Why Crossed Ideal Polarisers Transmit No Light | HSC Physics

Put two ideal polarisers in a beam of light and rotate the second until its transmission axis is perpendicular to the first. The first filter can transmit plenty of light, but the ideal prediction after the second filter is exactly zero.

Why? The first polariser has already restricted the surviving light to one direction of electric-field oscillation. The second polariser only allows the perpendicular direction through. There is no component left that it can transmit.

The important idea is that the second filter does not simply remove another half of the light. It removes all of it.

01Why does the second polariser block all the light?

Start with the electric field.

Light is a transverse electromagnetic wave. Its electric field oscillates perpendicular to the direction in which the light travels.

An ideal linear polariser has a transmission axis. It allows the component of the electric field parallel to this axis through and blocks the perpendicular component.

Suppose unpolarised light reaches a first polariser, P1.

After passing through P1, the transmitted light is linearly polarised. Its electric field now oscillates along P1’s transmission axis.

This is the crucial change.

The first polariser has selected one direction of oscillation. The light reaching the second polariser is no longer unpolarised.

Now place a second polariser, P2, with its transmission axis at 90° to P1.

The electric field entering P2 oscillates along P1’s axis. Since P2’s axis is perpendicular to P1’s axis, the incoming electric field is also perpendicular to P2’s transmission axis.

There is no electric-field component parallel to P2’s axis.

Therefore, no light passes through.

The mathematical explanation

The transmitted electric-field amplitude depends on the component parallel to the second polariser’s transmission axis.

If the incoming electric-field amplitude is \(E\), the parallel component is:

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

Here, \(\theta\) is the angle between the incoming polarisation direction and the second polariser’s transmission axis.

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

\[ I\propto E^2 \]

Consequently, the transmitted intensity is:

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

This is Malus’ law, where:

• \(I_1\) is the intensity immediately before the second polariser.
• \(I_2\) is the intensity immediately after the second polariser.
• \(\theta\) is the angle between the incoming polarisation direction and P2’s transmission axis.

For crossed polarisers, the relative angle is \(90^\circ\).

Substituting:

\[ \begin{aligned} I_2&=I_1\cos^2 90^\circ\\ &=I_1(0)^2\\ &=0 \end{aligned} \]

That’s why two crossed ideal polarisers transmit no light.

Notice that we only needed the angle between the two transmission axes. Their absolute orientations don’t matter.

For a broader explanation of the electric-field projection and the relationship between intensity and polariser angles, see Malus’ Law for HSC Physics: Intensity and Polariser Angles.

02Track the intensity before each polariser

A common mistake is to assume that every polariser halves the incoming intensity.

That is incorrect.

The first ideal polariser transmits half the intensity when the incoming light is unpolarised. Once the light has passed through it, the light is already polarised.

The second polariser must therefore be treated using the angle between its axis and the existing polarisation direction.

Let’s track the intensity through the two filters.

Assume the incoming light is unpolarised and has intensity \(I_0\).

The first polariser gives:

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

The second polariser receives this reduced intensity, not the original intensity.

Because the filters are crossed:

\[ I_2=I_1\cos^2 90^\circ=0 \]

The distinction is important:

• The first polariser halves initially unpolarised light.
• The second polariser transmits only the component parallel to its own axis.
• When the axes are crossed, that component is zero.

What if the incoming light is already polarised?

The first polariser does not automatically halve the intensity.

Suppose the incoming light is already linearly polarised at an angle \(\phi\) to P1’s transmission axis.

The intensity after P1 is:

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

The surviving light is now polarised along P1’s transmission axis.

If P2 is crossed with P1, the second polariser still transmits zero:

\[ I_2=I_1\cos^2 90^\circ=0 \]

This is a useful result: two crossed ideal polarisers produce zero transmitted intensity regardless of the initial polarisation state of the incoming light.

The intensity reaching the second polariser can vary, but the final intensity remains zero.

03What would an experiment actually show?

We can test the prediction using a simple optical arrangement.

Place a steady light source on the left. Direct its beam through a fixed polariser, P1, followed by a second polariser, P2, mounted so that it can rotate.

Place a photodetector behind P2. The detector measures the transmitted light intensity, or a signal proportional to it.

To measure the intensity between the filters, take a separate reading with the detector positioned immediately after P1.

Keep the source steady and shield the apparatus from surrounding light.

What should happen as P2 rotates?

Before taking measurements, predict the detector reading at three positions.

1. When P1 and P2 are parallel, the second polariser transmits all the light reaching it.
2. When their axes are at 45°, the second polariser transmits half the incoming intensity.
3. When their axes are crossed at 90°, the second polariser transmits no light in the ideal model.

