Why a Third Polariser Can Increase Transmission Between Crossed Filters
Discover why adding a middle polariser at 45° allows light to pass through two crossed filters. Learn to track intensity after each filter using Malus' law, interpret experimental results, and solve challenging HSC Physics questions.
Two polarising filters can be placed at right angles so that almost no light passes through both. Now put a third polariser between them, rotate it to 45°, and suddenly some light gets through.
That sounds impossible. Surely adding another filter can only block more light?
The key is that a polariser does two things: it reduces the light's intensity, and it changes the direction of its polarisation. The middle filter prepares the light in a new direction, allowing the final filter to transmit some of it.
**Adding a third polariser increases the transmitted intensity because it rotates the light's polarisation direction in stages.** It does not create extra light or make any individual filter increase the intensity passing through it.
Let's track exactly what happens, using the intensity before and after each filter.
## Why crossed polarisers block light
Imagine unpolarised light entering a polarising filter. The filter only transmits the component of the light's electric field that oscillates parallel to its transmission axis.
For ideal polarised light, the transmitted intensity depends on the angle between the light's polarisation direction and the filter's axis. This is described by Malus' law:
\[
I = I_{\text{in}}\cos^2\theta
\]
Here:
- \(I\) is the transmitted intensity, measured in watts per square metre (W m\(^{-2}\)).
- \(I_{\text{in}}\) is the intensity immediately before the filter.
- \(\theta\) is the angle between the incoming polarisation direction and the filter's transmission axis.
Notice that the angle is relative to the incoming polarisation, not necessarily relative to the horizontal.
Consider two filters:
- The first filter establishes a polarisation direction.
- The second filter has its transmission axis at 90° to the first.
The filters are called *crossed polarisers*. Their transmission axes are perpendicular.
After the first filter, the light is polarised along its transmission axis. The second filter only transmits the component parallel to its own axis. Since the axes are perpendicular, that component is zero.
Therefore,
\[
I_2 = I_1\cos^2 90^\circ = 0
\]
No light passes through the second ideal filter.
[IMAGE: Show a beam of unpolarised light entering a vertical transmission-axis polariser, followed by a second polariser with a horizontal axis at 90 degrees. Illustrate the vertical electric-field oscillation between the filters and zero transmitted beam after the second filter. Label the axes and the crossed arrangement.]
## What changes when a third polariser is added?
Now place a third filter between the original two. Keep the first and last filters crossed, but rotate the middle filter so that its axis is at an angle \(\theta\) relative to the first.
The important detail is that the middle filter is no longer perpendicular to the incoming polarisation. It can therefore transmit some light.
That transmitted light is now polarised along the middle filter's axis. The final filter receives light with a different polarisation direction from the one it would have received without the middle filter.
It can transmit some of this light, too.
[IMAGE: Annotated side-view diagram of three polarising filters in sequence. The first axis is vertical, the middle axis is tilted by angle theta relative to the first, and the third axis is horizontal, perpendicular to the first. Show the transmitted beam between each filter, with polarisation directions aligned to the preceding filter's axis. Label theta and the 90-degree angle between the first and third axes.]
### Track the intensity through all three filters
Assume:
- The incoming light is unpolarised, with initial intensity \(I_0\).
- All three filters are ideal, so absorption and reflection losses beyond the polarisation effect are ignored.
- The first and third transmission axes are perpendicular.
- The middle axis is at an angle \(\theta\) to the first axis, where \(0^\circ \leq \theta \leq 90^\circ\).
We will use relative angles rather than assigning absolute directions such as horizontal or vertical. Set the first axis as the reference direction.
**After the first polariser**
Unpolarised light passing through an ideal polariser has half its original intensity:
\[
I_1 = \frac{I_0}{2}
\]
The transmitted light is now polarised along the first filter's axis.
**After the middle polariser**
The angle between the incoming polarisation and the middle filter's axis is \(\theta\).
Applying Malus' law:
\[
I_2 = I_1\cos^2\theta
\]
Substitute the intensity after the first filter:
\[
I_2 = \frac{I_0}{2}\cos^2\theta
\]
The intensity decreases, but the transmitted light is now polarised along the middle filter's axis.
