9478 / 2027
Wave Motion overview

Full chapter

Wave Motion

All 6 topics and the revision summary on one page.

01

Propagation and particle motion

A progressive wave carries a disturbance through space. Identify what oscillates locally and distinguish that oscillation from the direction in which the pattern and energy travel.

What oscillates?

A mechanical wave involves particles of a material medium, such as a string or a fluid. The particles oscillate about equilibrium positions. Neighbouring parts exert forces on one another; as they move, work transfers energy through the medium.

An electromagnetic wave involves oscillating electric and magnetic fields in space and time. It can propagate through a vacuum, without a material medium. A curve representing a field does not show particles following that curve.

Classify the wave by comparing oscillation with propagation:

  • Transverse: the oscillation is perpendicular to propagation, as for the displacement of an ideal transverse string wave.
  • Longitudinal: the particle oscillation is parallel to propagation, as for sound in a fluid.

A sinusoidal graph alone does not tell you which type it represents. First read which quantity is plotted and how its physical direction relates to propagation.

Track a particle separately from a crest

In the ideal progressive mechanical-wave model, particles oscillate locally while energy is transferred through the medium. The medium does not advance with each crest. Other bulk flows can occur in real fluids or surface waves; they are not the oscillatory particle motion represented by this model.

Consider a transverse wave travelling right, with amplitude 4.0 mm, wavelength 0.80 m and period 0.20 s. A crest is the position of maximum positive displacement at an instant, not the identity of one particle.

The pattern moves right; the tagged particle stays at x = 0.10 m

The filled circle marks the same particle in all three frames. The hollow diamond tracks the same crest. The small downward arrow gives the tagged particle's instantaneous direction of motion.

t = 0.000 s; tagged displacement +2.83 mm

Right-moving transverse wave at 0.000 secondsThe fixed tagged equilibrium coordinate is 0.10 metres in every frame. Its displacement here is 2.83 millimetres, and its instantaneous velocity is downward. The same positive crest has advanced to 0.20 metres. A vertical dashed reference keeps the tagged horizontal position fixed. The rightward propagation arrow and the downward particle arrow represent different motions. Horizontal position and vertical displacement use different displayed scales, so the apparent profile angle is not a physical deflection angle or a phase angle.0.000.200.400.600.80-40+4TagWave rightDisplacement y / mmEquilibrium position x / m

t = 0.025 s; tagged displacement 0.00 mm

Right-moving transverse wave at 0.025 secondsThe fixed tagged equilibrium coordinate is 0.10 metres in every frame. Its displacement here is 0.00 millimetres, and its instantaneous velocity is downward. The same positive crest has advanced to 0.30 metres. A vertical dashed reference keeps the tagged horizontal position fixed. The rightward propagation arrow and the downward particle arrow represent different motions. Horizontal position and vertical displacement use different displayed scales, so the apparent profile angle is not a physical deflection angle or a phase angle.0.000.200.400.600.80-40+4TagWave rightDisplacement y / mmEquilibrium position x / m

t = 0.050 s; tagged displacement -2.83 mm

Right-moving transverse wave at 0.050 secondsThe fixed tagged equilibrium coordinate is 0.10 metres in every frame. Its displacement here is -2.83 millimetres, and its instantaneous velocity is downward. The same positive crest has advanced to 0.40 metres. A vertical dashed reference keeps the tagged horizontal position fixed. The rightward propagation arrow and the downward particle arrow represent different motions. Horizontal position and vertical displacement use different displayed scales, so the apparent profile angle is not a physical deflection angle or a phase angle.0.000.200.400.600.80-40+4TagWave rightDisplacement y / mmEquilibrium position x / m

The horizontal and vertical display scales differ. Across these 0.050 s, the crest moves 0.20 m right; the tag moves transversely through equilibrium. A crest is a travelling pattern, not a particle carried along the string.

The filled marker follows one particle at the same equilibrium horizontal position. The hollow marker follows the travelling crest. Propagation and particle-velocity arrows have different meanings; horizontal and vertical display scales are different.

At times 0, 0.025 and 0.050 s, the tagged particle at equilibrium position x = 0.10 m has displacements about +2.83, 0 and -2.83 mm. It moves down in all three shown states. Meanwhile, the corresponding crest advances from x = 0.20 to 0.30 to 0.40 m.

