Why Two 45° Polarising Filters Do Not Always Transmit Half the Light

Imagine sending 100 units of unpolarised light through two ideal polarising filters whose transmission axes are 45° apart. How much intensity reaches the other side?

You might predict 50 units because a filter at 45° transmits half the incoming intensity. That reasoning is only half right. The second filter does transmit half, but the first filter has already reduced the intensity from 100 units to 50 units. The second therefore transmits 25 units, not 50.

There is an important exception. If the incoming light is already linearly polarised parallel to the first filter, the first filter transmits all of it (in the ideal model). The second then transmits half the original intensity.

The key is to track the intensity immediately before each filter, rather than repeatedly comparing the output with the original beam.

01The answer depends on the incoming light

Consider two ideal filters, P1 and P2, with transmission axes separated by 45°.

The result depends on the polarisation of the light entering P1.

All values are illustrative, idealised intensity values.

Notice that the second filter halves the intensity in every example. What changes is the intensity that reaches it.

For unpolarised light, the first filter has already removed half the original intensity. For light aligned with P1, the first filter causes no reduction in the ideal model.

This is why saying that two filters at 45° transmit half the original light is not always correct.

02Why does a filter at 45° transmit half the incoming intensity?

To understand this, consider what a polarising filter actually allows through.

A linear polarising filter has a property called its transmission axis. It allows the component of the incoming electric field parallel to this axis to pass.

The perpendicular component is blocked in the ideal polariser model.

Suppose linearly polarised light reaches a filter with its electric-field direction at an angle \(\theta\) to the transmission axis.

The incoming electric-field amplitude is \(E\). Only its parallel component passes through:

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

But intensity is proportional to the square of the electric-field amplitude:

\[ I\propto E^2 \]

Therefore, the transmitted intensity is:

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

This is Malus’ law.

Here:

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

At 45°:

\[ \begin{aligned} I_{\mathrm{out}}&=I_{\mathrm{in}}\cos^2 45^\circ\\ &=I_{\mathrm{in}}\times 0.5\\ &=\frac{I_{\mathrm{in}}}{2} \end{aligned} \]

The filter transmits 50% of the intensity that reaches it.

A useful way to picture this is to imagine projecting an arrow onto a straight line. At 45°, the projected length is approximately 0.707 of the original length.

Because intensity depends on the square of the amplitude, the transmitted intensity is approximately \(0.707^2=0.5\) of the incoming intensity.

The arrow is just a vector analogy. An electric field is an oscillating field, not a physical arrow passing through the filter.

If you want a more detailed explanation of what the electric-field direction represents in polarisation diagrams, see What Electric-Field Direction Means in Polarisation Diagrams.

03The first filter behaves differently for unpolarised light

There is one extra detail when working with two filters.

Malus’ law describes what happens when linearly polarised light reaches a polarising filter. However, the light entering the first filter might be unpolarised.

Unpolarised light has no single fixed direction of linear polarisation. Its electric-field oscillations have randomly distributed orientations in the plane perpendicular to the direction of propagation.

An ideal polarising filter allows the component along its transmission axis through. Averaged over the random incoming orientations, this gives half the original intensity.

Therefore:

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

Here:

• \(I_0\) is the intensity immediately before P1.
• \(I_1\) is the intensity immediately after P1.

Something important has also happened to the transmitted light.

After passing through P1, the surviving light is now linearly polarised along P1’s transmission axis.

This means the second filter receives linearly polarised light, not unpolarised light.

We can now apply Malus’ law to P2.

If the filters are separated by 45°:

\[ I_2=I_1\cos^2 45^\circ \]

Substitute \(I_1=I_0/2\):

\[ \begin{aligned} I_2&=\frac{I_0}{2}\times\frac{1}{2}\\ &=\frac{I_0}{4} \end{aligned} \]

So, for unpolarised incoming light:

\[ \boxed{I_2=0.25I_0} \]

Only 25% of the original intensity emerges.

04Track the intensity and use relative angles

A reliable way to avoid mistakes is to assign an intensity to each stage.

For two filters, use:

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

The calculation follows the actual path of the light.