**After the final polariser**
The first and final axes are 90° apart. Since the middle axis is at angle \(\theta\) to the first, the angle between the middle and final axes is:
\[
90^\circ-\theta
\]
Apply Malus' law again:
\[
I_3 = I_2\cos^2(90^\circ-\theta)
\]
Using the trigonometric identity \(\cos(90^\circ-\theta)=\sin\theta\):
\[
I_3 = \frac{I_0}{2}\cos^2\theta\sin^2\theta
\]
This is the intensity that emerges from all three filters.
The crucial result is that both factors can be non-zero. The middle filter allows some light through the first stage and changes its polarisation direction so that the final filter can transmit some of that light.
Why 45° gives the greatest transmission
The equation for the final intensity is:
\[ I_3 = \frac{I_0}{2}\cos^2\theta\sin^2\theta \]
We want to find the angle that makes this expression as large as possible.
At first, you might think the middle filter should be closer to the first filter. That would allow more light through the middle.
But there’s a catch: the closer the middle filter is to the first, the closer it is to being perpendicular to the final filter. The final filter then blocks more of the light.
The best arrangement balances the two stages.
At \(45^\circ\):
\[ \cos^2 45^\circ = \sin^2 45^\circ = 0.5 \]
Therefore,
\[ \begin{aligned} I_3 &= \frac{I_0}{2}(0.5)(0.5)\\\ &= \frac{I_0}{8} \end{aligned} \]
Maximum transmission
The final intensity is one-eighth of the original unpolarised intensity.
This is the maximum because the product \(\cos^2\theta\sin^2\theta\) cannot exceed \(1/4\).
Compare the intensity at different angles
The following table uses the ideal model, with all intensities expressed as fractions of the original unpolarised intensity \(I_0\).
| Middle filter angle \(\theta\) | After first \(I_1\) | After middle \(I_2\) | After final \(I_3\) |
|---|---|---|---|
| 0° | 0.500 \(I_0\) | 0.500 \(I_0\) | 0 |
| 30° | 0.500 \(I_0\) | 0.375 \(I_0\) | 0.09375 \(I_0\) |
| 45° | 0.500 \(I_0\) | 0.250 \(I_0\) | 0.125 \(I_0\) |
| 60° | 0.500 \(I_0\) | 0.125 \(I_0\) | 0.09375 \(I_0\) |
| 90° | 0.500 \(I_0\) | 0 | 0 |
Notice two things.
First, the intensity after the final filter is greatest at 45°. Second, the results at 30° and 60° are identical.
Why?
At 30°, the middle filter transmits 75% of the light reaching it, but the final filter transmits only 25% of that amount.
At 60°, those percentages swap. The middle filter transmits 25%, and the final filter transmits 75%.
Both arrangements give the same overall transmission.
A useful distinction: more transmitted light does not mean more light at every stage
There is a subtle trap here.
Suppose the original first polariser transmits \(50\%\) of the unpolarised light. At the optimum middle-filter angle, the final output is \(12.5\%\) of the original intensity.
The middle filter has reduced the intensity from \(50\%\) to \(25\%\). The final filter reduces it again, from \(25\%\) to \(12.5\%\).
So how can we say that adding a filter increases transmission?
We are comparing two different arrangements:
- Without the middle filter, the crossed outer filters transmit zero light.
- With the middle filter at 45°, the complete arrangement transmits \(12.5\%\) of the original intensity.
The middle filter does not increase the intensity arriving at the next filter. It changes the polarisation direction, allowing the final filter to transmit a component that was previously blocked.
This distinction is essential: a passive polariser cannot increase the intensity of the light passing through it, but changing the orientation of a sequence of polarisers can increase the final transmitted intensity.
01Testing the explanation with an experiment
How could we check that the middle polariser really causes this change?
We can use a light source, three filters, and a light sensor. The experiment tests whether the final intensity depends on the orientation of the middle filter, while the two outer filters remain crossed.
Setting up the apparatus
Arrange the equipment in a straight line:
- Place a lamp producing unpolarised light at one end.
- Put the first polariser in the beam.
- Place a rotatable middle polariser after it.
- Place the final polariser after the middle one, with its transmission axis fixed at 90° to the first.
- Position a light sensor behind the final polariser to measure the transmitted intensity.
Only the middle polariser should rotate. Keep the source, first polariser, final polariser, and sensor fixed.
What should the experiment show?
Before rotating the middle filter, predict the sensor reading.