Positive displacement means above equilibrium, not necessarily upward velocity. At the first instant, the particle is above equilibrium but moving down. Comparing the successive frames reveals its motion.

Optional check A transverse-wave profile moves right. At the marked particle, the profile rises as you look from left to right, and the particle is above equilibrium. What is happening there in the ideal string-wave model?
A transverse-wave profile moves right. At the marked particle, the profile rises as you look from left to right, and the particle is above equilibrium. What is happening there in the ideal string-wave model?

Electromagnetic fields and propagation

In the plane electromagnetic-wave model, the electric and magnetic fields oscillate perpendicular to one another and to propagation. The electric-field direction is the direction used to describe the wave's polarisation.

Electromagnetic waves involve fields, not a material particle path

Electric field, magnetic field and propagation are mutually perpendicularAt one point and one instant in a plane electromagnetic wave, the electric-field arrow points up, the magnetic field points out of the page and the wave propagates right. The magnetic-field direction is represented by a dot inside a circle at the common reference point. The dot is a field-direction symbol, not a material particle or a current. A right-angle mark separates the in-plane electric-field and propagation directions; the magnetic field is perpendicular to the page. The arrow lengths do not use a common magnitude scale.Electric field EUpPropagationRightMagnetic field BOut of the page

The arrows show directions at one instant. The electric and magnetic fields oscillate; an electromagnetic wave can propagate without a material medium. The circled dot here denotes magnetic field out of the page.

At the indicated instant, electric field is upward, magnetic field is out of the page and propagation is rightward. The out-of-page symbol marks a field direction, not a current or material particle.

The fields reverse during the cycle, while the progressive wave continues in its propagation direction. These field arrows describe local oscillating quantities; they are not paths followed by matter through the diagram.

02

Transverse graphs and phase

A displacement-position graph is one instant across many positions. A displacement-time graph follows one fixed point through time. Read the horizontal axis before interpreting a spacing or gradient.

Name the quantities

Displacement is the signed offset of a particle from equilibrium. For the transverse string here, y is positive upwards. Amplitude is the maximum magnitude of that displacement.

Wavelength λ is the distance between neighbouring points in the same phase at one instant, such as consecutive crests. Period T is the time for a complete repeated state at one fixed point. Frequency f is cycles per second, with f = 1/T.

Phase identifies the stage of an oscillation, expressed as an angle. Phase difference compares two stages. Wave speed is the rate at which a phase feature such as a crest progresses, not the particle's up-and-down speed. The oscillation quantities use the same cycle and phase meanings.

Worked space and time graphs

One wave, two different horizontal axes

The supplied right-moving transverse model has amplitude 0.0040 m = 4.0 mm, wavelength 0.80 m and period 0.20 s.

Read the horizontal axis before interpreting a curve

Both graphs have amplitude 4.0 mm = 0.0040 m. The first compares different positions at one instant. The second follows one fixed point through time.

One instant across the string: t = 0

One instant across the string: t = 0At time zero, displacement is plotted in millimetres against equilibrium position in metres. It is zero at positions zero, 0.40 and 0.80 metres, positive four millimetres at 0.20 metres and negative four at 0.60 metres. The states at zero and 0.80 metres have the same phase. Their separation, one wavelength, is 0.80 metres. This is a view of many different particles at one instant, not one particle moving along the curve.0.000.200.400.600.80-40+4λ = 0.80 mDisplacement y / mmEquilibrium position x / m

The points x = 0 and x = 0.80 m have the same phase, one wavelength apart. This graph's gradient compares displacement with position.

One fixed point through time: x = 0

One fixed point through time: x = 0At fixed equilibrium position zero, displacement is plotted in millimetres against elapsed time in seconds. The trace starts at zero and initially decreases. At times zero, 0.050, 0.100, 0.150 and 0.200 seconds, the displacements are zero, negative four, zero, positive four and zero millimetres. The first and last states have the same downward velocity, so their separation is one full period, 0.20 seconds. The phase is deliberately different from that of the time-zero spatial graph.0.000.050.100.150.20-40+4T = 0.20 sDisplacement y / mmElapsed time t / s

At x = 0, the point moves downward just after t = 0. Its displacement-time gradient is particle velocity; it is not the rightward propagation speed.