The second filter’s angle is measured relative to the first filter’s transmission axis because P1 has already set the direction of the transmitted polarisation.

You do not need to measure either filter from the horizontal or another external reference.

For convenience, you can define P1’s transmission axis as 0° and P2’s as +45°. Take positive angles as anticlockwise in the plane perpendicular to the beam.

The important angle is the separation between the axes.

Remember that a transmission axis is a line, not an arrow. An axis at 0° is physically identical to one at 180°.

What if the incoming light is already polarised?

Suppose the incoming light is already linearly polarised at an angle \(\alpha\) to P1.

The first filter now follows Malus’ law:

\[ I_1=I_0\cos^2\alpha \]

The second filter is still 45° from P1, so:

\[ I_2=I_1\cos^2 45^\circ \]

Combining the equations:

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

This equation explains why the final intensity depends on the initial polarisation direction.

For example:

• If \(\alpha=0^\circ\), the final intensity is \(0.5I_0\).
• If \(\alpha=30^\circ\), the final intensity is \(0.375I_0\).
• If \(\alpha=90^\circ\), the ideal first filter blocks the light completely.

The 45° separation between P1 and P2 determines the second reduction. The angle between the incoming polarisation and P1 determines the first reduction.

A quick decision process

When you see a two-filter question, work through these decisions in order:

1. Identify the incoming light. Is it unpolarised or already linearly polarised?
2. Calculate the intensity after P1. Use \(I_0/2\) for unpolarised light or Malus’ law for already polarised light.
3. Identify the direction of the light leaving P1. It is now aligned with P1’s transmission axis.
4. Find the angle between P1 and P2.
5. Apply Malus’ law using \(I_1\), not \(I_0\).

This process works regardless of the absolute orientations of the filters.

05How an experiment tests the prediction

We can test this reasoning using a light source, two polarising filters, and a light detector.

Arrange the apparatus in the direction of light travel:

1. Place a steady source of approximately unpolarised light on the left.
2. Position a fixed polarising filter, P1, in front of the source.
3. Place a second filter, P2, after P1. Mount P2 so that its transmission axis can rotate.
4. Place a light detector behind P2 to measure the transmitted intensity.

What should the experiment show?

Suppose the unpolarised source produces an initial intensity of 100 arbitrary units.

First, place P1 in the beam.

The detector should record approximately 50 units after P1 because an ideal polariser transmits half of unpolarised light.

Now rotate P2 while keeping P1 fixed.

The predicted readings are:

These are illustrative theoretical predictions, not measurements from an actual experiment.

Notice what happens at 45°.

The detector reads 25 units, which is half the 50 units entering P2. It is also one-quarter of the original 100 units.

Both statements are correct. They describe different comparisons.

Why measuring each stage matters

Imagine the original experimental design only measures the intensity before P1 and after P2.

If the detector records 25 units from an initial 100, you know the complete arrangement transmits 25% of the original intensity.

However, you cannot directly establish how much intensity was transmitted by each filter separately.

A better design records three readings:

• Before P1: \(I_0\).
• After P1: \(I_1\).
• After P2: \(I_2\).

You can then calculate the individual transmission ratios:

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

and

\[ \frac{I_2}{I_1} \]

For the unpolarised source and 45° separation, both ratios should equal 0.5.

This directly tests the two separate predictions: the first filter halves unpolarised light, and the second transmits half of the linearly polarised light reaching it.

In a real experiment, you would also account for background detector readings, source fluctuations, and imperfect filters. Real polarising filters can absorb some light even along their transmission axes and may allow a small amount of unwanted polarisation through.

The exact fractions in HSC calculations assume ideal filters unless the question specifies otherwise.

08Questions and solutions

Question 1

An unpolarised beam has an intensity of \(64\ \mathrm{W\,m^{-2}}\) before reaching two ideal polarising filters.

P1 has a transmission axis at 20° to the horizontal. P2 has a transmission axis at 65° to the horizontal.

Calculate:

(a) The intensity immediately after P1.

(b) The intensity immediately after P2.

(c) The percentage of the original intensity transmitted by both filters.

Solution 1

The intensity after P1 is \(32\ \mathrm{W\,m^{-2}}\), and the final intensity after P2 is \(16\ \mathrm{W\,m^{-2}}\). The two filters transmit 25% of the original intensity.