With only the first and final filters in the beam, their crossed axes should block the transmitted light. The sensor should register approximately its background reading.
Now insert the middle filter.
As it rotates from 0° towards 45°, the final intensity should increase. It should reach a maximum at 45°, then decrease as the filter approaches 90°.
At 90°, the middle filter is perpendicular to the first. Ideally, no light reaches the final filter.
The expected pattern is:
| Middle filter angle | Predicted final intensity |
|---|---|
| 0° | 0 |
| 15° | \(0.03125I_0\) |
| 30° | \(0.09375I_0\) |
| 45° | \(0.125I_0\) |
| 60° | \(0.09375I_0\) |
| 75° | \(0.03125I_0\) |
| 90° | 0 |
Illustrative theoretical values for ideal polarisers, not recorded experimental measurements.
A small intensity calculation
Suppose the incoming unpolarised light has an intensity of \(80\ \mathrm{W\\,m^{-2}}\).
At the optimum middle-filter angle of 45°:
\[ I_1=\frac{80}{2}=40\ \mathrm{W\\,m^{-2}} \]
After the middle filter:
\[ I_2=40\cos^2 45^\circ =20\ \mathrm{W\\,m^{-2}} \]
After the final filter:
\[ I_3=20\cos^2 45^\circ =10\ \mathrm{W\\,m^{-2}} \]
The sensor should therefore receive \(10\ \mathrm{W\\,m^{-2}}\) in the ideal model.
That is a useful result: the original crossed arrangement transmits zero, while the three-filter arrangement transmits \(10\ \mathrm{W\\,m^{-2}}\).
What could go wrong with the experiment?
A sensor reading is not automatically a measurement of the light transmitted by the filters.
For example, ambient light could reach the sensor even when the crossed filters ideally block the lamp. A lamp whose output changes during the investigation could also make the readings misleading.
A better design would:
- Enclose the optical path to reduce ambient light.
- Measure the sensor’s background reading with the lamp blocked.
- Keep the lamp output and sensor position constant.
- Rotate only the middle polariser.
- Take readings at several angles, including both endpoints and 45°.
The experiment should then show a rise and fall in transmitted intensity, with a maximum near 45°.
Real filters are not perfect. They may absorb some light even when their axes are aligned, and crossed filters may allow a small amount of light through. Reflections, sensor sensitivity, and source stability can also affect the results.
The ideal calculation predicts the underlying pattern, not necessarily the exact reading from a real apparatus.
02Worked examples
Worked example: Calculate the intensity after each filter
Unpolarised light with an initial intensity of \(96\ \mathrm{W\\,m^{-2}}\) passes through three ideal polarisers. The first and third axes are perpendicular, and the middle axis is at 30° to the first.
Calculate the intensity after each filter.
Step 1: Calculate the intensity after the first polariser.
The incoming light is unpolarised, so the first ideal polariser transmits half its intensity.
\[ I_1=\frac{I_0}{2} =\frac{96}{2} =48\ \mathrm{W\\,m^{-2}} \]
Step 2: Calculate the intensity after the middle polariser.
The angle between the first and middle axes is 30°.
Using Malus’ law:
\[ \begin{aligned} I_2&=I_1\cos^2 30^\circ\\\ &=48(0.75)\\\ &=36\ \mathrm{W\\,m^{-2}} \end{aligned} \]
The surviving light is now polarised along the middle filter’s axis.
Step 3: Calculate the intensity after the final polariser.
The outer axes are perpendicular, so the angle between the middle and final axes is:
\[ 90^\circ-30^\circ=60^\circ \]
Apply Malus’ law again:
\[ \begin{aligned} I_3&=I_2\cos^2 60^\circ\\\ &=36(0.25)\\\ &=9\ \mathrm{W\\,m^{-2}} \end{aligned} \]
The final transmitted intensity is \(9\ \mathrm{W\\,m^{-2}}\).
Without the middle polariser, the crossed outer filters would ideally transmit zero. The middle filter allows some light to pass by changing its polarisation direction.
Worked example: Find the possible middle-filter angles
Unpolarised light with an initial intensity of \(160\ \mathrm{W\\,m^{-2}}\) passes through two fixed crossed polarisers with a rotatable middle polariser between them.
