Wave Intensity and the Inverse Square Law
The CSIRO Parkes radio telescope 'Murriyang' (64 m diameter) detects radio signals obeying the inverse square law. When Voyager 1 (23.7 billion km from Earth in 2024) transmits at 23 W, the power received per m² = $23/(4\pi \times (2.37\times10^{13})^2) = 3.3\times10^{-26}$ W/m² — 22 orders of magnitude fainter than when launched.
Practise this lesson
Three printable worksheets that build from foundations to mastery.
If you stand twice as far from a speaker, will the sound intensity become half as large, one quarter as large, or something else? Predict the change and explain why.
Warm-up — intensity is measured in which unit?
Know
- The inverse square relationship $I \propto 1/r^2$
- The ratio form $I_1/I_2 = r_2^2/r_1^2$
- The relationship $I \propto A^2$
Understand
- Why intensity falls with distance from a point source
- Why doubling distance gives one quarter the intensity
- Why amplitude changes have a squared effect on intensity
Can Do
- Compare intensities at two distances using the ratio form
- Connect amplitude changes to intensity changes
- Avoid common inverse-square mistakes
Core Content
Why intensity falls as $1/r^2$ — pure geometry of spreading
In 2024, Voyager 1 was 23.7 billion km from Earth — the most distant human-made object in existence. Its radio transmitter puts out 23 W of power, roughly the same as a small LED light globe. Yet the Parkes radio telescope 'Murriyang' in NSW can still detect it, receiving a signal with intensity of just $3.3 \times 10^{-26}$ W/m². That is not a rounding error — that is 100 trillion trillion times fainter than what left the spacecraft. The reason is purely geometric: Voyager's signal spreads over an ever-expanding spherical wavefront, and by the time it reaches Earth that sphere has a surface area of $4\pi r^2 \approx 7 \times 10^{27}$ m² — every square metre receives only a vanishingly small fraction of the original 23 W.
$I \propto \dfrac{1}{r^2}$ · $\dfrac{I_1}{I_2} = \dfrac{r_2^2}{r_1^2}$
| Distance change | Area change | Intensity change |
|---|---|---|
| Double distance | 4× area | 1/4 intensity |
| Triple distance | 9× area | 1/9 intensity |
| Half distance | 1/4 area | 4× intensity |
At 2 m from a source, intensity = 36 units. What is intensity at 6 m?
- $I_1/I_2 = r_2^2/r_1^2 = 6^2/2^2 = 36/4 = 9$
- $36/I_2 = 9 \Rightarrow I_2 = 4$ units
The inverse square law states that intensity from a point source is proportional to $1/r^2$: doubling the distance reduces intensity to one quarter. Ratio form: $I_1/I_2 = r_2^2/r_1^2$ (W/m²).
Pause — copy the highlighted law and formula into your book before moving on.
A wave from a point source obeys the inverse square law. If distance doubles, intensity becomes:
A bigger oscillation carries more energy — and the relationship is squared
We just saw that intensity falls as $1/r^2$ with distance from a point source. That raises a question: if distance is fixed, what controls the intensity? This card answers it → amplitude determines energy per oscillation, and the relationship is squared ($I \propto A^2$).
When amplitude increases, particles or fields oscillate more strongly. The energy of an oscillator is proportional to the square of its displacement (like a spring: $E \propto x^2$). Since intensity is energy per unit time per unit area, it inherits this squared dependence on amplitude.
This means: to double perceived loudness you need to quadruple intensity, which means doubling the amplitude. Engineers use this when designing amplifiers and speakers.
Intensity is proportional to the square of amplitude ($I \propto A^2$): doubling amplitude quadruples intensity; tripling amplitude gives nine times the intensity.
Add the highlighted amplitude-intensity relationship to your notes before the check below.
If amplitude triples, intensity becomes 9 times as large.
The inverse square law causes the source to lose energy as distance increases.
For a point source, compare the intensity at:
- 1 m and 2 m
- 2 m and 6 m
- 3 m and 1.5 m
A student says, "The torch must be producing less light energy when I move away from the wall." Correct this statement using surface area and intensity.
For each change below, state the factor by which intensity changes:
- Distance from a point source is doubled.
- Distance from a point source is halved.
- Amplitude is doubled while distance stays the same.
- Distance is tripled AND amplitude is also tripled.
Three of these change intensity. Which one does NOT directly affect intensity from a point source at a fixed distance?
A source is moved from 2 m to 6 m. The new intensity is $1/\_\_$ of the original.
Inverse Square Law
- $I \propto 1/r^2$
- Double distance → 1/4 intensity
- Ratio: $I_1/I_2 = r_2^2/r_1^2$
Amplitude and Intensity
- $I \propto A^2$
- Double amplitude → 4× intensity
Point Sources
- Energy spreads over sphere area $= 4\pi r^2$
- Does not apply to lasers or directed beams
Common Traps
- Do not halve intensity when distance doubles
- Do not double intensity when amplitude doubles
If amplitude triples, intensity becomes:
Five questions drawn from the lesson bank.
UnderstandBand 3(3 marks) 1. Explain why intensity from a point source decreases with distance, even if the source power stays constant.
ApplyBand 4(3 marks) 2. At 2 m from a source the intensity is 36 units. What is the intensity at 6 m?
AnalyseBand 6(4 marks) 3. Two speakers are the same distance from a listener, but speaker B produces waves with twice the amplitude of speaker A. Compare their intensities and explain the reasoning.
Show all answers
Activity 1 — Distance Comparisons
1 m to 2 m: double distance → 1/4 intensity. 2 m to 6 m: distance triples → 1/9 intensity. 3 m to 1.5 m: distance halves → 4× intensity.
Activity 4 — Double or Half?
1. 1/4 2. 4× 3. 4× 4. Distance ×3 gives 1/9; amplitude ×3 gives 9; combined = 1. Intensity unchanged.
Short Answer — Model Answers
Q1 (3 marks): Intensity decreases because a point source spreads its energy over a larger wavefront area as distance increases. That wavefront area grows with the square of the radius, so the same source output is distributed across more area. Energy per unit area each second becomes smaller.
Q2 (3 marks): $I_1/I_2 = r_2^2/r_1^2 \Rightarrow 36/I_2 = 36/4 = 9 \Rightarrow I_2 = 4$ units.
Q3 (4 marks): Since distance is the same, only amplitude matters. $I \propto A^2$. If B has twice the amplitude of A, then $I_B/I_A = 2^2 = 4$. Speaker B produces 4 times the intensity of speaker A.
The Parkes/Voyager example makes the answer concrete: Voyager 1 (23.7 billion km away in 2024) transmits 23 W and the received intensity at Earth is $3.3 \times 10^{-26}$ W/m² — a factor of $r^2 = (2.37 \times 10^{13})^2 \approx 5.6 \times 10^{26}$ times weaker than at 1 metre. When you stood at the start and asked "what happens to intensity when distance doubles?" — the full answer is it becomes one quarter as large, because the same power spreads over four times the spherical area. This $1/r^2$ relationship applies to sound, light, radio waves, and gravity equally.