For example, suppose the intensity immediately before P2 is \(100\ \mathrm{W\,m^{-2}}\).

The ideal predictions are:

These are ideal theoretical values, not experimental measurements.

Why might a real detector still register some light?

Here’s where the distinction between an ideal model and an actual experiment matters.

Real polarising filters are not perfect. Some unwanted light can pass through, even when the filters are intended to be crossed.

Other sources of a non-zero detector reading include:

• Slight misalignment of the transmission axes.
• Light entering around the edges of the filters.
• Reflections within the apparatus.
• Background light reaching the detector.
• A non-zero detector reading when no light reaches it.

Imagine an illustrative experiment in which the detector reads 2% of its maximum reading when the filters are crossed.

That does not mean the ideal prediction is wrong. It means the actual apparatus does not perfectly satisfy the assumptions of the ideal model.

The original experimental design also needs to distinguish between light leakage and background detector readings.

A revised experiment would:

1. Block the source to measure the detector’s background reading.
2. Shield the apparatus to reduce surrounding light.
3. Check that the beam passes through both filters without bypassing their edges.
4. Align the filters carefully, identifying the minimum transmitted intensity rather than relying entirely on printed angle markings.
5. Repeat the measurements using better-quality polarisers if necessary.

If the remaining signal is still non-zero after correcting for background, the filters may have a finite ability to block the unwanted polarisation.

The ideal model predicts exactly zero. A real experiment tests how closely the apparatus approaches that prediction.

A small alignment error can matter

Suppose the intensity immediately before P2 is \(100\ \mathrm{W\,m^{-2}}\), but the filters are not quite crossed.

Their axes are at \(87^\circ\), meaning they are 3° away from perfect crossing.

Using Malus’ law:

\[ \begin{aligned} I_2&=100\cos^2 87^\circ\\ &=100(0.002739)\\ &=0.274\ \mathrm{W\,m^{-2}} \end{aligned} \]

A small angular error produces a small but non-zero transmitted intensity.

At exactly \(90^\circ\), the ideal prediction returns to zero.

This is why careful alignment matters when testing crossed polarisers experimentally.

04Worked examples

05Questions and solutions

Question 1

An unpolarised beam of light has an intensity of \(180\ \mathrm{W\,m^{-2}}\).

It passes through two ideal polarisers with crossed transmission axes.

Calculate:

a) The intensity immediately before the first polariser.

b) The intensity between the two polarisers.

c) The intensity immediately after the second polariser.

Solution 1

The three intensities are \(180\ \mathrm{W\,m^{-2}}\), \(90\ \mathrm{W\,m^{-2}}\), and \(0\ \mathrm{W\,m^{-2}}\), respectively.

Part a)

The intensity before P1 is the original intensity:

\[ I_0=180\ \mathrm{W\,m^{-2}} \]

Part b)

The incoming light is unpolarised, so the first ideal polariser transmits half:

\[ \begin{aligned} I_1&=\frac{I_0}{2}\\ &=\frac{180}{2}\\ &=90\ \mathrm{W\,m^{-2}} \end{aligned} \]

Part c)

The second polariser is crossed with the first.

The angle between the incoming polarisation direction and P2’s transmission axis is \(90^\circ\).

\[ \begin{aligned} I_2&=I_1\cos^2 90^\circ\\ &=90(0)\\ &=0\ \mathrm{W\,m^{-2}} \end{aligned} \]

The first filter halves the unpolarised light. The second blocks all the light that reaches it.

Question 2

A linearly polarised beam has an intensity of \(150\ \mathrm{W\,m^{-2}}\).

Its electric-field oscillation direction is at \(40^\circ\) to the transmission axis of P1.

The beam passes through P1, followed by P2, whose transmission axis is perpendicular to P1.

Calculate:

a) The intensity immediately before P2.

b) The intensity immediately after P2.

A student claims that the intensity before P2 must be \(75\ \mathrm{W\,m^{-2}}\) because the first polariser always halves the intensity. Explain the error.

Solution 2

The intensity before P2 is approximately \(88.0\ \mathrm{W\,m^{-2}}\), and the intensity after P2 is zero.

Part a)

The incoming light is already linearly polarised, so we use Malus’ law directly.

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

Substitute the known values:

\[ \begin{aligned} I_1&=150\cos^2 40^\circ\\ &=150(0.5868)\\ &=88.0\ \mathrm{W\,m^{-2}} \end{aligned} \]

Part b)

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

P2 is perpendicular to that direction.