A sensor measures a final transmitted intensity of \(15\ \mathrm{W\\,m^{-2}}\).
Find the possible angles of the middle polariser relative to the first.
Step 1: Start with the expression for final intensity.
For crossed outer filters:
\[ I_3=\frac{I_0}{2}\cos^2\theta\sin^2\theta \]
Substitute the known intensities:
\[ 15=80\cos^2\theta\sin^2\theta \]
Step 2: Rearrange using a double-angle identity.
Recall that:
\[ \sin 2\theta=2\sin\theta\cos\theta \]
Therefore,
\[ \sin^2\theta\cos^2\theta =\frac{1}{4}\sin^2 2\theta \]
Substitute:
\[ \begin{aligned} 15&=20\sin^2 2\theta\\\ \sin^2 2\theta&=0.75 \end{aligned} \]
Step 3: Find both possible angles.
Within the interval \(0^\circ\leq\theta\leq90^\circ\):
\[ 2\theta=60^\circ \quad\text{or}\quad 120^\circ \]
Therefore,
\[ \boxed{\theta=30^\circ\text{ or }60^\circ} \]
Both angles produce the same final intensity.
This is not a calculation error. The arrangement is symmetrical about 45°. At 30°, the middle filter transmits more light and the final filter transmits less. At 60°, the opposite happens.
A final sensor reading alone cannot distinguish between these two arrangements.
If you need to practise applying Malus’ law to arbitrary sequences of filter angles, see How to Apply Malus’ Law to Three Polarising Filters. The important extra idea here is that the order and orientation of the filters determine how much light survives the complete arrangement.
## When does adding a third polariser NOT increase transmission?
The middle polariser must have a suitable orientation. Simply inserting another filter does not guarantee a greater final intensity.
Consider the two extreme cases.
**Case 1: The middle filter is parallel to the first.**
Here, \(\theta=0^\circ\).
The first and middle filters have parallel axes, so the middle filter transmits all the light that reaches it:
\[
I_2=I_1\cos^2 0^\circ=I_1
\]
However, the final filter is perpendicular to the middle filter. It blocks all the transmitted light.
Therefore, \(I_3=0\).
**Case 2: The middle filter is parallel to the final filter.**
Here, \(\theta=90^\circ\).
The middle filter is now perpendicular to the first filter. It receives light polarised at 90° to its transmission axis, so it blocks everything before the light can reach the final filter.
Again, \(I_3=0\).
[IMAGE: Two contrasting three-polariser arrangements. In the first, P1 and P2 have parallel axes, while P3 is perpendicular. In the second, P2 and P3 have parallel axes, while P2 is perpendicular to P1. Show where the light is blocked in each arrangement.]
The important conclusion is that the middle filter must sit at an intermediate angle to allow light through both stages.
At 45°, the two stages are equally favourable. At either endpoint, one stage blocks everything.
This is why the statement "adding another filter always reduces transmitted intensity" is incomplete. It is true when comparing the intensity immediately before and after an individual passive filter. It is not necessarily true when comparing the final outputs of two different filter arrangements.
## Questions and solutions
Try these questions without looking at the solutions first. They progress from straightforward applications of Malus' law to problems involving experimental reasoning and competing explanations.
### Question 1
Unpolarised light with an initial intensity of \(120\ \mathrm{W\,m^{-2}}\) passes through three ideal polarising filters.
The first and third filters have perpendicular transmission axes. The middle filter is positioned at 30° relative to the first.
Calculate the intensity:
a. After the first filter.
b. After the middle filter.
c. After the final filter.
Explain why the final intensity is greater than zero.
[IMAGE: Three-filter arrangement for Question 1. Show unpolarised incoming light, P1 with a reference axis, P2 at 30 degrees to P1, and P3 at 90 degrees to P1. Include the light propagation direction but do not show calculated intensity values.]
### Solution 1
The final transmitted intensity is \(11.25\ \mathrm{W\,m^{-2}}\). The middle filter changes the polarisation direction, allowing some light through the final filter.
**Step 1**: Calculate the intensity after the first filter.
The incoming light is unpolarised, so an ideal polariser transmits half its intensity.
\[
\begin{aligned}
I_1&=\frac{I_0}{2}\\
&=\frac{120}{2}\\
&=60\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
**Step 2**: Calculate the intensity after the middle filter.
The angle between the first and middle axes is 30°.
