’ law.

c) Explain why the reading at \(90^\circ\) is not exactly zero.

d) Explain why using the intensity before the analyser is more reliable than comparing every reading with the original source intensity.

Solution 4

The normalised transmission ratios are approximately \(1.00\), \(0.750\), \(0.250\), and \(0.0323\). The first three closely match the cosine-squared predictions. The small residual at \(90^\circ\) is consistent with imperfections in real filters or background detector readings.

Step 1: Calculate the transmission ratio at each angle.

Use

\[ T=\frac{I_{\text{after}}}{I_{\text{before}}} \]

At \(0^\circ\):

\[ T_0=\frac{94.0}{100.0}=0.940 \]

At \(30^\circ\):

\[ T_{30}=\frac{73.3}{104.0}=0.7048 \]

At \(60^\circ\):

\[ T_{60}=\frac{22.6}{96.0}=0.2354 \]

At \(90^\circ\):

\[ T_{90}=\frac{3.1}{102.0}=0.0304 \]

Step 2: Normalise the ratios to the aligned-filter reading.

Divide each ratio by \(T_0=0.940\).

\[ \begin{aligned} T_{\text{relative},0}&=\frac{0.940}{0.940}=1.00\\ T_{\text{relative},30}&=\frac{0.7048}{0.940}=0.750\\ T_{\text{relative},60}&=\frac{0.2354}{0.940}=0.250\\ T_{\text{relative},90}&=\frac{0.0304}{0.940}=0.0323 \end{aligned} \]

Compare these with the ideal predictions:

The agreement is close at the first three angles.

Step 3: Explain the non-zero reading at \(90^\circ\).

For ideal crossed polarisers, the transmitted intensity is zero.

Real filters can allow a small amount of light through, even when their transmission axes are perpendicular. The detector may also register background light.

The data alone cannot establish which effect is responsible for the residual reading.

Step 4: Explain why the intensity before the analyser matters.

The intensity before A varies between \(96.0\) and \(104.0\ \mathrm{W\,m^{-2}}\).

If the student compared every downstream reading with a single original source reading, changes in source brightness could be mistaken for changes caused by rotating A.

Using the intensity immediately before A accounts for variations in the light actually reaching it.

Normalising to the aligned-filter reading also compensates for a constant transmission loss in the real apparatus.

The experiment therefore provides evidence for the cosine-squared relationship without requiring real filters to behave perfectly.

Question 5

This final problem tests a subtle distinction between the orientation of the filters and the orientation of the incoming light.

A beam has an intensity of \(160\ \mathrm{W\,m^{-2}}\). It can be prepared in either of two ways:

• Case U: The beam is unpolarised.
• Case P: The beam is plane-polarised, with its electric-field direction at \(10^\circ\) above the horizontal.

In both cases, the beam passes through two ideal filters:

• P1 has its transmission axis at \(40^\circ\).
• P2 has its transmission axis at \(100^\circ\).

A technician then rotates the entire two-filter assembly \(30^\circ\) anticlockwise. The light source and incoming beam remain unchanged.

Answer the following:

a) Calculate the final intensity in Case U before and after rotating the assembly.

b) Calculate the final intensity in Case P before and after rotating the assembly.

c) A student argues that the final intensity must be unchanged in both cases because the angle between P1 and P2 has not changed. Explain precisely why this argument works for one case but fails for the other.

d) In Case P, what additional physical adjustment would preserve both relative angles and restore the original final intensity?

Solution 5

The final intensities are:

• Case U: \(20.0\ \mathrm{W\,m^{-2}}\) before and after rotation.
• Case P: \(30.0\ \mathrm{W\,m^{-2}}\) before rotation and \(10.0\ \mathrm{W\,m^{-2}}\) afterwards.

The student’s argument is correct for unpolarised input but incorrect for already plane-polarised input. The missing consideration is the angle between the incoming electric field and the first filter.

Step 1: Calculate the original intensity in Case U.

