9478 / 2027
Wave Motion overview

Lesson 5 of 6

Intensity and spreading

  • I can use intensity as power transferred per unit area and its proportionality to amplitude squared under the same wave conditions.Syllabus

    Syllabus 9478, 10(g). Use intensity as power transferred per unit area and its proportionality to amplitude squared under the same wave conditions.

  • I can apply inverse-square intensity for lossless point-source spreading with the stated radiation geometry.Syllabus

    Syllabus 9478, 10(h). Apply inverse-square intensity for lossless point-source spreading with the stated radiation geometry.

Make a guess. You are not marked.

A small loudspeaker radiates sound equally in all directions. You walk from 2 m away to 4 m away. What happens to the intensity of the sound reaching you?

  1. It halves, because you are now twice as far from the loudspeaker.
  2. It stays the same, because the loudspeaker power is unchanged.
  3. It falls to one quarter, because the energy spreads over four times the area.
Show the answer

It falls to one quarter, because the energy spreads over four times the area.

The same power spreads over a sphere of area 4 pi r2, so intensity follows an inverse square law. Doubling the distance divides the intensity by four.

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.

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. 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.

Try it step by step

Work it out in steps: Ratio of distances r2 / r1, then Intensity ratio I2 / I1 = (r1 / r2)2, then Amplitude ratio A2 / A1 = r1 / r2.

  1. Problem 1. A point source radiates equally in all directions. Compare the intensity and the amplitude at 6.0 m with those at 2.0 m.
  2. Problem 2. A point source radiates equally in all directions. Compare the intensity and the amplitude at 3.0 m with those at 1.5 m.
  3. Problem 3. A point source radiates equally in all directions. Compare the intensity and the amplitude at 10 m with those at 4.0 m.
Show the answer

Problem 1.

Ratio of distances r2 / r1: r2 / r1 = 6.0 / 2.0 = 3

Intensity ratio I2 / I1 = (r1 / r2)2: I2 / I1 = (1/3)2 = 0.111

Amplitude ratio A2 / A1 = r1 / r2: Intensity is proportional to amplitude squared, so A2 / A1 = 1/3 = 0.333

Problem 2.

Ratio of distances r2 / r1: r2 / r1 = 3.0 / 1.5 = 2

Intensity ratio I2 / I1 = (r1 / r2)2: I2 / I1 = (1/2)2 = 0.25

Amplitude ratio A2 / A1 = r1 / r2: A2 / A1 = 1/2 = 0.50

Problem 3.

Ratio of distances r2 / r1: r2 / r1 = 10 / 4.0 = 2.5

Intensity ratio I2 / I1 = (r1 / r2)2: I2 / I1 = (1/2.5)2 = 0.16

Amplitude ratio A2 / A1 = r1 / r2: A2 / A1 = 1/2.5 = 0.40

Check your understanding

An unchanged isotropic point source radiates without loss. Intensity is 1.00 W/m2 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?

  1. Intensity 0.160 W/m2; amplitude ratio 0.160.
  2. Intensity 0.400 W/m2; amplitude ratio 0.160.
  3. Intensity 0.160 W/m2; amplitude ratio 0.400.
Show the answer

Intensity 0.160 W/m2; amplitude ratio 0.400.

The same power crosses a larger spherical area. I_5/I_2 = (2/5)2 = 0.160, and the amplitude ratio is its square root, 0.400.

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.

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.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Define the intensity of a wave. At 2.0 m from a small source that emits uniformly in all directions, the intensity is 36 mW m-2 and the amplitude is 0.90 mm. Calculate the intensity and amplitude at 6.0 m. [3]

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