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How does polarisation reveal that light is a transverse wave?

Plane polarisation of transverse waves, the action of polarisers and Malus's law, polarisation by reflection and Brewster's angle, and applications.

An SQA Advanced Higher Physics answer on polarisation, covering plane polarisation as a property of transverse waves, the action of polarisers and Malus's law, polarisation by reflection and Brewster's angle, and practical applications.

Generated by Claude Opus 4.813 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. Plane polarisation and transverse waves
  3. Polarisers and Malus's law
  4. Polarisation by reflection: Brewster's angle
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to know that plane polarisation is restricting a transverse wave to one plane of vibration, use polarisers and Malus's law to calculate transmitted intensity, calculate Brewster's angle for polarisation by reflection, and describe practical applications of polarisation.

Plane polarisation and transverse waves

Ordinary (unpolarised) light vibrates in all planes perpendicular to its direction of travel. The fact that light can be polarised is decisive evidence that it is a transverse wave, since a longitudinal wave has no sideways vibration to restrict. Passing unpolarised light through a polariser leaves only the vibrations in one plane.

Polarisers and Malus's law

When θ=0\theta = 0 the axes are aligned and all the light passes (cos20=1\cos^2 0 = 1); when θ=90\theta = 90^{\circ} the axes are crossed and no light passes (cos290=0\cos^2 90^{\circ} = 0). For unpolarised light entering the first polariser, the transmitted intensity is halved, since on average half the vibration is aligned with the axis. Rotating one of two crossed polarisers between dark and bright demonstrates Malus's law directly.

Polarisation by reflection: Brewster's angle

At Brewster's angle the reflected and refracted rays are at right angles, and the reflected light vibrates parallel to the surface. Because reflected glare off water, glass and roads is largely horizontally polarised, a polariser aligned to block it removes glare. This is the physics behind polarising sunglasses and camera filters.

Examples in context

Polarising sunglasses cut glare by blocking the horizontally polarised light reflected from water and roads. Photographers use polarising filters to darken skies and remove reflections from glass and water. Stress analysis sends polarised light through transparent models or plastics, where stress patterns rotate the polarisation and show up as coloured fringes. Liquid-crystal displays (LCDs) sandwich a liquid crystal between crossed polarisers and switch pixels by rotating the polarisation electrically.

Try this

Q1. State which type of wave can be polarised and why this matters for light. [1 mark]

  • Cue. Only transverse waves; that light can be polarised shows it is transverse.

Q2. Write Malus's law for plane-polarised light through an analyser at angle θ\theta. [1 mark]

  • Cue. I=I0cos2θI = I_0\cos^2\theta.

Q3. State the property of light reflected at Brewster's angle. [1 mark]

  • Cue. It is completely plane polarised.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style4 marksLight of intensity I0I_0 passes through a polariser and then a second polariser (analyser) whose axis is at 3030^{\circ} to the first. Calculate the transmitted intensity as a fraction of I0I_0, assuming the incident light is already plane polarised by the first polariser.
Show worked answer →

After the first polariser the intensity is I0I_0 (taking that as the reference) and the light is plane polarised.

Malus's law gives the intensity through the analyser: I=I0cos2θI = I_0\cos^2\theta.

Substitute θ=30\theta = 30^{\circ}: I=I0cos230=I0×(0.866)2=0.75I0I = I_0\cos^2 30^{\circ} = I_0 \times (0.866)^2 = 0.75\,I_0.

So 75%75\% of the intensity is transmitted.

Markers reward using Malus's law, squaring the cosine, and the final fraction. A common error is to use cosθ\cos\theta rather than cos2θ\cos^2\theta.

SQA AH style4 marksCalculate Brewster's angle for light reflecting off water of refractive index 1.331.33, and state the property of the reflected light at this angle.
Show worked answer →

Brewster's angle satisfies tanip=n\tan i_p = n.

So ip=tan1(n)=tan1(1.33)i_p = \tan^{-1}(n) = \tan^{-1}(1.33).

Evaluate: ip=53i_p = 53^{\circ} (to two significant figures).

At Brewster's angle the reflected light is completely plane polarised, with its electric field vibrating parallel to the reflecting surface.

Markers reward the relationship tanip=n\tan i_p = n, the angle, and the statement that the reflected light is fully plane polarised.

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