(a) Intensity after P1

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

\[ \begin{aligned} I_1&=\frac{I_0}{2}\\ &=\frac{64}{2}\\ &=32\ \mathrm{W\,m^{-2}} \end{aligned} \]

(b) Intensity after P2

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

The relative angle between the two filters is:

\[ \theta=65^\circ-20^\circ=45^\circ \]

Apply Malus’ law:

\[ \begin{aligned} I_2&=I_1\cos^2 45^\circ\\ &=32\times0.5\\ &=16\ \mathrm{W\,m^{-2}} \end{aligned} \]

(c) Percentage transmitted

\[ \begin{aligned} \text{Percentage}&=\frac{16}{64}\times100\\ &=25\% \end{aligned} \]

The absolute angles of 20° and 65° are not directly substituted into Malus’ law for P2. The relevant angle is their 45° separation.

Question 2

An unpolarised beam has an initial intensity of \(240\ \mathrm{W\,m^{-2}}\). It passes through an ideal fixed polarising filter, P1, followed by a rotatable analyser, P2.

Calculate the transmitted intensity when the angle between the two transmission axes is:

(a) 0°.

(b) 30°.

(c) 45°.

(d) 90°.

For each result, assume the intensity entering P1 remains unchanged.

Explain why the 45° result is not 50% of the original intensity.

Solution 2

The predicted intensities are \(120\), \(90\), \(60\), and \(0\ \mathrm{W\,m^{-2}}\), respectively. At 45°, the final intensity is 25% of the original, even though P2 transmits 50% of the intensity entering it.

Step 1: Calculate the intensity after P1.

The incoming beam is unpolarised:

\[ \begin{aligned} I_1&=\frac{240}{2}\\ &=120\ \mathrm{W\,m^{-2}} \end{aligned} \]

This is the intensity entering P2 for every angle.

Step 2: Calculate the intensity for each analyser angle.

Use Malus’ law:

\[ I_2=I_1\cos^2\theta \]

For 0°:

\[ I_2=120\cos^2 0^\circ =120\ \mathrm{W\,m^{-2}} \]

For 30°:

\[ I_2=120\cos^2 30^\circ =90\ \mathrm{W\,m^{-2}} \]

For 45°:

\[ I_2=120\cos^2 45^\circ =60\ \mathrm{W\,m^{-2}} \]

For 90°:

\[ I_2=120\cos^2 90^\circ =0\ \mathrm{W\,m^{-2}} \]

Step 3: Explain the 45° result.

The final intensity is:

\[ \frac{60}{240}\times100=25\% \]

However, the second filter’s transmission ratio is:

\[ \frac{60}{120}\times100=50\% \]

There is no contradiction.

P1 has already reduced the intensity by 50%. P2 then reduces the remaining intensity by another 50%.

Question 3

A student investigates two ideal polarising filters using an unpolarised light source.

The detector has a constant background reading of 4 arbitrary intensity units, even when the source is blocked.

The following readings are recorded:

Assume the background reading remains constant and the source intensity is stable.

(a) Calculate the corrected intensity at each measurement position.

(b) Calculate the fraction of intensity transmitted by P1.

(c) Calculate the fraction transmitted by P2 at 45°.

(d) Determine the overall fraction transmitted by both filters at 45°.

(e) Explain why the raw detector reading of 4 units at 90° does not contradict the ideal prediction.

Solution 3

After correcting for the background, the intensities are 200, 100, 50, and 0 units. The first filter transmits 50%, the second transmits 50% at 45°, and the complete arrangement transmits 25%.

(a) Correct the readings

Subtract the 4-unit background from every reading.

(b) Transmission through P1

\[ \begin{aligned} \text{Fraction}&=\frac{100}{200}\\ &=0.5 \end{aligned} \]

P1 transmits 50% of the unpolarised light.

(c) Transmission through P2 at 45°

Use the intensity immediately before P2:

\[ \begin{aligned} \text{Fraction}&=\frac{50}{100}\\ &=0.5 \end{aligned} \]

P2 also transmits 50% of its incoming intensity.

