Physics › Waves › Longitudinal, transverse and polarisation

Longitudinal, transverse and polarisation

Waves come in two kinds, distinguished by whether the particles oscillate along the direction of travel or across it. Only transverse waves can be polarised, so a pair of filters is a direct test of which kind a wave is, and it is how light was shown to be transverse.

Year 12AQA 3.3.1.2 CIE 7.2.1, 7.2.2, 7.4.1, 7.4.2, 7.4.3, 7.5.1, 7.5.2 OCR A 4.4.1(a), 4.4.1(e), 4.4.1(f)(i), 4.4.2(a), 4.4.2(b), 4.4.2(c)EDEXCEL 61, 62, 63, 82

Builds on Progressive waves.

IN THIS TOPIC

Classify a wave as transverse or longitudinal from the direction of its oscillations.
Mark the compressions and rarefactions on the displacement-distance graph of a longitudinal wave, and say why they are not at the crests.
Describe what a polarising filter does to unpolarised light, and what a second, crossed filter does next.
Explain why polarisation is evidence that light is transverse, and why sound, longitudinal in air, cannot be polarised.
CIE only: use Malus's law on plane-polarised light passing one filter, and then a series of them.

COMMON MISCONCEPTION

Any wave can be polarised if you build the right filter.

Two ways to oscillate

In a transverse wave, the oscillations are at right angles to the direction the energy travels. Waves on a rope are transverse, and so are all electromagnetic waves, from radio to gamma rays, every one of them travelling at the same speed c in a vacuum.

FIG. 1Two media carry the same wave speed and wavelength. Above, transverse, where each particle bobs across the travel direction. Below, longitudinal, where each shuttles along it, so the crowd bunches and spreads and those compressions (amber markers) sweep to the right while every particle stays near its home. The coral particle in each channel wears its oscillation direction as a double arrow, set against the amber travel arrow.

In a longitudinal wave, the oscillations are parallel to the direction of energy transfer. The particles bunch together and spread apart as the wave passes, forming compressions, where the pressure is highest, and rarefactions, where it is lowest. Sound is the example that matters. Air particles shuffle back and forth along the very line the sound is travelling.

Graphing a longitudinal wave

Both kinds of wave are graphed the same way, and that is where the marks go missing. Plot the displacement of every particle at one instant against its rest position and you have the displacement-distance graph from the last lesson. For a transverse wave that graph is a portrait of the rope, so reading it feels like looking at the wave itself. For a longitudinal wave it is nothing of the kind. The sound is a pattern of bunching and spreading along a single line, and the sine curve on the page is a piece of bookkeeping about that line, not a picture of it.

Everything then turns on what the sign of a displacement means. A longitudinal particle has only two ways to move, forwards or backwards along the line the wave travels, and the graph gives forwards the positive half. A particle plotted at +0.2 mm sits 0.2 mm ahead of its rest position, in the direction the wave is going, and one plotted at −0.2 mm sits the same distance behind.

Now look for a compression. The air is bunched there, which can only happen if the particles just behind it have moved forwards and the particles just ahead of it have moved backwards, closing in from both sides. The particle at the very centre of the crowd is pressed equally from either side and has not moved at all. So a compression sits where the curve crosses the axis on its way down, from positive displacement to negative. A rarefaction is the same argument reversed: the particles behind have dropped back and those ahead have run on, the air between them is stretched thin, and the curve crosses the axis on its way up.

FIG. 2The same wave twice. Above, the particles themselves, bunched at each C and spread at each R. Below, the displacement-distance graph of that row, with forwards counted positive. Every dashed line drops from a compression or a rarefaction onto a point where the curve cuts the axis, never onto a crest or a trough, and the amber bracket spans one wavelength from compression to compression.

The crests and troughs are the places to leave alone. At a crest every particle nearby has moved forwards by much the same amount, so nobody is crowding anybody: the spacing is normal there and so is the pressure. A crest marks maximum displacement, and a question asking you to label C and R on a graph is asking whether you know that displacement and pressure peak in different places.

The spacings follow from the picture. Adjacent compressions are one wavelength apart, adjacent rarefactions likewise, and a compression sits half a wavelength from its nearest rarefaction. Plot pressure instead of displacement and the graph peaks at every compression and dips at every rarefaction, which puts the pressure curve a quarter of a cycle out of step with the displacement curve.