The incoming light is unpolarised, so P1 transmits half its intensity.

\[ I_1=\frac{160}{2}=80.0\ \mathrm{W\,m^{-2}} \]

The separation between P1 and P2 is

\[ \theta=100^\circ-40^\circ=60^\circ \]

Therefore,

\[ \begin{aligned} I_2&=80.0\cos^2(60^\circ)\\ &=80.0(0.25)\\ &=20.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

Step 2: Calculate the rotated intensity in Case U.

After rotating both filters anticlockwise by \(30^\circ\):

\[ \alpha_1=70^\circ \]

\[ \alpha_2=130^\circ \]

The incoming light is still unpolarised, so the first filter still transmits half:

\[ I_1'=\frac{160}{2}=80.0\ \mathrm{W\,m^{-2}} \]

The separation between the filters remains

\[ \theta=130^\circ-70^\circ=60^\circ \]

Therefore,

\[ I_2'=80.0\cos^2(60^\circ) =20.0\ \mathrm{W\,m^{-2}} \]

The intensity is unchanged.

This happens because unpolarised light has no preferred electric-field direction. Rotating the first ideal polariser does not change the average intensity it transmits.

Step 3: Calculate the original intensity in Case P.

This time, the incoming beam is already polarised at \(10^\circ\).

The first filter is at \(40^\circ\), so the relative angle is

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

Apply Malus’ law:

\[ \begin{aligned} I_1&=160\cos^2(30^\circ)\\ &=160(0.75)\\ &=120\ \mathrm{W\,m^{-2}} \end{aligned} \]

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

The angle between P1 and P2 is \(60^\circ\).

Therefore,

\[ \begin{aligned} I_2&=120\cos^2(60^\circ)\\ &=120(0.25)\\ &=30.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

Step 4: Calculate the intensity after rotating the assembly in Case P.

The incoming electric field remains at \(10^\circ\), but P1 has moved to \(70^\circ\).

The new first-filter angle is

\[ \theta_1'=70^\circ-10^\circ=60^\circ \]

Therefore,

\[ \begin{aligned} I_1'&=160\cos^2(60^\circ)\\ &=160(0.25)\\ &=40.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

P2 is now at \(130^\circ\), so the angle between the filters is still \(60^\circ\).

The final intensity becomes

\[ \begin{aligned} I_2'&=40.0\cos^2(60^\circ)\\ &=40.0(0.25)\\ &=10.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

The final intensity has fallen from \(30.0\) to \(10.0\ \mathrm{W\,m^{-2}}\).

Step 5: Identify the missing adjustment.

To preserve both relative angles, the incoming electric-field direction must also rotate anticlockwise by \(30^\circ\).

Its new orientation would be \(40^\circ\).

The angle between the incoming polarisation direction and P1 would then remain \(30^\circ\), while the separation between P1 and P2 would remain \(60^\circ\).

The original intensity calculation would therefore be restored:

\[ I_1=160\cos^2(30^\circ)=120\ \mathrm{W\,m^{-2}} \]

\[ I_2=120\cos^2(60^\circ)=30.0\ \mathrm{W\,m^{-2}} \]

This is the important distinction: keeping the angle between two filters constant preserves the transmission fraction at the second filter, but it does not necessarily preserve the intensity reaching that filter.

The first filter still interacts with the incoming light’s original polarisation direction.

05What to remember in your next HSC question

Before using Malus’ law, identify the light reaching the filter you are considering.

The most useful habit is to keep an intensity ledger:

• Write down the intensity immediately before each filter.
• Identify the polarisation direction of the incoming light.
• Calculate the angle between that direction and the next transmission axis.
• Apply the appropriate intensity rule.
• Update the polarisation direction after the filter.

Once this is clear, questions involving three or more filters become much easier. Each new filter acts on the intensity and polarisation direction left by the previous one. You can apply the same reasoning in How to Apply Malus’ Law to Three Polarising Filters.