9478 / 2027
Wave Motion overview

Lesson 6 of 6

Polarisation and filter axes

  • I can explain why polarisation is associated with transverse waves.Syllabus

    Syllabus 9478, 10(i). Explain why polarisation is associated with transverse waves.

  • I can use Malus' law and field projection to calculate transmitted intensity and amplitude of plane-polarised electromagnetic waves.Syllabus

    Syllabus 9478, 10(j). Use Malus' law and field projection to calculate transmitted intensity and amplitude of plane-polarised electromagnetic waves.

Make a guess. You are not marked.

Unpolarised light of intensity 40 W/m2 passes through one ideal polarising filter. What intensity emerges?

  1. 40 W/m2, because an ideal filter absorbs none of the light that reaches it.
  2. 0 W/m2, because unpolarised light cannot pass a polariser.
  3. 20 W/m2, half, because the filter passes one component of the random field.
Show the answer

20 W/m2, half, because the filter passes one component of the random field.

Unpolarised light has its electric field in random directions. On average, an ideal polariser passes the component along its axis, which carries half the intensity.

A polarising filter selects a transverse oscillation direction. To calculate what passes through, use the angle between the current incident polarisation and that filter's transmission axis.

Why the transverse direction matters

A transverse wave can oscillate in different directions perpendicular to propagation. Polarisation selects a direction within that transverse plane. For plane-polarised light, the electric field oscillates along one fixed transverse direction.

In a longitudinal wave, particle oscillation is along propagation. There is no equivalent choice of transverse vibration direction for a polarising filter to select. Polarisation is therefore a phenomenon associated with transverse waves.

Project the electric-field amplitude

Let the incident plane-polarised electric-field amplitude be E0. An ideal filter transmits its component along the filter axis. If θ is the angle between the incident polarisation and that axis:

Transmitted amplitude magnitude = E0|cos θ|

For angles from 0° to 90°, the amplitude ratio is simply cos θ. The emerging polarisation lies along the filter axis. The angle is measured between two directions in the transverse plane, not between the ray and the filter surface.

At the same frequency and in the same medium, intensity is proportional to electric-field amplitude squared. Squaring the amplitude ratio gives Malus' law:

I = I0 cos2 θ

I0 is the intensity incident on this filter, not necessarily the original source intensity if earlier filters are present.

Worked single ideal filter

Amplitude and intensity have different ratios

Use plane-polarised input intensity I0 = 80 W/m2. At θ = 30°:

Etransmitted/E0 = cos 30° = √3/2 ≈ 0.866
I = 80 cos2 30° = 80 × 3/4
= 60 W/m2

Transmit the electric-field component along the filter axis

This is a head-on transverse view: the page is perpendicular to propagation. The incident electric-field direction is vertical. The brown dashed line is the filter's transmission axis, 30° from that direction.

A thirty-degree filter transmits 0.866 times the incident field amplitudeIn the plane perpendicular to propagation, an incident electric-field amplitude arrow points vertically upward from O. A separate dashed transmission axis runs through O at thirty degrees clockwise from the incident field. The transmitted amplitude arrow lies along that axis and ends at the perpendicular projection of the incident arrow's tip. The right-angle mark identifies the projection, not an incidence angle. The incident arrow length is 130 drawing units; the transmitted length is 130 times cosine thirty degrees, with rightward and upward components 130 times 0.433013 and 130 times 0.750000. Thus the transmitted amplitude ratio is approximately 0.8660 and the intensity ratio is 0.7500. The arrows are electric-field amplitude vectors, not optical ray paths or forces.Electric-field amplitudesE030°OFilter axisEoutEout = E0 cos 30°= 0.8660 E0

From a plane-polarised input of 80 W/m2, I = 80 cos230° = 60 W/m2. The emerging polarisation lies along the filter axis. The 30° angle is between field direction and axis, not between a ray and a surface.