WORKED EXAMPLE

Finding the compressions on a graph

A sound wave travels in the direction of increasing x. Its displacement-distance graph crosses zero at x = 0, 0.25 m, 0.50 m, 0.75 m and 1.00 m, and its first crest is at x = 0.125 m. Give the wavelength, and the positions of the first compression and the first rarefaction.

Two zeros fit into each cycle, so the pattern repeats every 0.50 m: λ = 0.50 m. Sketch it before going further, because the rest of the question is read off the sketch.

The curve rises from x = 0 to the crest at 0.125 m and comes back down through the axis at 0.25 m. That is a crossing from positive to negative, so the first compression is at x = 0.25 m.

It carries on down to a trough at 0.375 m and rises back through the axis at 0.50 m, a crossing from negative to positive, so the first rarefaction is at x = 0.50 m. The two sit half a wavelength apart, as they must.

Answering 0.125 m for the compression is the standard slip. That is the crest, where the particles are furthest forward and the air is at ordinary pressure.

Polarisation

An ordinary lamp sends out light whose oscillations point in every direction perpendicular to the ray, changing randomly and rapidly. This is unpolarised light. A polarising filter transmits only the component of the oscillation along one direction, its transmission axis. What comes out is plane polarised, its oscillations now confined to a single plane containing the ray.

FIG. 3Unpolarised light meets a vertical filter and leaves oscillating in one plane. A second filter at right angles to the first transmits nothing.

Hold up a second filter and rotate it. The transmitted intensity falls as the angle between the two transmission axes grows, and at 90°, with the filters crossed, the light is blocked completely. The first filter left only vertical oscillations, and a horizontal slot passes none of a vertical oscillation.

The argument runs in three steps. Filtering by oscillation direction only means anything if the oscillations are perpendicular to the ray in the first place. A longitudinal wave oscillates along its direction of travel, the one direction no filter orientation can distinguish. So the fact that light can be polarised is direct evidence that light is a transverse wave, and sound, which travels through air as a longitudinal wave, cannot be polarised. (Transverse vibrations do exist in solids, but the sound this course deals with is the longitudinal kind.)

How far the intensity falls between those two extremes is CIE's question and nobody else's. The other boards want that variation described in words and print no equation for it. CIE wants it as an equation, Malus's law. Take light that is already plane polarised, of intensity $I_0$ , and send it through a filter whose transmission axis makes an angle $\theta$ with the plane of polarisation. What emerges has intensity

I = I_0\cos^2\theta

which no board prints, so on 9702 it is a recall item. The square is worth understanding rather than memorising. The filter passes only the component of the oscillation lying along its own axis, and that cuts the amplitude by a factor of cos $\theta$ , while intensity goes with amplitude squared. Test it at the ends. Parallel axes give $\theta = 0$ and $\cos^2 0 = 1$ , so the beam passes untouched, and crossed axes give $\theta = 90°$ and zero, the blackout described above.

A series of filters is handled one filter at a time, because each filter the light survives leaves it polarised along that filter's own axis. The angle to use at any filter is therefore the angle between its axis and the axis of the filter before it, not the angle back to the first one.

Take plane-polarised light of intensity $I_0$ , polarised vertically, and send it through three filters whose transmission axes are at 0°, 45° and 90° to the vertical. The first is aligned with the light and passes all of it. The second turns 45° from the first, so it passes $I_0\cos^2 45° = 0.50\,I_0$ . The third turns another 45° from the second, so it passes $0.50\,I_0\cos^2 45° = 0.25\,I_0$ , a quarter of what went in. The first and last filters are genuinely crossed at 90°, and on their own they pass nothing at all. Adding a filter has let light through, which sounds impossible until you notice that the middle filter re-polarises the beam halfway across, so the last filter never sees the vertical light the first one made.

One line of the syllabus draws a firm boundary. Malus's law is for light that is already plane polarised. You are not asked to calculate what an unpolarised beam loses at the first filter it meets, so when a question opens with unpolarised light, that first filter's job is simply to polarise it, and the intensity leaving it is whatever the question gives you.

The electromagnetic family

Polarisation also supports a bigger claim, that light belongs to one family of transverse waves, the electromagnetic spectrum, every member travelling at 3.00 × 10⁸ m s⁻¹ in a vacuum and differing only in wavelength.