Therefore:

\[ \begin{aligned} I_2&=88.0\cos^2 90^\circ\\ &=0\ \mathrm{W\,m^{-2}} \end{aligned} \]

Why the student’s prediction is wrong

The student has confused the behaviour of unpolarised light with already polarised light.

An ideal polariser transmits half the intensity of unpolarised incoming light. It does not halve every beam that reaches it.

Here, the incoming beam is already polarised. Its intensity reduction depends on the angle between its electric-field direction and P1’s transmission axis.

Question 3

An unpolarised beam has an intensity of \(240\ \mathrm{W\,m^{-2}}\).

It passes through two ideal polarisers.

The angle between their transmission axes is \(87^\circ\), rather than the \(90^\circ\) required for perfect crossing.

Calculate:

a) The intensity after the first polariser.

b) The intensity after the second polariser.

c) The change in final intensity if P2 is adjusted to exactly \(90^\circ\).

Explain why a small deviation from the crossed position allows some light through.

Solution 3

The intensity after P1 is \(120\ \mathrm{W\,m^{-2}}\). The intensity after P2 is approximately \(0.33\ \mathrm{W\,m^{-2}}\). Adjusting P2 to exactly \(90^\circ\) reduces the final intensity to zero.

Part a)

The incoming light is unpolarised.

Therefore, P1 transmits half the original intensity:

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

Part b)

The angle between the two transmission axes is \(87^\circ\).

Apply Malus’ law:

\[ \begin{aligned} I_2&=I_1\cos^2 87^\circ\\ &=120(0.002739)\\ &=0.329\ \mathrm{W\,m^{-2}} \end{aligned} \]

Therefore:

\[ I_2\approx0.33\ \mathrm{W\,m^{-2}} \]

Part c)

If P2 is adjusted to exactly \(90^\circ\), the transmitted intensity becomes:

\[ I_2=120\cos^2 90^\circ=0 \]

The final intensity decreases by approximately \(0.33\ \mathrm{W\,m^{-2}}\).

Explanation

At \(87^\circ\), P2 is not quite perpendicular to the incoming electric-field direction.

There is a small component of the electric field parallel to P2’s transmission axis. This component can pass through.

At exactly \(90^\circ\), the parallel component disappears, so no light is transmitted.

Question 4

A student investigates crossed polarisers using a light source, two real polarising filters, and a photodetector.

The detector produces the following readings:

Assume:

• The detector response is linear.
• Background readings add to the light signal.
• The source intensity remains constant between measurements.

a) Calculate the corrected detector reading for the parallel filters.

b) Calculate the corrected detector reading for the crossed filters.

c) Express the corrected crossed reading as a percentage of the corrected parallel reading.

d) Explain why the non-zero crossed reading does not contradict the ideal polariser model. Identify two possible experimental limitations.

Solution 4

The corrected parallel reading is 100 units, while the corrected crossed reading is 2 units. The crossed reading is 2.0% of the parallel reading.

Part a)

Subtract the background reading from the parallel-filter reading:

\[ \begin{aligned} I_{\text{parallel}}&=102-2\\ &=100\text{ units} \end{aligned} \]

Part b)

Subtract the same background reading from the crossed-filter reading:

\[ \begin{aligned} I_{\text{crossed}}&=4-2\\ &=2\text{ units} \end{aligned} \]

Part c)

Calculate the percentage:

\[ \begin{aligned} \text{Percentage} &=\frac{2}{100}\times100\\ &=2.0\% \end{aligned} \]

Part d)

The ideal model predicts zero transmitted intensity because the crossed filters have perpendicular transmission axes.

However, real filters are not perfectly ideal.

Two possible explanations for the remaining signal are:

• Finite extinction ratio: the real second polariser allows a small amount of the unwanted polarisation component through.
• Angular misalignment: the filters may not be exactly perpendicular, leaving a small component of the incoming electric field parallel to P2’s transmission axis.

Other possibilities include light bypassing the filters or reflections within the apparatus.

Subtracting the background reading removes the detector offset, but it does not remove these other experimental limitations.

The correct conclusion is that the real apparatus does not perfectly reproduce the ideal model. The ideal prediction remains zero.

Question 5

An unknown beam of light is either:

• Completely unpolarised, or
• Completely linearly polarised.

Its initial intensity and, if applicable, its polarisation direction are unknown.

Two ideal polarisers, P1 and P2, are mounted together so that their transmission axes always remain crossed. The entire pair can rotate without changing the angle between the filters.

A calibrated detector is used to measure the intensity immediately after P1. Measurements are taken in separate runs, with the detector moved between the filters and the source kept constant.

