How to Calculate Intensity After Rotating an Analyser

A beam of light passes through a fixed polariser and then a rotatable analyser. You turn the analyser by 20° towards the first polariser. Does the transmitted intensity decrease by a factor of \(\cos^2 20^\circ\), because you rotated the analyser by 20°?

Not necessarily. The most common mistake in these questions is using the angle the analyser has rotated instead of the angle between its final transmission axis and the polarisation direction of the incoming light.

The reliable method is to track the intensity before and after every filter, then use the relative angle between adjacent transmission axes. This works for two polarising filters, as well as longer sequences of polariser-analyser combinations.

01The method: track intensity through each filter

Imagine shining unpolarised light through two filters. The first filter is called the polariser. The second is called the analyser.

The first filter produces plane-polarised light: the electric field of the transmitted light oscillates in one particular direction. The analyser only transmits the component of that electric field parallel to its own transmission axis.

This means that the amount of light transmitted depends on the angle between the incoming polarisation direction and the analyser’s axis.

Malus’ law

The relationship between the intensity entering an analyser and the intensity leaving it is:

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

where:

• \(I_{\text{in}}\) is the light intensity immediately before the analyser, measured in \(\mathrm{W\,m^{-2}}\).
• \(I_{\text{out}}\) is the intensity immediately after the analyser, in the same units.
• \(\theta\) is the angle between the incoming light’s polarisation direction and the analyser’s transmission axis.

The important detail is that \(\theta\) is a relative angle. It is not automatically the number of degrees through which you rotated the analyser.

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

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

Here, \(I_0\) is the initial unpolarised intensity, and \(I_1\) is the intensity after the first polariser.

You then use Malus’ law for the analyser:

\[
I_2=\frac{I_0}{2}\cos^2\theta
\]

If the incoming light is already plane-polarised, however, you do not automatically halve its intensity. Apply Malus’ law directly to the intensity entering the next filter.

A reliable decision process

For every filter, follow the same sequence:

1. Identify the intensity immediately before the filter.
2. Identify the polarisation direction of the light arriving at it.
3. Find the angle between that direction and the filter’s transmission axis.
4. Apply the appropriate intensity rule.
5. Use the resulting intensity as the input for the next filter.

For an ideal first polariser receiving unpolarised light, use \(I_{\text{out}}=I_{\text{in}}/2\). For each subsequent analyser, use \(I_{\text{out}}=I_{\text{in}}\cos^2\theta\).

Do not keep returning to the original intensity when working through multiple filters. Each new filter acts on the light that actually reaches it.

Why is the cosine squared?

The cosine comes from resolving the electric field into the component that can pass through the analyser.

Imagine the incoming electric field as an arrow. The analyser only transmits the part of that arrow parallel to its transmission axis.

The parallel component is:

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

Light intensity is proportional to the square of the electric field amplitude:

\[
I\propto E^2
\]

Therefore, the transmitted intensity is proportional to:

\[
I_{\text{out}}\propto (E\cos\theta)^2
\]

Dividing by the incoming intensity gives Malus’ law:

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

This gives you three useful reference cases:

At \(0^\circ\), the axes are parallel, so the entire electric field component passes through an ideal analyser.

At \(90^\circ\), the axes are perpendicular, so no component passes through. This arrangement is called crossed polarisers.

At \(45^\circ\), half the incoming intensity is transmitted. Notice that the electric field amplitude is reduced by a factor of \(\cos45^\circ\), but the intensity is reduced by a factor of \(\cos^245^\circ\).

02How to find the angle after rotating an analyser

This is the step that causes the most trouble in exam calculations.

Suppose the analyser is initially at an angle of \(55^\circ\) anticlockwise from the preceding polariser. You rotate it \(25^\circ\) clockwise, towards the preceding polariser.

What angle should you use in Malus’ law?

The final separation is:

\[
\theta=55^\circ-25^\circ
\]

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

You use \(30^\circ\), not \(25^\circ\).