Using Malus' law:
\[
\begin{aligned}
I_2&=I_1\cos^2 30^\circ\\
&=60(0.75)\\
&=45\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
**Step 3**: Calculate the intensity after the final filter.
The angle between the middle and final axes is:
\[
90^\circ-30^\circ=60^\circ
\]
Apply Malus' law again:
\[
\begin{aligned}
I_3&=I_2\cos^2 60^\circ\\
&=45(0.25)\\
&=11.25\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
The final intensity is therefore \(11.25\ \mathrm{W\,m^{-2}}\).
Without the middle filter, the first and final filters would be crossed and the ideal transmitted intensity would be zero. The middle filter allows a non-zero component of the light to pass through the final filter.
### Question 2
Unpolarised light with an initial intensity of \(200\ \mathrm{W\,m^{-2}}\) passes through three ideal polarisers.
The first and final polarisers are crossed. The middle polariser is rotatable, with its transmission axis between 0° and 90° relative to the first.
A sensor measures a final transmitted intensity of \(12.5\ \mathrm{W\,m^{-2}}\).
a. Calculate the maximum possible final intensity.
b. Determine the two possible orientations of the middle polariser.
c. Explain why the sensor reading does not identify a unique angle.
[IMAGE: Three crossed outer and intermediate polariser arrangement for Question 2. Show P1 and P3 perpendicular, P2 rotatable through an unknown angle theta, and a light sensor after P3. Do not indicate the two possible angles.]
### Solution 2
The maximum possible final intensity is \(25\ \mathrm{W\,m^{-2}}\), and the middle filter could be positioned at either \(22.5^\circ\) or \(67.5^\circ\).
**Step 1**: Calculate the maximum transmitted intensity.
The maximum occurs when the middle filter is at 45°.
\[
I_{3,\max}=\frac{I_0}{8}
\]
Substitute:
\[
I_{3,\max}=\frac{200}{8}
=25\ \mathrm{W\,m^{-2}}
\]
**Step 2**: Use the measured intensity to find the angle.
The final intensity equation is:
\[
I_3=\frac{I_0}{8}\sin^2 2\theta
\]
Substitute the measured values:
\[
12.5=25\sin^2 2\theta
\]
Therefore,
\[
\sin^2 2\theta=0.5
\]
Within the specified angular range, this occurs when:
\[
2\theta=45^\circ
\quad\text{or}\quad
135^\circ
\]
Hence:
\[
\boxed{\theta=22.5^\circ\text{ or }67.5^\circ}
\]
**Step 3**: Explain the two possible answers.
The two orientations are symmetrical about 45°.
At 22.5°, the middle filter transmits a relatively large fraction of the incoming light, but the final filter transmits a smaller fraction.
At 67.5°, the middle filter transmits a smaller fraction, but the final filter transmits a larger fraction.
The product of the two transmission fractions is identical.
A final intensity measurement alone cannot tell us which arrangement was used.
### Question 3
A student investigates whether adding a middle polariser can increase the transmitted intensity between two crossed polarisers.
They arrange the apparatus as follows:
- A lamp produces unpolarised light.
- The light passes through a fixed first polariser.
- It then passes through a rotatable middle polariser.
- A fixed final polariser has its transmission axis perpendicular to the first.
- A light sensor measures the intensity after the final polariser.
The student rotates the middle filter from 0° to 90° and observes that the sensor reading rises, reaches a maximum near 45°, and then falls towards its original background reading.
a. Explain why this pattern is expected.
b. Identify two variables that should be controlled during the investigation.
c. Explain how the student could distinguish genuine transmitted light from background light.
d. Explain why the increase in final transmitted intensity does not violate conservation of energy.
[IMAGE: Experimental arrangement showing a lamp, fixed first polariser, rotatable middle polariser, fixed crossed final polariser, and light sensor. Include a small inset showing the middle polariser rotating from 0 to 90 degrees relative to the first. Do not include the predicted graph.]
### Solution 3
The observed pattern is consistent with Malus' law: the final intensity is greatest when the middle polariser is at 45° relative to the first.
**Step 1**: Explain the change in transmitted intensity.
After the first filter, the light is polarised along its transmission axis.
When the middle filter is at 0°, its axis is parallel to the first filter. The light passes through the middle filter, but the final crossed filter blocks it.