The view is along propagation. The incident electric-field direction, filter axis and transmitted projection are distinct. The projected amplitude is shorter than the incident amplitude; the intensity ratio is the square of that length ratio.
Separate ideal-filter settings for the same 80 W/m2 plane-polarised input
Angle / °Amplitude ratioTransmitted intensity / (W/m2)
300.86660
600.50020
9000

At 60°, amplitude halves but intensity becomes one quarter, 20 W/m2. At 90°, both vanish in the ideal model. Without a supplied E0, these data determine amplitude ratios, not an absolute electric-field amplitude.

Apply the law again after each filter

Start with the same plane-polarised 80 W/m2 input. The first filter axis is 45° to the original polarisation; the second is 90° to the original. After the first filter, the light is polarised at 45°, so the second filter's local angle is 45°.

Use the incoming polarisation at each filter

Read these successive head-on views from top to bottom. The named filter angles are measured from the original vertical direction. In each filter view, a thin grey dashed line marks its incoming polarisation; the brown line is its transmission axis. Green amplitude arrows share one scale.

Two successive forty-five-degree projections leave one quarter of the initial intensityThree stacked head-on views represent the plane-polarised input, output from filter one and output from filter two, in that order. The incoming amplitude E0 is vertical and has length seventy drawing units, with intensity eighty watts per square metre. Filter one has its axis forty-five degrees clockwise from the original vertical direction. Its output amplitude is 0.707107 E0, drawn with length seventy divided by square root two along that axis; intensity is forty watts per square metre. Filter two has a horizontal axis, ninety degrees to the original vertical but only forty-five degrees to its own incident polarisation. Its output is 0.5000 E0, drawn with length thirty-five units horizontally; intensity is twenty watts per square metre. Grey dashed references and angle arcs identify each local forty-five-degree comparison. These are filter-plane views, not three bends in the ray's travel.Plane-polarised inputAmplitude E080 W/m2Original: 0°Filter 1: 45 degrees to original45°E1 = 0.7071 E040 W/m2Filter 2: 90 degrees to original45°E2 = 0.5000 E020 W/m2

The intensity sequence is 80 → 40 → 20 W/m2. A single ideal filter at 90° to the original input would give zero. With the intermediate filter present, the second filter receives a different polarisation direction and must be evaluated from that new incident state.

Each filter view shows the incident polarisation for that stage and its filter axis. The first output becomes the second input, so both local projection angles are 45°.
I1 = 80 cos2 45° = 40 W/m2
I2 = 40 cos2 45° = 20 W/m2
E1/E0 = 1/√2
E2/E0 = (1/√2)2 = 1/2

The final amplitude is half the original, so final intensity is one quarter of 80 W/m2. A single filter at 90° to the original input would transmit zero. The intermediate filter changes the polarisation reaching the final filter; the complete two-filter output still contains less intensity than the original input.

Check your understanding

Plane-polarised light of intensity 80 W/m2 passes through ideal filter axes at 45 degrees and then 90 degrees to its original polarisation. What intensity emerges from the second filter?

  1. 40 W/m2, because the first filter alone changes the intensity.
  2. 20 W/m2, because each filter in turn passes the 45-degree projection of the light arriving at it.
  3. 0 W/m2, because the final axis is at 90 degrees to the original polarisation direction.
Show the answer

20 W/m2, because each filter in turn passes the 45-degree projection of the light arriving at it.

The first transmits 80 cos2 45 = 40 W/m2 and polarises it along its axis. The second transmits 40 cos2 45 = 20 W/m2.

Distinguish an unpolarised input

Unpolarised light has no single fixed transverse electric-field direction. An ideal first polariser transmits half its time-averaged incident intensity and produces a plane-polarised output. Later filters use Malus' law on that output.

An already plane-polarised input is not automatically halved by every filter: its transmitted fraction depends on the angle. Real absorption and incomplete extinction are additional effects beyond the ideal filter calculations here.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Unpolarised light of intensity 40 W m-2 passes through a polarising filter and then through a second filter whose transmission axis is at 60° to the first. Calculate the intensity after each filter. [3]

Saved on this browser only. Anyone using this browser can see it.