The spatial snapshot is at t = 0. The separate time trace follows the point with fixed equilibrium position x = 0, which initially moves down. The time trace therefore does not have the same starting phase as the spatial profile.

At t = 0, the snapshot has zeros at x = 0, 0.40 and 0.80 m, a positive crest at 0.20 m and a trough at 0.60 m. Consecutive crests are 0.80 m apart. The 0.40 m separation between neighbouring zeros is only half a wavelength in this sinusoidal model.

At fixed x = 0, the displacements at t = 0, 0.050, 0.100, 0.150 and 0.200 s are 0, -4.0, 0, +4.0 and 0 mm. The first and last zero crossings are both downward, so they are one full period apart. The intervening upward crossing is only half a period after the first.

The time graph's gradient gives the transverse particle velocity, in displacement per time. The spatial graph's gradient describes how displacement varies along the string. Neither is automatically the propagation speed. To measure propagation speed, track the advance of the same phase feature across position and time.

Connect phase to distance and delay

For this sinusoidal travelling wave, use positive distance or time separations to find the phase-difference magnitude:

|Δφ| = 2πΔx/λ for a spatial separation
|Δφ| = 2πΔt/T for a time separation

A separation of one wavelength or one period gives a phase change of 2π rad, returning to the same phase. Phase describes the oscillation stage, not a physical deflection angle of the drawn string.

For Δx = 0.20 m:
|Δφ| = 2π(0.20/0.80) = π/2 rad
Corresponding time delay = T/4 = 0.050 s

The time interval 0.050 s corresponds to a phase difference of π/2 rad; they are different quantities with different units. State which point reaches the corresponding motion state first.

For this right-moving wave, the downstream point lags

Solid blue: x = 0. Dashed brown: x = 0.20 m. The filled markers identify the corresponding positive peaks at 0.150 s and 0.200 s.

A point 0.20 metres downstream reaches the corresponding state 0.050 seconds laterTwo time traces have amplitude four millimetres and period 0.20 seconds. The solid trace at position zero starts at zero with downward velocity. The dashed downstream trace at position 0.20 metres starts at positive four millimetres and reaches its corresponding downward zero crossing at 0.050 seconds. Matching positive peaks are explicitly marked at 0.150 seconds for position zero and 0.200 seconds downstream, joined by a 0.050-second time bracket. The positive peak at the downstream trace's initial time belongs to the preceding cycle, not the marked event. Thus the downstream point lags by one quarter period, corresponding to a phase difference of pi over two radians.0.000.050.100.150.20-40+40.050 sDisplacement y / mmElapsed time t / s

The downstream point reaches the matching state 0.050 s later. That time lag corresponds to π/2 rad of phase. The direction of travel is needed to decide which point leads or lags.

The traces follow x = 0 and x = 0.20 m. Corresponding downward zero crossings, or the marked corresponding positive peaks, show the later arrival at the downstream point.

The downward zero crossing at x = 0 occurs at t = 0 and reaches x = 0.20 m at t = 0.050 s. Similarly, corresponding positive peaks occur at 0.150 and 0.200 s. The point to the right lags by 0.050 s in this right-moving wave.

Equal displacement at one instant does not prove equal phase: points may have opposite velocity directions. Use corresponding states and the propagation direction when deciding lead or lag.

Optional check A space snapshot has successive crests 0.80 m apart. A fixed-point time trace repeats after 0.20 s. For this right-moving wave, how does a point 0.20 m downstream compare with the first point?
A space snapshot has successive crests 0.80 m apart. A fixed-point time trace repeats after 0.20 s. For this right-moving wave, how does a point 0.20 m downstream compare with the first point?

Measure a graph with a known scale and reference

Use a calibrated spatial image for wavelength, or known sample times for a fixed-point period. Measure several wavelength or cycle intervals where possible and divide by their number. Count intervals between matching phase points, not simply the number of visible crests.

Keep the camera or observation point fixed when claiming a fixed-point time trace. Record the sample interval, spatial scale and displacement zero. A moving observation point changes which part of the wave is sampled, and a smooth-looking drawn curve does not establish fine measurement resolution.