Remember that polariser axes are lines

A transmission axis has no arrowhead. An axis at \(150^\circ\) is physically equivalent to one at \(30^\circ\), because both describe the same line.

For Malus’ law, use the smaller separation between the two axes, between \(0^\circ\) and \(90^\circ\).

A simple convention helps:

• Treat anticlockwise rotations as positive.
• Treat clockwise rotations as negative.
• Calculate the final relative orientation.
• Use the smaller angle between the final axes in Malus’ law.

You do not need to know the analyser’s absolute orientation relative to the room, the table, or the light source. Only the relative orientation matters.

What if the question gives the original intensity reading?

Sometimes a question gives you the intensity measured at the analyser’s initial position, then asks for the intensity after rotating it.

You can calculate the new intensity directly using a ratio.

At the initial position:

\[
I_{\text{old}}=I_{\text{before}}\cos^2\theta_{\text{old}}
\]

After rotating:

\[
I_{\text{new}}=I_{\text{before}}\cos^2\theta_{\text{new}}
\]

Divide the equations:

\[
\boxed{
I_{\text{new}}=
I_{\text{old}}
\frac{\cos^2\theta_{\text{new}}}
{\cos^2\theta_{\text{old}}}
}
\]

This shortcut works when the incoming intensity stays constant and the original transmission is non-zero.

The key is that the intensity immediately before the analyser has not changed. You are recalculating how much of that same incoming light passes through at the new angle.

03The experiment behind Malus’ law

Why should intensity follow a cosine-squared relationship in the first place?

A simple polarisation experiment allows us to test the prediction.

Setting up the apparatus

Arrange the equipment in the order that the light travels:

1. A steady light source produces an initially unpolarised beam.
2. The beam passes through a fixed polariser.
3. It then passes through an analyser mounted on a rotating stage with an angular scale.
4. A light sensor behind the analyser measures the transmitted intensity.

The sensor stays in position while the analyser rotates.

What should happen?

First, align the two transmission axes so that they are parallel.

The analyser should transmit the maximum possible intensity from the polarised beam.

Now rotate the analyser gradually.

Malus’ law predicts that the intensity will decrease according to \(\cos^2\theta\), reaching zero when the axes are crossed at \(90^\circ\).

For example, suppose the intensity immediately after the first polariser is 80 arbitrary intensity units.

The predicted readings are:

Illustrative idealised predictions, not measurements from a real experiment.

The predicted curve is not a straight line. Rotating the analyser through equal angular intervals does not produce equal decreases in intensity.

What could go wrong with the first experimental design?

Simply rotating the analyser and recording the sensor reading is not enough to guarantee a reliable result.

The sensor might detect background light, the light source might fluctuate, or the sensor might have a non-zero reading even when the beam is blocked.

These effects can make crossed polarisers appear to transmit light.

A better experimental procedure is to:

• Record the sensor’s background reading with the beam blocked.
• Subtract that background reading from subsequent measurements.
• Record a reference reading with the axes parallel.
• Keep the source, filters, and sensor aligned.
• Compare readings at different angles with the parallel-axis reference.

Real polarising filters are also imperfect. Some light can leak through even when the axes are crossed. Scattering and imperfect alignment can contribute to the remaining signal.

The experiment therefore tests whether the measured intensity approximately follows the predicted cosine-squared relationship, rather than expecting every real measurement to be exactly zero at \(90^\circ\).

04Calculating intensity through multiple filters

The same method works when light passes through three or more polarising filters.

The important rule is that each filter is compared with the polarisation direction immediately before it.

Consider three filters:

• \(P_1\): the first polariser.
• \(P_2\): a second polarising filter.
• \(A\): the final analyser.

If the incoming light is unpolarised, the calculation proceeds as follows:

Here, \(I_1\) is the intensity after the first polariser, and \(I_2\) is the intensity after the second polariser.

