7.5 - Polarisation and Malus's law

7.5 - Polarisation and Malus's law

Explain why polarisation identifies transverse waves, use Malus's law for initially plane-polarised light, and track intensity through successive polarising filters. Measure each angle from the polarisation direction entering that particular filter.

Selecting a transverse oscillation direction

Can adding another polarising filter let more light through? Between two crossed filters, a suitably oriented extra filter can do exactly that. To understand why, track the direction of the transmitted oscillation rather than treating filters as simple dimming sheets.

A transverse wave has oscillations perpendicular to propagation. There are different possible directions within that perpendicular plane. In a rope model, vertical and horizontal oscillations can travel along the same horizontal rope. A narrow vertical slot favours vertical motion while preventing horizontal motion, illustrating selection of an oscillation direction.

Plane polarisation

A plane-polarised transverse wave has its oscillations restricted to one fixed direction perpendicular to propagation. For light, this describes the direction of the oscillating electric field; that direction and the propagation direction define a plane.

A polarising filter has a transmission axis. For an ideal filter, the transmitted electric-field component is along this axis. A plane-polarised beam aligned with the axis is fully transmitted in the ideal model; one perpendicular to the axis is blocked. Intermediate angles transmit a smaller amplitude along the filter's axis.

Unpolarised light has no single fixed transverse polarisation direction. A filter can select a plane-polarised transmitted beam. To apply Malus's law using one angle, the incident beam must already be plane-polarised with a specified direction.

Longitudinal waves cannot be polarised in this sense. Their oscillations already lie along propagation, so rotating a filter about the propagation direction cannot select different transverse oscillation directions. Observing polarisation is therefore evidence for a transverse wave, not a longitudinal one.

Watch out

A transverse wave can be unpolarised or plane-polarised. “It can be polarised” identifies its transverse nature; it does not mean every transverse wave already has one fixed oscillation direction.

Malus's law for one filter

When the incoming oscillation is at an angle to a filter's axis, resolve its amplitude along that axis. The component has amplitude proportional to cos⁡θ\cos\theta. Because intensity is proportional to amplitude squared, the intensity factor is cos⁡2θ\cos^{2}\theta, rather than cos⁡θ\cos\theta.

Malus's law

I=I0cos⁡2θI=I_0\cos^{2}\theta

Here I0I_0 is the intensity of plane-polarised light immediately before the filter, II is the transmitted intensity, both in W m−2\mathrm{W\,m^{-2}}, and θ\theta is the angle between the incident polarisation direction and the filter's transmission axis. Use an ideal polarising filter and a beam incident normally on it.

Not on the data sheet: learn it.

At 0∘0^\circ, the directions align and I=I0I=I_0. At 90∘90^\circ, the directions are crossed and I=0I=0. At 45∘45^\circ, cos⁡245∘=1/2\cos^{2}45^\circ=1/2, so half the incident plane-polarised intensity is transmitted. Rotating an axis through 180∘180^\circ returns the same physical orientation; the intensity pattern repeats every 180∘180^\circ.

[DIAGRAM: asset_name: Malus-law intensity curve; asset_slug: malus-law-curve; caption: Relative angle, rather than a filter's absolute orientation, determines transmission; description: A graph of transmitted intensity divided by incident intensity against relative angle from 0 to 360 degrees follows cosine squared. Maxima of one occur at 0,180 and360 degrees; zeros occur at90 and270 degrees; the45-degree point is marked at one half.]
Relative angle, rather than a filter's absolute orientation, determines transmission

Worked example 1: intensity at a stated angle

Question. Plane-polarised light of intensity 24 W m−224\,\mathrm{W\,m^{-2}} meets an ideal filter whose axis is 30∘30^\circ from the incident polarisation direction. Calculate the transmitted intensity.

Plan. Square the cosine before multiplying by intensity.

I=24cos⁡230∘=24×0.75=18 W m−2I=24\cos^{2}30^\circ=24\times0.75=18\,\mathrm{W\,m^{-2}}

Answer. 18 W m−218\,\mathrm{W\,m^{-2}}.

Check. The intensity lies between zero and the incident value. Using cos⁡30∘\cos30^\circ without squaring would give the amplitude factor instead.

For an unknown angle, take the square root of the intensity ratio before using the inverse cosine. If I/I0=0.36I/I_0=0.36 and 0∘≤θ≤90∘0^\circ\le\theta\le90^\circ, then cos⁡θ=0.36=0.60\cos\theta=\sqrt{0.36}=0.60, so θ=53.1∘\theta=53.1^\circ. Specify the angle range when choosing a unique solution: different axis rotations can give the same squared cosine.

A graph's angle may be measured from the filter's initial orientation instead. If it begins crossed, transmitted intensity starts at zero and reaches a maximum after a 90∘90^\circ turn. Do not copy a cosine-squared curve starting at maximum unless the angle is measured from initial alignment.

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