Initially, P1 is at orientation A. The pair is then rotated together by \(45^\circ\), followed by a further \(45^\circ\).

The detector produces the following readings:

A second detector positioned after P2 records zero intensity at all three orientations.

Determine:

a) Whether the incoming beam is unpolarised or linearly polarised.

b) The initial intensity of the beam.

c) The incoming polarisation direction relative to P1 when the pair is at orientation A.

d) Why the measurements after P2 alone would not be sufficient to identify the original beam.

Assume the source intensity is constant, the detectors have no background offset, and the beam must belong to one of the two stated polarisation categories.

Solution 5

The incoming beam is completely linearly polarised, with an initial intensity of \(40\ \mathrm{W\,m^{-2}}\). Its polarisation direction is parallel to P1’s transmission axis when the pair is at orientation A.

This conclusion requires considering all three intermediate measurements, not just the final detector readings.

Part a)

First, consider the possibility that the incoming beam is unpolarised.

An ideal polariser transmits half the intensity of unpolarised light, regardless of its orientation.

Rotating P1 would therefore leave the intermediate intensity unchanged.

If the first reading of \(40\ \mathrm{W\,m^{-2}}\) came from unpolarised light, the intermediate readings would have to be:

\[ 40,\quad40,\quad40 \]

But the actual readings are:

\[ 40,\quad20,\quad0 \]

The unpolarised model cannot explain the changing readings.

Therefore, the incoming beam must be completely linearly polarised.

Part b)

Now consider the fully polarised beam.

Let:

• \(I_0\) be its initial intensity.
• \(\phi\) be the angle between the incoming polarisation direction and P1’s transmission axis at orientation A.

Apply Malus’ law at the first polariser:

\[ I_A=I_0\cos^2\phi \]

The first measurement gives:

\[ I_0\cos^2\phi=40 \]

We cannot determine both \(I_0\) and \(\phi\) from this measurement alone.

We need the other two readings.

Part c)

When the pair rotates by \(90^\circ\), P1’s transmission axis is perpendicular to its original direction.

The intensity after P1 becomes:

\[ I_{A+90}=I_0\cos^2(\phi-90^\circ) \]

Using the identity:

\[ \cos(\phi-90^\circ)=\sin\phi \]

we obtain:

\[ I_{A+90}=I_0\sin^2\phi \]

The third measurement is zero:

\[ I_0\sin^2\phi=0 \]

Since the incoming beam has non-zero intensity:

\[ \sin\phi=0 \]

Therefore, the incoming polarisation direction is parallel to P1’s transmission axis at orientation A.

This means:

\[ \phi=0^\circ \]

Substitute this into the first measurement:

\[ \begin{aligned} 40&=I_0\cos^2 0^\circ\\ 40&=I_0(1)\\ I_0&=40\ \mathrm{W\,m^{-2}} \end{aligned} \]

The original beam intensity is therefore \(40\ \mathrm{W\,m^{-2}}\).

We can check the middle measurement.

When the pair rotates by \(45^\circ\):

\[ \begin{aligned} I_{A+45}&=40\cos^2 45^\circ\\ &=40(0.5)\\ &=20\ \mathrm{W\,m^{-2}} \end{aligned} \]

This agrees with the experimental reading.

Part d)

The detector after P2 records zero for both possible incoming states.

This happens because P2 always remains perpendicular to P1. Whatever light passes through P1 becomes polarised along P1’s transmission axis, so P2 blocks all of it.

A single intermediate reading of \(40\ \mathrm{W\,m^{-2}}\) would also be ambiguous:

• An unpolarised beam with an initial intensity of \(80\ \mathrm{W\,m^{-2}}\) would produce an intermediate intensity of 40.
• A linearly polarised beam with an initial intensity of \(40\ \mathrm{W\,m^{-2}}\), aligned with P1, would also produce an intermediate intensity of 40.

Both would give zero after P2.

The changing intermediate readings resolve the ambiguity. Unpolarised light would produce the same intermediate intensity at every orientation, whereas linearly polarised light changes its transmitted intensity as P1 rotates.

This is the important distinction: the crossed pair always produces a dark output in the ideal model, but the intensity between the filters still contains information about the original beam.

06Where to go next

The key decision when solving polariser questions is identifying the state of the light immediately before each filter.

If it is unpolarised, the first ideal polariser transmits half its intensity. If it is already linearly polarised, use Malus’ law and the relative angle between its polarisation direction and the filter’s transmission axis.

Once you can track both intensity and polarisation direction after each filter, you can extend the same reasoning to three or more polarisers.

Continue with How to Apply Malus’ Law to Three Polarising Filters.