03

Read longitudinal waves

In a longitudinal wave, particle displacement is parallel to propagation. Compression describes neighbouring particle spacing, not the height of a displacement graph.

Let x label a particle's equilibrium position along a right-moving wave. Let ξ denote its signed displacement: positive means to the right of equilibrium and negative means to the left. Plotting ξ on a vertical graph axis does not make the actual particle motion transverse.

Distinguish displacement from compression

A compression is a region where neighbouring particles are closer together than at equilibrium; a rarefaction has greater spacing. What matters is how displacements differ between neighbours.

If particles farther right have smaller displacements than their neighbours to the left, their spacing decreases. A negative spatial displacement gradient therefore corresponds to compression in this small-displacement model. The most negative gradient marks maximum compression; a positive gradient corresponds to increased spacing.

Displacement is not the same quantity as compression

A positive longitudinal displacement ξ means a particle is to the right of its equilibrium position. Vertical height on the graph encodes that signed horizontal displacement; the particles do not move up and down.

Spatial snapshot at t = 0

Compression is greatest where the displacement gradient is most negativeThe upper graph plots signed longitudinal displacement in millimetres against equilibrium position in metres. At positions zero, 0.15, 0.30, 0.45 and 0.60 metres the values are zero, positive two, zero, negative two and zero millimetres. Beneath it, one horizontal row shows 21 equally spaced equilibrium positions and a separate row shows the same ordered particles at that instant. Only their longitudinal displacements are magnified twenty times to make the spacing change visible; equilibrium spacing stays on the same position scale. The actual-particle row is closest-spaced near 0.30 metres, where displacement is zero with the most negative spatial gradient. It is widest-spaced near zero and 0.60 metres, where the gradient is positive. Brown rings identify the same equilibrium-coordinate 0.30-metre particle in both rows. No particle crosses another. The rows are a position comparison, not two successive times.0.000.150.300.450.60-20+2Displacement ξ / mmEquilibrium position x / mEquilibrium positionsPositions at this instantCompression near x = 0.30 mDisplacements in the dot row: ×20

Near x = 0.30 m, particles on the left are displaced right and particles on the right are displaced left, reducing their spacing. The displacement maxima at 0.15 and 0.45 m do not locate maximum compression. The widest spacing is near 0 and 0.60 m.

Time trace of the particle at equilibrium x = 0.30 m

The longitudinal particle starts at zero displacement and moves rightThe same particle with equilibrium coordinate 0.30 metres is followed through time. Signed longitudinal displacement in millimetres is plotted against time in milliseconds. At zero, 0.50, 1.00, 1.50 and 2.00 milliseconds the values are zero, positive two, zero, negative two and zero millimetres. The initial positive graph slope means motion to the right, not upward motion in space. Its two-millisecond period combines with wavelength 0.60 metres to give propagation speed 300 metres per second. This curve is displacement, not pressure or density.0.000.501.001.502.00-20+2Displacement ξ / mmElapsed time t / ms

This is the same fixed equilibrium position throughout. The rising trace initially means the particle moves right. A longitudinal displacement graph must not be relabelled as pressure or density.

The spatial graph plots signed displacement against equilibrium position, with a separate particle-spacing row. Displacements in that row are magnified 20 times. The time graph instead follows the fixed equilibrium position x = 0.30 m and initially rises.

The supplied model has displacement amplitude 0.0020 m = 2.0 mm, wavelength 0.60 m and period 0.0020 s = 2.0 ms. In the t = 0 snapshot:

  • Displacement is most positive at x = 0.15 m and most negative at x = 0.45 m.
  • Maximum compression is near x = 0.30 m, where displacement is zero but its spatial gradient is most negative.
  • Maximum rarefaction is near x = 0 and 0.60 m, where displacement is also zero but the gradient is positive.

A particle at a compression centre need not have a large displacement. Its neighbours determine the local spacing. A pressure or density trace would be a different plotted quantity; do not relabel this displacement curve as pressure.