(d) Overall transmission

Compare the final intensity with the original intensity:

\[ \begin{aligned} \text{Fraction}&=\frac{50}{200}\\ &=0.25 \end{aligned} \]

The complete arrangement transmits 25% of the original intensity.

(e) Explain the reading at 90°

At 90°, the transmission axes are perpendicular.

The first filter produces linearly polarised light along its transmission axis. The second filter blocks the perpendicular electric-field component completely in the ideal model.

The raw reading of 4 units is simply the detector’s background reading. After correction, the transmitted intensity is zero.

This question highlights an important experimental skill: a detector reading is not automatically the same as the intensity of the light being measured. Background readings must be considered.

Question 4

A linearly polarised beam has an initial intensity of \(200\ \mathrm{W\,m^{-2}}\).

Its polarisation direction is 30° anticlockwise from P1’s transmission axis.

P2’s transmission axis is 45° anticlockwise from P1’s axis.

The two filters are ideal.

Calculate the final intensity in each of the following arrangements:

(a) P1 first, followed by P2.

(b) P2 first, followed by P1.

(c) Explain why reversing the order changes the final intensity, even though the angle between the filters remains 45°.

Solution 4

The final intensity is \(75\ \mathrm{W\,m^{-2}}\) when P1 comes first, and approximately \(93.3\ \mathrm{W\,m^{-2}}\) when P2 comes first.

The reason is that the first filter’s orientation relative to the incoming polarisation determines how much intensity enters the second filter.

(a) P1 followed by P2

The incoming polarisation is 30° from P1.

Apply Malus’ law:

\[ \begin{aligned} I_1&=200\cos^2 30^\circ\\ &=200\times0.75\\ &=150\ \mathrm{W\,m^{-2}} \end{aligned} \]

The light leaving P1 is now polarised along P1’s axis.

P2 is 45° from P1:

\[ \begin{aligned} I_2&=150\cos^2 45^\circ\\ &=150\times0.5\\ &=75\ \mathrm{W\,m^{-2}} \end{aligned} \]

(b) P2 followed by P1

Now the incoming light encounters P2 first.

The incoming polarisation is at +30°, while P2 is at +45°.

The first relative angle is therefore:

\[ \theta_1=45^\circ-30^\circ=15^\circ \]

Apply Malus’ law:

\[ \begin{aligned} I_1'&=200\cos^2 15^\circ\\ &=200\times0.9330\\ &=186.6\ \mathrm{W\,m^{-2}} \end{aligned} \]

The surviving light is now polarised along P2’s axis.

P1 is 45° from P2, so:

\[ \begin{aligned} I_2'&=186.6\cos^2 45^\circ\\ &=186.6\times0.5\\ &=93.3\ \mathrm{W\,m^{-2}} \end{aligned} \]

(c) Explain the difference

The tempting prediction is that reversing the filters should make no difference because their relative angle remains 45°.

That overlooks the incoming polarisation direction.

When P1 comes first, the initial angle is 30°. When P2 comes first, the initial angle is only 15°.

Consequently, the first filter transmits different amounts of intensity in the two arrangements.

After the first filter, the transmitted light is aligned with that filter’s axis. The second filter then applies its own 45° reduction.

The order matters because the initial light is already polarised.

Question 5

An unknown light source produces an intensity of 100 arbitrary units immediately before P1.

Two ideal filters are arranged with P1 first and P2 second. Their transmission axes are initially:

• P1: 0°.
• P2: +45°.

A detector records 50 units immediately after P1 and 25 units after P2.

A student concludes that the incoming light must have been unpolarised.

Another student argues that the original beam could instead have been fully linearly polarised at +45° to P1.

Both students claim their explanation is consistent with the measurements.

(a) Show mathematically that both incoming-light models produce the recorded intensities.

(b) Both filters are now rotated anticlockwise by 45° together, keeping their relative separation unchanged. The incoming beam remains fixed. Predict the new readings after P1 and P2 for each model.

(c) Explain why the original measurements cannot distinguish the two incoming-light states.

(d) Explain why rotating both filters together is a better test than rotating P1 alone while leaving P2 fixed.