Region	Typical wavelength	A source
Radio	10³ m down to 0.1 m	transmitters
Microwave	10 cm to 1 mm	ovens, satellite links
Infrared	1 mm to 700 nm	warm objects
Visible	700 nm to 400 nm	the Sun, red to violet
Ultraviolet	400 nm to 10 nm	the Sun, arc lamps
X-ray	10 nm to 0.01 nm	X-ray tubes
Gamma	below 0.01 nm	excited nuclei

Carry the orders of magnitude in your head, visible light above all, from 400 nm at the violet end to 700 nm at the red. Questions ask for them directly, and they anchor every c = fλ calculation you will do. Every member of the family is transverse, so every member can be polarised, and that shared behaviour is how the family was assembled in the first place.

Where you meet it

Sunlight reflected from water or wet road is partially polarised horizontally. Polarising sunglasses mount their filters with a vertical transmission axis, so they remove that glare while passing most other light.

Television and radio signals are transmitted plane polarised. An aerial receives best when its rods lie along the plane of polarisation of the incoming wave. Look along the rooftops of any town and every aerial points the same way, matched to the local transmitter.

GUIDED PRACTICE

Who can be polarised at all?

Sound diffracts around an open door; light does not noticeably do so, but light can be polarised and sound cannot. Explain both facts from the nature of each wave.

Show the working

A doorway is about a metre wide, comparable to sound's wavelength, so sound diffracts strongly; light's wavelength is millions of times smaller than the gap, so its spreading is imperceptible.

Polarisation needs oscillations across the travel direction, so that a filter has planes to choose between. Light is transverse and qualifies. Sound in air oscillates along its own travel direction, leaving a polariser with nothing to select.

ASSESSMENT FOCUS

Definitions earn their marks from the comparison. Oscillations lie perpendicular to the direction of energy transfer in a transverse wave and parallel to it in a longitudinal one. Name both directions or the mark goes.
On a displacement-distance graph of a longitudinal wave, a compression is a zero crossing running positive to negative and a rarefaction is the crossing the other way. Crests and troughs are maximum displacement at ordinary pressure. Every board sets this graph, and Edexcel names it in the specification, so learn the two crossings rather than guessing at the peaks.
“Why can sound not be polarised?” wants two steps. Sound is longitudinal, and only transverse waves can be polarised. One step alone is half an answer.
Rotating one filter above another, the intensity is greatest with the axes parallel and zero at 90°. AQA and Edexcel want that variation in words and set no equation for it. CIE wants Malus's law, $I = I_0\cos^2\theta$ , with $\theta$ measured between the transmission axis and the plane of polarisation of the light arriving at that filter.
Say plane polarised, and name the plane where you can. Aerial questions are alignment questions, so state that the rods lie parallel to the plane of polarisation. “The light is filtered” describes the apparatus rather than the physics asked for.

CHECK YOURSELF

Ultrasound is used to image a foetus, and light is used to read a barcode. One of these waves could in principle be polarised. Which one, and why?

Show a hint

Classify each wave first. Which way does each one oscillate compared with its direction of travel?

Show the answer

The light. Light is an electromagnetic wave, so it is transverse, and its oscillations sit perpendicular to the ray. A filter can select one of those oscillation directions, and that selection is what polarisation means.

Ultrasound is sound, so it is longitudinal. Its particles oscillate along the direction of travel, leaving no perpendicular directions to choose between, and no orientation of any filter can polarise it.

Transverse oscillates across the travel direction.

Longitudinal oscillates along it.

Only transverse waves can be polarised, and light can.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

17 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the longitudinal, transverse and polarisation questions page.

9 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Trigonometry and resolving vectors is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Classify a wave as transverse or longitudinal from the direction of its oscillations.
Mark the compressions and rarefactions on the displacement-distance graph of a longitudinal wave, and say why they are not at the crests.
Describe what a polarising filter does to unpolarised light, and what a second, crossed filter does next.
Explain why polarisation is evidence that light is transverse, and why sound, longitudinal in air, cannot be polarised.
CIE only: use Malus's law on plane-polarised light passing one filter, and then a series of them.

Open the full revision checklist to track your progress across the whole unit.