Year 11 PhysicsModule 3⏱ ~40 min5 MC · 3 Short AnswerLesson 3 of 18
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.
Today's hook: The CSIRO Parkes radio telescope 'Murriyang' (64 m diameter) receives signals from Voyager 1, now 23.7 billion km from Earth. Voyager's transmitter puts out only 23 W, about the same power as a refrigerator light bulb. By the time that signal reaches Earth, its intensity is $3.3 \times 10^{-26}$ W/m², 22 orders of magnitude below what left the spacecraft. This extreme attenuation is a direct consequence of the inverse square law: $I \propto 1/r^2$. The fact that Parkes can detect it at all is a triumph of the 64 m dish collecting an enormous area of that faint wavefront.
0/5TASKS
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Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
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?
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Learning Intentions
Learning Intentions
goals
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
Key Terms
vocab
Intensity (I)Power per unit area of a wave; W/m².
Inverse square law$I \propto 1/r^2$; doubling distance reduces intensity to one quarter.
Point sourceSource small enough that waves radiate equally in all directions; spherical wavefronts.
Cross-lesson links: The inverse square law built in this lesson underpins L10's decibel scale (sound intensity), L03 also connects to Module 5 where gravity itself follows an inverse square law with distance. The Parkes/Voyager 1 example (23 W transmitter, $3.3 \times 10^{-26}$ W/m² received at 23.7 billion km) is the most extreme real-world demonstration of $I = P/(4\pi r^2)$ in any year-11 physics course.
Misconceptions to fix
✗ Wrong: Doubling distance halves intensity.
✓ Right: $I \propto 1/r^2$. Doubling distance reduces intensity to one quarter.
✗ Wrong: Doubling amplitude doubles intensity.
✓ Right: $I \propto A^2$. Doubling amplitude makes intensity four times as large.
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The Inverse Square Law
Core Content
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The Inverse Square Law
+5 XP
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.
Each circle is a cross-section of a spherical wavefront. The source power is unchanged; intensity falls because the area grows as r².
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
Worked example, inverse square law
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:
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Why Intensity Depends on Amplitude ($I \propto A^2$)
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Why Intensity Depends on Amplitude ($I \propto A^2$)
+5 XP
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.
Exam trap
Do not confuse "double amplitude" with "double intensity." Doubling amplitude gives four times the intensity ($2^2 = 4$).
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.
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Activity 1, Distance Comparisons
Activity 1, Distance Comparisons
ApplyBand 3
For a point source, compare the intensity at:
1 m and 2 m
2 m and 6 m
3 m and 1.5 m
Activity 2, Explain the Torch
UnderstandBand 3
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.
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Activity 3, Practical Investigation: Light Intensity and Distance
Activity 3, Practical Investigation: Light Intensity and Distance
ApplyBand 4
A practical method for the NESA investigation into the inverse square law using light
Apparatus: a small point-like light source (e.g. a single LED or a torch bulb, not a wide-beam lamp), a light sensor or lux meter (or a data logger with a light probe), a metre ruler or trundle wheel, and a darkened room.
Set up the light source on a bench and mount the light sensor facing it, at the same height, so the sensor's flat face stays perpendicular to the light path at every distance.
Darken the room as much as possible to minimise ambient light reaching the sensor, this is a controlled variable, along with the sensor's angle to the source and the source's power output.
Position the sensor at a starting distance (e.g. 0.20 m) from the source and record the intensity reading.
Move the sensor to increasing distances (e.g. 0.20 m, 0.40 m, 0.60 m, 0.80 m, 1.00 m), recording the intensity at each distance while keeping the source power and sensor angle unchanged.
Repeat each measurement twice more and record an average intensity at each distance to reduce random error.
Tabulate distance $r$, average intensity $I$, and calculate $1/r^2$ for each row.
Plot two graphs: $I$ against $r$ (a curve, steep near the source and flattening further away) and $I$ against $1/r^2$ (a straight line through the origin, if the inverse square law holds).
Independent variable: distance $r$ from the source. Dependent variable: intensity $I$. Controlled variables: ambient light, sensor angle, source power.
Complete your table and describe your graph's shape below.
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Activity 4, Double or Half?
Activity 4, Double or Half?
ApplyBand 4
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.
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Summary, Copy into your books
Summary, Copy into your books
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:
✓
Show what you have learned
Multiple Choice, intensity and inverse square
+5 XP
Five questions drawn from the lesson bank.
Short Answer, 10 marks
+5 XP
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.
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.
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.
✓
Retrieve, reflect and finish
Check what actually stuck
How did your thinking change?
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.