Notice that the angle for the analyser is measured relative to \(P_2\), not automatically relative to \(P_1\).

This is because the second filter has changed the polarisation direction. The light leaving \(P_2\) is polarised along the transmission axis of \(P_2\).

05The common mistake: treating rotation as another filter

There is one more situation worth understanding.

Suppose an intensity of \(40\ \mathrm{W\,m^{-2}}\) reaches an analyser. Initially, the analyser is positioned at \(60^\circ\) to the incoming polarisation direction.

It is then rotated \(30^\circ\) towards the parallel position.

What happens to the transmitted intensity?

First, calculate the original reading:

\[
I_{\text{old}}=40\cos^2 60^\circ
\]

\[
I_{\text{old}}=40(0.25)
\]

\[
I_{\text{old}}=10\ \mathrm{W\,m^{-2}}
\]

The analyser is now at:

\[
\theta_{\text{new}}=60^\circ-30^\circ
\]

\[
\theta_{\text{new}}=30^\circ
\]

The new transmitted intensity is:

\[
I_{\text{new}}=40\cos^2 30^\circ
\]

\[
I_{\text{new}}=40(0.75)
\]

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

Intensity increases

The intensity has increased from 10 to \(30\ \mathrm{W\,m^{-2}}\).

Why?

The analyser is now closer to being parallel with the incoming polarisation direction. A greater proportion of the electric field passes through.

A tempting but incorrect calculation would be:

\[
I_{\text{new}}=10\cos^2 30^\circ
\]

This gives \(7.5\ \mathrm{W\,m^{-2}}\), which is wrong.

The mistake is treating the rotation as though another filter has been added after the original analyser. It hasn’t. You have simply changed the orientation of the same analyser.

Always return to the intensity immediately before that analyser when calculating its new transmitted intensity.

06Assumptions and limitations

The calculations in this article use the ideal polariser model normally used for introductory Malus’ law problems.

Keep these assumptions in mind:

• The first polariser receives unpolarised light when you use the half-intensity rule.
• Each ideal polariser transmits light polarised along its transmission axis.
• The light source produces a steady beam.
• The distance between the source, filters, and detector stays constant.
• Real filters are imperfect and may transmit a small amount of light even when their axes are crossed.

One final distinction is worth remembering.

Intensity can change because of the orientation of a polariser, but it can also change because light spreads out as it travels away from a source.

These are different physical effects. If a question changes the distance between a light source and detector, you may need the inverse-square law for light intensity.

For a rotating-analyser question, however, keep the source and detector geometry fixed unless the question tells you otherwise.

07Questions and solutions

Try each question before reading its solution. The questions move from straightforward Malus’ law calculations towards more demanding problems involving changing polarisation directions and experimental evidence.

Question 1

An initially unpolarised beam of light has an intensity of \(96\ \mathrm{W\,m^{-2}}\). It passes through an ideal polariser, followed by an analyser whose transmission axis is at \(60^\circ\) to the first polariser.

Calculate:

a) The intensity after the first polariser.

b) The intensity after the analyser.

Solution 1

The final transmitted intensity is \(12\ \mathrm{W\,m^{-2}}\). The first polariser halves the unpolarised beam, and the analyser then applies Malus’ law.

Step 1: Calculate the intensity after the first polariser.

Because the incoming light is unpolarised:

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

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

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

Step 2: Apply Malus’ law to the analyser.

The relative angle between the two transmission axes is \(60^\circ\).

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

\[
I_2=48(0.25)
\]

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

The final intensity is \(12\ \mathrm{W\,m^{-2}}\).

The analyser transmits only one-quarter of the intensity reaching it. Remember that this is one-quarter of \(48\), not one-quarter of the original \(96\).

Question 2

A beam of light is already plane-polarised and has an intensity of \(64\ \mathrm{W\,m^{-2}}\).

