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How does polarisation show that light is a transverse wave?

Explain polarisation of light and how it provides evidence that light is a transverse wave

A focused answer to the WACE Year 12 Physics Unit 4 content point on polarisation. Why only transverse waves can be polarised, how polarising filters work, what happens when two filters are crossed, and everyday applications like polarising sunglasses.

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What this dot point is asking

WACE wants you to explain what polarisation is, how filters produce and detect it, and why it proves light is transverse. This is the one wave property that distinguishes transverse from longitudinal waves.

Transverse waves and polarisation

In a transverse wave the oscillation is perpendicular to the direction of travel, so it has a definite orientation in the plane across the beam. Unpolarised light from a typical source contains electric-field oscillations in every direction perpendicular to travel, mixed randomly. Polarised light has its oscillation confined to a single direction.

How a polarising filter works

A polarising filter has a transmission axis and passes only the component of the electric field aligned with that axis, absorbing the perpendicular component. Sending unpolarised light through one filter therefore produces polarised light and roughly halves the intensity, because on average half the oscillation lies along the axis.

Crossed filters

If polarised light then meets a second filter (an analyser) whose axis is parallel to the first, it passes through. If the second axis is at right angles to the first, no component of the already-polarised light lies along it, so the light is completely blocked. Rotating one filter between these positions smoothly dims the light from bright to dark, which is the standard demonstration that the light has been polarised.

Why only transverse waves can be polarised

A longitudinal wave, such as sound, oscillates along its direction of travel and has no sideways orientation to select, so it cannot be polarised. The fact that light can be polarised therefore shows it must be transverse. This was important historical evidence pinning down the nature of light waves.

Everyday applications

Polarising sunglasses have a vertical transmission axis that blocks the strongly horizontal glare reflected off roads and water. Photographers use polarising filters to deepen sky contrast and cut reflections, and stress patterns in transparent plastics show up between crossed polarisers because the material rotates the plane of polarisation.

Malus's law and partial angles

For angles between parallel and crossed, the transmitted intensity follows Malus's law: if already-polarised light of intensity I1I_1 meets a filter whose axis is at angle θ\theta to the light's polarisation, the transmitted intensity is I=I1cos2θI=I_1\cos^2\theta. This gives the full range smoothly: at θ=0\theta=0 the factor cos20=1\cos^2 0=1 and all the light passes; at θ=45\theta=45^\circ the factor is 0.50.5 so half passes; at θ=90\theta=90^\circ the factor is 00 and the light is extinguished. Remember the first filter acting on unpolarised light is a special case: it always transmits half, regardless of orientation, because the source contains all directions equally. Only once the light is polarised does the cos2θ\cos^2\theta rule apply to subsequent filters.

Explaining the evidence

When asked why polarisation matters, state clearly that it can occur only for transverse waves and therefore shows light is transverse. Describe the two-filter test, noting that the second filter dims the light as it rotates and blocks it completely when crossed.

Exam-style practice questions

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

WACE 20226 marksUnpolarised light of intensity I0I_0 passes through two polarising filters. The first has a vertical transmission axis; the second is at 3030^\circ to the vertical. (a) Determine the intensity of the light after the first filter. (b) Using Malus's law I=I1cos2θI=I_1\cos^2\theta, calculate the intensity after the second filter in terms of I0I_0. (c) State what happens to the transmitted intensity as the second filter is rotated to 9090^\circ.
Show worked answer →

A 6 mark calculation rewards the half-intensity step, Malus's law and the crossed-filter result.

(a) After the first filter
Unpolarised light is halved and polarised: I1=12I0I_1=\tfrac{1}{2}I_0.
(b) After the second filter
Apply Malus's law with θ=30\theta=30^\circ:
I2=I1cos2θ=12I0cos230=12I0(0.866)2=12I0(0.75)=0.375I0.I_2=I_1\cos^2\theta=\tfrac{1}{2}I_0\cos^2 30^\circ=\tfrac{1}{2}I_0(0.866)^2=\tfrac{1}{2}I_0(0.75)=0.375\,I_0.
(c) At 9090^\circ
cos290=0\cos^2 90^\circ=0, so I2=0I_2=0: the filters are crossed and no light is transmitted.

Markers reward I1=12I0I_1=\tfrac{1}{2}I_0, the correct use of cos230\cos^2 30^\circ giving 0.375I00.375\,I_0, and zero transmission at 9090^\circ.

WACE 20215 marksExplain how the polarisation of light provides evidence that light is a transverse wave, and explain why sound cannot be polarised.
Show worked answer →

A 5 mark explanation needs the transverse-orientation argument applied to both light and sound.

Light is transverse. A transverse wave oscillates perpendicular to its direction of travel, so the oscillation has a definite orientation across the beam that a polariser can select. Light can be polarised: passing it through one filter confines the electric-field oscillation to one direction, and a second crossed filter then blocks it. The ability to select and block oscillation directions shows light has sideways oscillations, so it must be transverse.

Sound cannot. Sound is a longitudinal wave, oscillating along its direction of travel (compressions and rarefactions). There is no sideways orientation to select, so a polariser has nothing to filter and sound cannot be polarised.

Markers reward polarisation requiring a transverse oscillation, the filter demonstration, and sound being longitudinal with no transverse component.

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