Solution 5

Both explanations are consistent with the original measurements. The crucial test is to rotate both filters together while keeping their relative angle constant. The two possible incoming-light states then produce different results.

(a) Test the two possible incoming states

Model 1: Unpolarised incoming light

The first ideal polariser transmits half the original intensity:

\[ \begin{aligned} I_1&=\frac{100}{2}\\ &=50\text{ units} \end{aligned} \]

P2 is 45° from P1:

\[ \begin{aligned} I_2&=50\cos^2 45^\circ\\ &=25\text{ units} \end{aligned} \]

This agrees with both detector readings.

Model 2: Fully linearly polarised light at +45° to P1

The incoming light is already polarised at 45° to P1.

Therefore:

\[ \begin{aligned} I_1&=100\cos^2 45^\circ\\ &=50\text{ units} \end{aligned} \]

After P1, the surviving light is polarised along P1’s 0° transmission axis.

P2 is still at +45°:

\[ \begin{aligned} I_2&=50\cos^2 45^\circ\\ &=25\text{ units} \end{aligned} \]

This also agrees with both detector readings.

Neither explanation can be rejected using the original measurements.

(b) Rotate both filters by 45°

The new orientations are:

• P1: +45°.
• P2: +90°.

Their relative separation remains 45°.

Consider each incoming-light model again.

Model 1: Unpolarised incoming light

Rotating P1 does not change the fraction transmitted by an ideal polariser when the incoming light is unpolarised.

Therefore:

\[ I_1=100\times\frac{1}{2}=50\text{ units} \]

P2 still has a 45° separation from P1:

\[ I_2=50\cos^2 45^\circ =25\text{ units} \]

The new readings remain 50 units and 25 units.

Model 2: Incoming light polarised at +45°

P1 is now aligned with the incoming polarisation direction.

The first filter transmits all the intensity in the ideal model:

\[ I_1=100\cos^2 0^\circ =100\text{ units} \]

P2 remains 45° from P1:

\[ I_2=100\cos^2 45^\circ =50\text{ units} \]

The new readings are 100 units and 50 units.

The predictions can be compared:

The two models now give different results.

(c) Why couldn’t the original measurements distinguish the models?

Here is the subtle point.

In the original arrangement, both possible incoming beams produce exactly the same transmitted intensity and polarisation after P1.

In the unpolarised case, P1 transmits half the intensity and produces light polarised along its transmission axis.

In the initially polarised case, P1 also transmits half the intensity because the incoming polarisation is at 45°. The surviving light is again polarised along P1’s transmission axis.

Both cases therefore leave P1 with:

• An intensity of 50 units.
• A linear polarisation direction of 0°.

From this point onwards, the two beams behave identically when passing through P2.

The first filter has removed the information needed to distinguish these two particular input states using downstream intensity measurements alone.

This is why the original experiment cannot prove that the source was unpolarised.

(d) Why rotate both filters together?

If you rotate P1 but leave P2 fixed, you also change the angle between the filters.

That changes two things at once:

1. The angle between the incoming polarisation and P1.
2. The angle between P1 and P2.

Any change in the final intensity could therefore result from either effect.

Rotating both filters together avoids this problem. Their relative separation stays at 45°, so the second filter always transmits half the intensity entering it.

The experiment isolates the effect of changing P1’s orientation relative to the incoming light.

This is a useful example of experimental control: change one relevant relationship while keeping the other conditions fixed.

09The main idea to carry into your next question

Two ideal polarising filters separated by 45° always transmit half the intensity that leaves the first filter.

They do not necessarily transmit half the intensity that originally entered the first filter.

For unpolarised incoming light, the first filter transmits half, and the second transmits half of that remaining intensity. The final result is 25% of the original.

For incoming light already polarised parallel to P1, the first filter transmits all the intensity, and the second leaves 50%.

Whenever you encounter a sequence of polarising filters, update two things after each filter:

• The intensity of the transmitted light.
• Its new polarisation direction.

That same reasoning extends to three or more filters. The next step is to apply Malus’ law repeatedly, using the intensity and transmission axis of the immediately preceding filter.

For that extension, see How to Apply Malus’ Law to Three Polarising Filters.