As the middle filter rotates, it changes the polarisation direction of the transmitted light.
The final filter can now transmit a component of that light.
At 45°, the two successive transmission angles are both 45°. This produces the maximum final intensity.
As the middle filter approaches 90°, less light reaches it, so the final transmitted intensity decreases.
**Step 2**: Identify controlled variables.
Two suitable controlled variables are:
- The intensity of the incoming light, which should remain constant.
- The positions and orientations of the first and final filters, which should remain fixed.
The distance between the source and sensor should also remain constant.
If these variables change, the student cannot confidently attribute the change in sensor readings to the rotation of the middle filter.
**Step 3**: Account for background light.
The student should measure the sensor reading with the lamp blocked or switched off.
This gives the background reading, which can be subtracted from the readings taken during the experiment.
An enclosed optical path would also reduce interference from ambient light.
**Step 4**: Explain conservation of energy.
The middle polariser does not create additional energy.
Each filter absorbs or otherwise rejects part of the incoming light. The middle filter changes the polarisation direction of the surviving light, changing how much of it passes through the final filter.
The final intensity is still lower than the intensity after the first filter.
The increase is only relative to the original two-filter arrangement, which ideally transmits no light.
### Question 4
A beam of already plane-polarised light has an intensity of \(100\ \mathrm{W\,m^{-2}}\).
Its polarisation direction is 20° relative to the transmission axis of the first polariser.
The light then passes through three ideal polarisers:
- The first polariser is at the reference direction.
- The middle polariser is at 35° relative to the first.
- The final polariser is perpendicular to the first.
Calculate the intensity after each polariser.
Explain why the first polariser does not automatically halve the incoming intensity.
[IMAGE: Three-filter arrangement for Question 4. Show an incoming linearly polarised beam with its electric-field direction at 20 degrees to P1, P2 at 35 degrees to P1, and P3 at 90 degrees to P1. Clearly distinguish the incoming polarisation direction from the first filter's transmission axis.]
### Solution 4
The final transmitted intensity is approximately \(19.5\ \mathrm{W\,m^{-2}}\). The first polariser does not halve the intensity because the incoming light is already plane-polarised.
**Step 1**: Calculate the intensity after the first polariser.
The incoming polarisation direction is 20° from the first filter's axis.
Apply Malus' law directly:
\[
\begin{aligned}
I_1&=I_0\cos^2 20^\circ\\
&=100(0.8830)\\
&=88.3\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
The surviving light is now polarised along the first filter's axis.
**Step 2**: Calculate the intensity after the middle polariser.
The relative angle between the first and middle filters is 35°.
\[
\begin{aligned}
I_2&=I_1\cos^2 35^\circ\\
&=88.3(0.6710)\\
&=59.3\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
The surviving light is now polarised along the middle filter's axis.
**Step 3**: Calculate the intensity after the final polariser.
The first and final axes are perpendicular.
Therefore, the angle between the middle and final axes is:
\[
90^\circ-35^\circ=55^\circ
\]
Apply Malus' law:
\[
\begin{aligned}
I_3&=I_2\cos^2 55^\circ\\
&=59.3(0.3290)\\
&\approx19.5\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
The final intensity is approximately \(19.5\ \mathrm{W\,m^{-2}}\).
The common mistake is to halve the intensity at the first filter simply because it is the first polariser. The half-intensity rule applies to initially unpolarised light. Already polarised light must be treated using Malus' law, with the actual angle between its incoming polarisation and the filter's axis.
### Question 5
An unpolarised light source produces an initial intensity of \(160\ \mathrm{W\,m^{-2}}\).
The light passes through three ideal polarisers. The first and final polarisers have fixed, perpendicular transmission axes. The middle polariser is rotatable, and its angle \(\theta\) is between 0° and 90° relative to the first.
A sensor placed after the final polariser measures \(15\ \mathrm{W\,m^{-2}}\).
A student claims that the middle polariser must be at 30°.
A second sensor is then placed immediately after the middle polariser. It measures an intensity of \(60\ \mathrm{W\,m^{-2}}\).