Optional check In the longitudinal snapshot, signed displacement is largest positive at x = 0.15 m, zero with its most negative spatial gradient at x = 0.30 m, and largest negative at x = 0.45 m. Where is maximum compression?
In the longitudinal snapshot, signed displacement is largest positive at x = 0.15 m, zero with its most negative spatial gradient at x = 0.30 m, and largest negative at x = 0.45 m. Where is maximum compression?

Follow one particle through time

At equilibrium position x = 0.30 m, the displacements at t = 0, 0.50, 1.00, 1.50 and 2.00 ms are 0, +2.00, 0, -2.00 and 0 mm. At the first instant its displacement is zero and increasing, so the particle is moving right. One millisecond later it crosses equilibrium moving left.

The period is 2.0 ms, giving f = 1/0.0020 = 500 Hz. The wavelength is the spatial repeat distance, 0.60 m; it is not obtained from the horizontal axis of this time trace.

The particle moves back and forth along the same direction as propagation, but does not travel steadily with a compression. Matching a compression between frames tracks the wave; following the same labelled particle tracks local motion.

When interpreting a measured or simulated representation, check whether horizontal position means equilibrium position or instantaneous particle position. Also check any magnification of particle displacement. Use stated scales and readings rather than treating an enlarged particle-spacing sketch as a physical ruler.

04

Wave speed, frequency and wavelength

Wave speed measures how quickly a phase feature progresses. In one period, that feature travels one wavelength, connecting speed to frequency and wavelength.

Wavelength λ is the distance between neighbouring points in the same phase at one instant. Period T is the repeat time at a fixed point, and frequency f = 1/T counts cycles per second.

v = distance travelled / time taken
= λ/T = fλ

Use λ in metres and f in Hz to obtain v in m/s. A metre per cycle multiplied by cycles per second gives metres per second.

Estimate the speed scale

Assume roughly 1 m between corresponding crests and about 5 cycles each second. The propagation speed is then of order 5 m/s. These rough inputs check the scale before using the exact supplied wave values.

Worked mechanical waves

Keep metres and seconds consistent

For the transverse model with λ = 0.80 m and T = 0.20 s:

f = 1/0.20 = 5.0 Hz
v = 5.0 × 0.80 = 4.0 m/s

This agrees with a crest advancing 0.20 m in 0.050 s. It does not mean a string particle travels right at 4.0 m/s. That particle oscillates transversely, with a changing local velocity.

For the separate longitudinal model, λ = 0.60 m and T = 2.0 ms = 0.0020 s:

f = 1/0.0020 = 500 Hz
v = 500 × 0.60 = 300 m/s

The 300 m/s is the propagation speed of its pattern, not a constant speed of its material particles. At a displacement extreme, a sinusoidally oscillating particle can be momentarily at rest while the wave continues to progress.

Use c for electromagnetic waves in vacuum

The symbol c denotes the speed of electromagnetic waves in a vacuum, with SI unit m/s. For vacuum propagation, v = c and c = fλ.

Worked vacuum wavelength

Use supplied c = 3.00 × 108 m/s and light frequency f = 6.00 × 1014 Hz:

λ = c/f
= (3.00 × 108)/(6.00 × 1014)
= 5.00 × 10-7 m = 500 nm

One nanometre is 10-9 m. In a material medium, use the appropriate propagation speed rather than automatically substituting the vacuum value c.

Changing the source frequency does not necessarily change propagation speed. In the same unchanged medium under a nondispersive approximation, speed stays fixed and wavelength adjusts according to λ = v/f. This condition matters: propagation speed can depend on frequency in other media or frequency ranges.

When obtaining v from data, keep wavelength and frequency from the same wave and medium. A spatial image provides λ; a fixed-point time record provides T or f. Read their scales and references before combining them.

05

Intensity and spreading

Intensity describes how much power a wave transfers per unit area. It can decrease as the same power spreads over a larger area, even when no energy is absorbed.

Power per area normal to propagation

The intensity I is the time-averaged power transferred per unit area normal to propagation. For power P distributed uniformly across that area A:

I = P/A
Unit: W/m2

Power is an energy-transfer rate, in watts; intensity also depends on the area over which that power is distributed. P/A gives the average over a chosen area when intensity varies across it.