It passes through two ideal filters:

• Filter \(P\) has its transmission axis at \(30^\circ\) to the incoming polarisation direction.
• Filter \(A\) has its transmission axis at \(60^\circ\) to the transmission axis of \(P\).

Calculate the intensity after each filter.

A student argues that the total angle is \(30^\circ+60^\circ=90^\circ\), so no light should emerge from the second filter.

Explain why this prediction is incorrect.

Solution 2

The final intensity is \(12\ \mathrm{W\,m^{-2}}\), not zero. The key is that each filter changes the polarisation direction of the transmitted light.

Step 1: Calculate the intensity after filter \(P\).

The incoming light is already plane-polarised.

Therefore, do not halve its intensity. Apply Malus’ law directly:

\[
I_P=64\cos^2 30^\circ
\]

\[
I_P=64(0.75)
\]

\[
\boxed{I_P=48\ \mathrm{W\,m^{-2}}}
\]

Step 2: Calculate the intensity after filter \(A\).

The light leaving \(P\) is now polarised along the transmission axis of \(P\).

Therefore, the relevant angle for \(A\) is \(60^\circ\).

\[
I_A=48\cos^2 60^\circ
\]

\[
I_A=48(0.25)
\]

\[
\boxed{I_A=12\ \mathrm{W\,m^{-2}}}
\]

Step 3: Explain the student’s mistake.

The student has added the two angles and treated the final filter as though it acts directly on the original polarisation direction.

That ignores what happens at the first filter.

Filter \(P\) changes the polarisation direction to match its own transmission axis. Filter \(A\) then acts on this newly polarised light.

The correct calculation is therefore:

\[
I_A=64\cos^2 30^\circ\cos^2 60^\circ
\]

It is not:

\[
I_A=64\cos^2 90^\circ
\]

The direction of polarisation is reset by each ideal filter. That is why you must track the intensity and relative angle at every stage.

Question 3

An analyser initially has its transmission axis at \(70^\circ\) to the incoming polarisation direction. A light sensor measures an intensity of \(4.68\ \mathrm{W\,m^{-2}}\) after the analyser.

The analyser is rotated \(40^\circ\) towards the parallel position. Assume the intensity reaching the analyser remains constant.

Calculate:

a) The new relative angle.

b) The new transmitted intensity.

c) The factor by which the transmitted intensity changes.

Solution 3

The new transmitted intensity is approximately \(30.0\ \mathrm{W\,m^{-2}}\), an increase by a factor of approximately 6.41.

Step 1: Find the new relative angle.

The analyser rotates towards the parallel position:

\[
\theta_{\text{new}}=70^\circ-40^\circ
\]

\[
\boxed{\theta_{\text{new}}=30^\circ}
\]

Step 2: Calculate the intensity immediately before the analyser.

We know the original transmitted intensity and the original angle.

Using Malus’ law:

\[
I_{\text{old}}=I_{\text{before}}\cos^2 70^\circ
\]

Rearrange:

\[
I_{\text{before}}=
\frac{I_{\text{old}}}{\cos^2 70^\circ}
\]

Substitute:

\[
I_{\text{before}}=
\frac{4.68}{0.11698}
\]

\[
I_{\text{before}}\approx40.0\ \mathrm{W\,m^{-2}}
\]

Step 3: Calculate the new transmitted intensity.

Use the new angle of \(30^\circ\):

\[
I_{\text{new}}=40.0\cos^2 30^\circ
\]

\[
I_{\text{new}}=40.0(0.75)
\]

\[
\boxed{I_{\text{new}}=30.0\ \mathrm{W\,m^{-2}}}
\]

Step 4: Calculate the factor of change.

\[
\text{Factor}=
\frac{I_{\text{new}}}{I_{\text{old}}}
\]

\[
\text{Factor}=\frac{30.0}{4.68}
\]

\[
\boxed{\text{Factor}\approx6.41}
\]

The intensity increases by a factor of approximately 6.41.