Assume that both sensors measure intensity without disturbing the light passing through the apparatus.
a. Determine all possible orientations of the middle polariser using the original final sensor reading.
b. Explain why the first student's conclusion is not justified by the final sensor reading alone.
c. Use the second sensor reading to determine the actual orientation of the middle polariser.
d. The middle polariser is rotated a further 10° away from the first polariser. Calculate the new final transmitted intensity and determine whether it increases or decreases.
[IMAGE: Detailed three-polariser apparatus for Question 5. Show an unpolarised source, fixed P1, rotatable P2 with an unknown angle theta, and fixed P3 perpendicular to P1. Include one intensity sensor immediately after P2 and a second sensor after P3. Label both measurement positions without supplying readings or the unknown angle.]
### Solution 5
The original final reading allows two possible angles: 30° and 60°. The second sensor establishes that the actual angle is 30°. Rotating the middle filter to 40° increases the final intensity to approximately \(19.4\ \mathrm{W\,m^{-2}}\).
**Step 1**: Find the possible middle-filter orientations.
The incoming light is unpolarised, so after the first polariser:
\[
I_1=\frac{160}{2}
=80\ \mathrm{W\,m^{-2}}
\]
For crossed outer filters:
\[
I_3=\frac{I_0}{2}\cos^2\theta\sin^2\theta
\]
Substitute the measured final intensity:
\[
15=80\cos^2\theta\sin^2\theta
\]
Using the double-angle identity:
\[
15=20\sin^2 2\theta
\]
Therefore,
\[
\sin^2 2\theta=0.75
\]
This gives:
\[
2\theta=60^\circ
\quad\text{or}\quad
120^\circ
\]
Hence:
\[
\boxed{\theta=30^\circ\text{ or }60^\circ}
\]
**Step 2**: Explain why the original measurement is insufficient.
Both orientations produce the same final intensity, but the intensities immediately after the middle polariser are different.
For a middle-filter angle of 30°:
\[
I_2=80\cos^2 30^\circ
\]
\[
I_2=60\ \mathrm{W\,m^{-2}}
\]
For a middle-filter angle of 60°:
\[
I_2=80\cos^2 60^\circ
\]
\[
I_2=20\ \mathrm{W\,m^{-2}}
\]
Both arrangements produce a final reading of \(15\ \mathrm{W\,m^{-2}}\).
The original final sensor cannot distinguish between them because it only measures the product of the two successive transmission fractions.
**Step 3**: Use the second sensor to identify the actual angle.
The second sensor measures:
\[
I_2=60\ \mathrm{W\,m^{-2}}
\]
This matches the 30° arrangement, not the 60° arrangement.
Therefore, the actual middle-filter angle is:
\[
\boxed{\theta=30^\circ}
\]
This demonstrates why measurements at different stages can provide information that a final measurement alone cannot.
**Step 4**: Calculate the new intensity after rotating the middle filter.
The middle filter is rotated a further 10° away from the first polariser.
Its new angle is:
\[
\theta=30^\circ+10^\circ=40^\circ
\]
The angle between the middle and final filters is now:
\[
90^\circ-40^\circ=50^\circ
\]
First, calculate the intensity after the middle filter:
\[
\begin{aligned}
I_2'&=80\cos^2 40^\circ\\
&=80(0.5868)\\
&=46.9\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
Now calculate the final transmitted intensity:
\[
\begin{aligned}
I_3'&=46.9\cos^2 50^\circ\\
&=46.9(0.4132)\\
&\approx19.4\ \mathrm{W\,m^{-2}}
\end{aligned}
\]
The final intensity increases from \(15\) to approximately \(19.4\ \mathrm{W\,m^{-2}}\).
This makes sense because the middle filter moves closer to the optimum angle of 45°.
There is another interesting detail. If the original middle-filter angle had been 60°, rotating it a further 10° would have taken it to 70°. That would move it further away from the optimum, and the final intensity would decrease instead.
The two original arrangements have identical final readings, but respond differently when the middle filter is rotated. This is why knowing the intermediate intensity matters.
## Where this leads next
The key skill is tracking two things after every polariser: the intensity of the surviving light and its new polarisation direction.
Once you can do that, the same reasoning applies when a single analyser is rotated, when the incoming light is already polarised, or when a question involves several filters at different relative angles.
For the next step, see [How to Calculate Intensity After Rotating an Analyser](https://tutorgum.com/learn/resources/hsc-physics-how-to-calculate-intensity-after-rotating-an-analyser/).