Intensity follows amplitude squared under the same conditions

For the same wave type, medium and relevant frequency conditions:

I ∝ (amplitude)2
I2/I1 = (amplitude2/amplitude1)2

Tripling amplitude multiplies intensity by 9. A fourfold intensity means twice the amplitude, not four times.

Specify which amplitude is meant. Mechanical displacement amplitude is measured in metres, whereas electromagnetic electric-field amplitude is a field quantity. Two different wave types or frequencies need not share the same proportionality constant, so the square-law ratio is not a licence to compare unspecified waves.

Derive inverse-square spreading

Model a point source radiating time-averaged power P equally in all directions, with no absorption or reflection. A spherical surface at centre distance r has area 4πr2. The same total power crosses every complete sphere:

I = P/(4πr2)
I ∝ 1/r2
I2/I1 = (r1/r2)2

The distance is measured from the source centre, and the point-source approximation requires distances large enough compared with the source's size. A collimated beam does not automatically spread over whole spheres. For a directional source, an inverse-square comparison must keep the same angular direction and unchanged radiation pattern.

Estimate the intensity scale

A source of order 50 W spreading uniformly over a sphere of radius about 2 m gives an area of order 50 m2. Its intensity is therefore of order 1 W/m2, assuming negligible loss. This is a rough scale check, not a detector reading.

Worked spherical spreading

Compare 2.0 m with 5.0 m

An isotropic point-source model radiates power P = 16π W without loss.

At r = 2.0 m:
I = 16π/(4π × 2.02) = 1.00 W/m2
At r = 5.0 m:
I = 16π/(4π × 5.02) = 0.160 W/m2

The same total power is spread over a larger area

An isotropic point-source model emits P = 16π W, with no absorption or reflection. The circles below are cross-sections of complete spherical surfaces, on one radius scale.

Equal total power crosses the two spheres while intensity decreasesTwo concentric circles show cross-sections of spheres centred on an isotropic point source O. The radii are 54 and 135 drawing units, representing two and five metres on one distance scale. The inner radius is measured vertically from the centre; the outer radius is measured horizontally. The same total power, sixteen pi watts, crosses each complete sphere in this lossless model. Their spherical areas are sixteen pi and one hundred pi square metres. Intensity is therefore one watt per square metre at two metres and 0.160 watt per square metre at five metres. The drawing represents spherical spreading rather than two flat discs or a collimated beam.O2.0 m5.0 mCross-section through source ORadii measured from the centreEach complete sphere: P = 16π WAt 2.0 m: I = 1.00 W/m2At 5.0 m: I = 0.160 W/m2

The larger sphere has 6.25 times the area, so its intensity is 0.160 times as large. Under the same wave conditions, its amplitude is √0.160 = 0.400 times the amplitude at 2.0 m.

The same total power crosses both complete spherical surfaces. The larger area has less power per unit area. The drawing represents spreading energy, not particles travelling outwards with the wave.

The intensity ratio is (2.0/5.0)2 = 0.160. With the same medium, wave type and frequency, the amplitude ratio is:

amplitude at 5.0 m / amplitude at 2.0 m
= √0.160 = 0.400

This is a dimensionless ratio, not an amplitude of 0.400 m. The source still radiates 16π W; spreading reduces intensity without removing power from the complete spherical account.

Optional check An unchanged isotropic point source radiates without loss. Intensity is 1.00 W/m^2 at 2.0 m. At 5.0 m, what are the intensity and the amplitude relative to its value at 2.0 m, under the same wave conditions?
An unchanged isotropic point source radiates without loss. Intensity is 1.00 W/m^2 at 2.0 m. At 5.0 m, what are the intensity and the amplitude relative to its value at 2.0 m, under the same wave conditions?

Investigate the distance relationship

Use background-corrected intensity readings at measured source-centre distances. Keep source output, detector orientation and the instrument's operating range controlled. Plot intensity against 1/r2; the ideal lossless model predicts a straight line under the stated geometry.

A detector's raw voltage is not automatically intensity. Establish its calibration or supplied proportionality first. If a signal is proportional to amplitude instead, squaring may be needed to compare intensities; use the actual instrument relationship rather than assuming one.

Absorption, reflections and background signals can change the observed distance relation. Measurements too close to an extended source can also violate the point-source approximation. Repeat readings to assess variation while checking these model conditions separately.