The important point is that the original analyser reading was already reduced by the original \(70^\circ\) orientation. You must account for that reduction before calculating the new reading.

Question 4

A student investigates Malus’ law using a fixed polariser, a rotating analyser, and a light sensor.

The student records the following illustrative readings:

When the beam is blocked, the sensor reads 4 units.

a) Calculate the background-corrected readings.

b) Use the parallel-axis reading to predict the ideal readings at \(45^\circ\) and \(60^\circ\).

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

d) Suggest one improvement to the experimental method.

Solution 4

The corrected readings are 100, 75, 50, 25, and 4 units. They follow the ideal cosine-squared pattern closely, except for the residual reading at \(90^\circ\).

Step 1: Subtract the background reading.

The background reading is 4 units.

Subtract 4 from every raw reading:

Step 2: Calculate the predicted readings.

The corrected intensity when the axes are parallel is 100 units.

At \(45^\circ\):

\[
I=100\cos^2 45^\circ
\]

\[
I=100(0.5)
\]

\[
\boxed{I=50\text{ units}}
\]

At \(60^\circ\):

\[
I=100\cos^2 60^\circ
\]

\[
I=100(0.25)
\]

\[
\boxed{I=25\text{ units}}
\]

Both predictions match the corrected readings.

Step 3: Explain the residual intensity at \(90^\circ\).

For ideal crossed polarisers, Malus’ law predicts zero transmitted intensity.

However, the corrected experimental reading is 4 units.

The dark reading has already been subtracted, so this residual is not explained by the sensor’s original background offset.

Possible explanations include:

• Imperfect polarising filters that allow some light through.
• Scattering of light around the filters.
• Small errors in the alignment of the transmission axes.

The result does not automatically invalidate Malus’ law. It shows that the ideal model does not account for every feature of a real apparatus.

Step 4: Suggest an improvement.

One improvement would be to enclose the optical path in a dark tube to reduce stray light.

Using higher-quality polarising filters and carefully checking their alignment would also improve the reliability of the measurements.

The experiment provides good evidence for the cosine-squared relationship over the measured angles, while also showing the limitations of the ideal filter model.

Question 5

An initially unpolarised beam of intensity \(I_0\) passes through four ideal polarising filters in this order:

• A first polariser \(P\).
• Two intermediate polarisers, \(M_1\) and \(M_2\).
• A final analyser \(A\).

The transmission axes of \(P\) and \(A\) are perpendicular, so they are crossed at \(90^\circ\).

The two intermediate filters can be rotated freely, but their axes must lie in order between the axes of \(P\) and \(A\), without reversing direction.

Let:

• \(\alpha\) be the relative angle between \(P\) and \(M_1\).
• \(\beta\) be the relative angle between \(M_1\) and \(M_2\).
• \(\gamma\) be the relative angle between \(M_2\) and \(A\).

Therefore:

\[
\alpha+\beta+\gamma=90^\circ
\]

Determine the arrangement of the intermediate filters that maximises the final transmitted intensity.

Then:

a) Calculate the maximum transmitted intensity as a fraction of \(I_0\).

b) Compare this with the maximum transmitted intensity when only one intermediate polariser is placed between \(P\) and \(A\).

c) Explain how adding another filter can increase the final transmitted intensity, even though every individual filter reduces the intensity reaching it.

Solution 5

The maximum transmitted intensity occurs when all three successive angular separations are equal to \(30^\circ\).

The maximum final intensity is:

\[
\boxed{I_{\text{max}}=\frac{27}{128}I_0}
\]

This is approximately \(21.1\%\) of the original unpolarised intensity.

Step 1: Express the final intensity in terms of the three relative angles.