06

Polarisation and filter axes

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.

Optional check Plane-polarised light of intensity 80 W/m^2 passes through ideal filter axes at 45 degrees and then 90 degrees to its original polarisation. What intensity emerges from the second filter?
Plane-polarised light of intensity 80 W/m^2 passes through ideal filter axes at 45 degrees and then 90 degrees to its original polarisation. What intensity emerges from the second filter?

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.

07

Revision summary

Identify the oscillating quantity, the graph's horizontal axis and the conditions of the model before selecting an equation.

Keep local motion separate from propagation

A mechanical wave involves particles oscillating in a medium; an electromagnetic wave involves oscillating electric and magnetic fields and can propagate through vacuum. In the ideal progressive mechanical-wave model, energy travels through the medium without its particles travelling along with each crest.

Transverse oscillation is perpendicular to propagation; longitudinal oscillation is parallel. A displacement graph drawn with a vertical axis does not by itself identify a transverse wave.

Recall the eight wave quantities

  • Displacement: signed offset from equilibrium. Amplitude: its maximum magnitude.
  • Period: time for one full repeated state at a fixed point. Frequency: cycles per second, f = 1/T.
  • Phase: stage of oscillation expressed as an angle. Phase difference: the difference between two stages.
  • Wavelength: distance between neighbouring points in the same phase at one instant. Wave speed: propagation rate of a phase feature.

A position snapshot gives wavelength; a fixed-point time trace gives period. The time-gradient of a displacement trace is particle velocity. To obtain wave speed, track a corresponding phase feature or use v = fλ.

v = λ/T = fλ
Vacuum electromagnetic waves: c = fλ
|Δφ| = 2πΔx/λ or 2πΔt/T

Use positive separations for these phase magnitudes, then establish which point leads or lags from the propagation direction. A quarter-period delay corresponds to π/2 rad of phase; time and phase are different quantities. Equal displacement alone does not establish equal phase.

For a longitudinal displacement-position snapshot, compression depends on neighbouring particle spacing. Maximum compression occurs where the displacement gradient is most negative in the small-displacement model, not where positive displacement is largest. Revisit the separate displacement and particle-spacing representations when a graph's meaning is unclear.

State intensity conditions

I = P/A
I2/I1 = (amplitude2/amplitude1)2
Isotropic lossless point source:
I = P/(4πr2)

Area is normal to propagation; P/A is an area average if intensity is not uniform. The amplitude ratio requires the same wave type, medium and relevant frequency conditions. Spherical inverse-square spreading additionally requires the stated point-source geometry and negligible losses, with r measured from the source centre.

In the supplied 2.0 m to 5.0 m comparison, intensity becomes 0.160 of its initial value and amplitude becomes 0.400. The power crossing each complete sphere stays the same.

Use each filter's own incident polarisation

Transmitted amplitude = E0|cos θ|
I = I0 cos2 θ

The ideal filter selects a transverse electric-field direction. θ is the angle between the incident polarisation and the filter axis; the output is polarised along that axis. For successive filters, update both the incident intensity and its polarisation at each stage.

With plane-polarised input of 80 W/m2 and axes at 45° then 90° to its original polarisation, each local angle is 45°: 80 → 40 → 20 W/m2. A first ideal polariser halves unpolarised intensity; that rule does not replace Malus' law for an already plane-polarised input.

Quantities and units

Symbols used in these wave models
QuantitySymbolUnit or meaning
Particle displacementy or ξm; signed from equilibrium
Displacement amplitudey0 or ξ0m; maximum magnitude
PeriodTs
FrequencyfHz = s-1
Wavelengthλm
Propagation speedvm/s
Vacuum electromagnetic speedcm/s; vacuum propagation
Phase and phase differenceφ, Δφrad, or explicitly stated degrees
Filter projection angleθrad or degrees; match the calculator mode
Electric-field amplitudeE0N/C; a field amplitude
PowerPW
AreaAm2
IntensityIW/m2

Use a calibrated spatial scale, a known timebase and the correct reference point when reading data. Count full wavelength or cycle intervals, and establish the detector's calibration before treating its signal as intensity.

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