The first polariser receives unpolarised light:

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

After \(M_1\):

\[
I_1=\frac{I_0}{2}\cos^2\alpha
\]

After \(M_2\):

\[
I_2=\frac{I_0}{2}
\cos^2\alpha\cos^2\beta
\]

After the final analyser:

\[
I_A=\frac{I_0}{2}
\cos^2\alpha\cos^2\beta\cos^2\gamma
\]

We need to maximise this expression, subject to:

\[
\alpha+\beta+\gamma=90^\circ
\]

Step 2: Determine how the angular separation should be distributed.

A tempting approach is to make one of the angular gaps very small. That filter would transmit almost all the light reaching it.

However, this forces a larger angular gap somewhere else. That larger gap produces a greater reduction in intensity.

The best arrangement balances the three angular gaps.

We can justify this using the trigonometric identity:

\[
\cos a\cos b=
\frac{\cos(a+b)+\cos(a-b)}{2}
\]

For a fixed value of \(a+b\), the first cosine term is constant.

The second term, \(\cos(a-b)\), is largest when \(a=b\), because its maximum value is 1.

Therefore, for a fixed combined angular separation, two unequal gaps transmit more efficiently when they are made equal.

Applying this balancing argument to the three successive gaps shows that the maximum occurs when:

\[
\alpha=\beta=\gamma
\]

Since their sum must be \(90^\circ\):

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

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

Step 3: Calculate the intensity after each filter.

After the first polariser:

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

After \(M_1\):

\[
I_1=\frac{I_0}{2}\cos^2 30^\circ
\]

\[
I_1=\frac{3I_0}{8}
\]

After \(M_2\):

\[
I_2=\frac{3I_0}{8}\cos^2 30^\circ
\]

\[
I_2=\frac{9I_0}{32}
\]

After the final analyser:

\[
I_A=\frac{9I_0}{32}\cos^2 30^\circ
\]

\[
\boxed{I_A=\frac{27I_0}{128}}
\]

Step 4: Compare this with only one intermediate polariser.

With one intermediate polariser, the total \(90^\circ\) rotation is divided into two equal angles to maximise transmission.

Each angle is:

\[
\theta=\frac{90^\circ}{2}=45^\circ
\]

The final intensity is:

\[
I_{\text{one}}=
\frac{I_0}{2}
\cos^2 45^\circ
\cos^2 45^\circ
\]

\[
I_{\text{one}}=
\frac{I_0}{2}(0.5)(0.5)
\]

\[
\boxed{I_{\text{one}}=\frac{I_0}{8}}
\]

Compare the two arrangements:

Adding a second intermediate polariser increases the final transmitted intensity by approximately \(68.75\%\) compared with the best arrangement using just one intermediate polariser.

Step 5: Explain the apparent contradiction.

It is true that every ideal polariser reduces or maintains the intensity of the light immediately before it.

However, the comparison here is between two different arrangements, not between the input and output of one additional filter.

With only the first and final polarisers, the axes are crossed. The light emerging from \(P\) is perpendicular to the transmission axis of \(A\), so the final intensity is zero.

The intermediate polarisers progressively change the polarisation direction. The final analyser therefore receives light whose electric field has a component parallel to its transmission axis.

The extra filters do not create light or increase the intensity at any individual stage. They change the polarisation direction so that more light can pass through the final analyser.

That is the important distinction: each filter reduces the intensity reaching the next stage, but changing the sequence and orientation of filters can increase the final transmitted intensity.

07What to remember in your next HSC question

When a question asks you to calculate intensity after rotating an analyser, use this sequence:

1. Find the intensity immediately before the first filter.
2. If that light is unpolarised, halve its intensity after the first ideal polariser.
3. Identify the polarisation direction entering the analyser.
4. Calculate the analyser’s final relative angle after rotation.
5. Apply Malus’ law using the intensity immediately before that analyser.
6. Repeat the process for any additional filters.

The biggest exam trap is confusing the angle of rotation with the final relative angle between transmission axes.

Once you keep those two quantities separate, even a long sequence of rotating polarisers becomes a straightforward intensity calculation.