A sound or light source feels weaker as you move away from it, not because it runs out of energy instantly, but because the same energy is spread over a larger area. That spreading gives us the inverse square law.
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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.
Type your prediction below. You will revisit it at the end.
Write your prediction in your book. You will revisit it at the end.
Wrong: Power is the same as energy.
Right: Power is the rate of energy transfer (energy per unit time); more power means faster energy use, not more total energy.
📚 Core Content
Intensity is how much wave energy passes through a given area each second. It is not just "how strong it feels."
For sound, intensity helps explain why a speaker sounds quieter as you move away. For light, intensity helps explain why a torch beam looks dimmer with distance. The source may keep producing energy, but that energy spreads over a larger and larger area as the wave moves outward. This spreading is a purely geometric consequence of three-dimensional space — it has nothing to do with the source getting weaker or the wave losing energy to the medium (though some absorption does occur in real media).
If the same energy is shared across more area, the energy received per square metre falls. That is the physical reason behind the inverse square law. Intensity is measured in watts per square metre (W/m²) because it is a power-per-area quantity. Even a very loud sound is only transferring a tiny fraction of a watt per square metre at your ear, which is why the eardrum is so sensitive.
For a point source, intensity falls with the square of the distance because spherical wavefront area grows with $r^2$.
Imagine the source at the centre of an expanding sphere. As the radius doubles, the surface area becomes four times larger. The source energy is now spread over four times the area, so the intensity becomes one quarter. This is not an approximation or a special case — it is a direct consequence of geometry in three-dimensional space. The surface area of a sphere is $4\pi r^2$, so any quantity distributed uniformly over that surface must scale as $1/r^2$.
The ratio form $\dfrac{I_1}{I_2} = \dfrac{r_2^2}{r_1^2}$ is particularly useful because it does not require you to know the source power. You only need the two distances. If you know the intensity at one distance, you can find it at any other distance using this ratio. This is a common exam calculation.
Doubling radius makes the wavefront area four times larger, so intensity becomes four times smaller.
| 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 |
A bigger oscillation carries more energy, so intensity increases with the square of amplitude.
When amplitude increases, particles or fields oscillate more strongly. That means more energy is transferred each second, so intensity rises. The relationship is not linear: $I \propto A^2$. This squared dependence arises because 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 has practical consequences. If you want to double the perceived loudness of a sound, you need to quadruple its intensity, which means doubling the amplitude. If you want to make a light source appear twice as bright, you similarly need to increase the amplitude of the electromagnetic oscillation by a factor of $\sqrt{2}$. Engineers use this relationship when designing amplifiers, speakers, and optical systems.
Intensity becomes $2^2 = 4$ times as large.
Intensity becomes $3^2 = 9$ times as large.
Sound and light use the same spreading logic, even though one is mechanical and the other is electromagnetic.
For sound, intensity helps describe how much acoustic energy reaches your ear. For light, intensity helps describe how much radiant energy reaches a surface. In both cases, a point source spreading outward leads to an inverse square drop with distance. This universality is one of the beautiful features of wave physics — the geometry of spreading does not care whether the wave is a compression of air or an oscillation of electric and magnetic fields.
There are differences too. Sound is longitudinal and requires a medium, so its intensity can be affected by wind, temperature gradients, and absorption in the air. Light is transverse and can travel through vacuum, so its intensity obeys the inverse square law perfectly in empty space. In materials, light intensity also drops due to absorption and scattering, but the inverse square law still governs the geometric spreading component.
Further from a speaker means less sound intensity because the wave energy is spread over a larger area.
Further from a bulb or torch means less light intensity for the same spreading reason.
Same source energy, larger wavefront area, smaller intensity.
The inverse square law is a direct result of energy conservation applied to expanding wavefronts in three dimensions.
Consider a tiny light bulb or a small speaker. In an ideal situation, it radiates energy equally in all directions. At distance $r$, that energy is distributed over a sphere of radius $r$ with surface area $4\pi r^2$. If you move to distance $2r$, the surface area is $4\pi (2r)^2 = 16\pi r^2$, which is four times larger. Because the total power passing through any spherical surface around the source must be the same (energy is conserved), the power per unit area — intensity — must be one quarter.
This reasoning fails if the source is not point-like or if the energy is not radiated equally in all directions. A laser pointer, for example, keeps its beam narrow over long distances because the light is collimated (all waves travel in nearly the same direction). A flat-panel LED array is an extended source, not a point source. But for stars, light bulbs, and small speakers, the point-source approximation works extremely well and the inverse square law is the right tool.
| Source type | Spreading behaviour | Law applies? |
|---|---|---|
| Small light bulb | Spherical spreading | Yes — inverse square |
| Laser pointer | Narrow collimated beam | No — much slower drop-off |
| Large panel display | Depends on viewing angle | Only approximately at large distance |
✏️ Worked Examples
Scenario: A student measures sound intensity at 2 m from a speaker. What happens to the intensity at 4 m?
The student moved from 2 m to 1 m instead? The distance would halve, so intensity would become 4 times as large.
Scenario: The amplitude of a wave is increased from 3 units to 6 units. Compare the new intensity to the original intensity.
If amplitude were halved instead, the intensity would become one quarter of the original.
Visual Break
🏃 Activities
For a point source, compare the intensity at:
A student says, "The torch must be producing less light energy when I move away from the wall." Write a short response correcting this statement using surface area and intensity.
If one sound wave has 3 times the amplitude of another in the same context, compare their intensities.
For each change below, state the factor by which the intensity changes (e.g. 2×, 4×, 1/4, etc.). Show your reasoning.
Earlier you were asked: If you stand twice as far from a speaker, will the sound intensity become half as large, one quarter as large, or something else?
The full answer: it becomes one quarter as large. The source energy spreads over a wavefront area that grows with the square of distance. Doubling the distance makes the area 4 times as large, so the intensity becomes 1/4 of the original.
Now revisit your prediction. What did you originally assume about distance and spreading?
Annotate your prediction in your book with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
✅ Check Your Understanding
1. A wave from a point source obeys the inverse square law. If distance doubles, intensity becomes:
2. A detector is moved from 3 m to 6 m from a point source. The new intensity is:
3. Which statement best explains the inverse square law?
4. If amplitude doubles, intensity becomes:
5. A light sensor is moved from 4 m to 2 m from a lamp. The intensity at 2 m is:
6. A student says, "If amplitude triples, intensity also triples." The best correction is:
7. Explain why intensity from a point source decreases with distance, even if the source power stays constant. 3 MARKS
8. At 2 m from a source the intensity is 36 units. What is the intensity at 6 m? 3 MARKS
9. 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. 4 MARKS
1 m to 2 m: doubling distance makes intensity one quarter.
2 m to 6 m: distance triples, so intensity becomes one ninth.
3 m to 1.5 m: distance halves, so intensity becomes 4 times as large.
Because $I \propto A^2$, if amplitude triples, intensity becomes $3^2 = 9$ times as large. The wave with 3× amplitude carries 9× the intensity.
1. Distance doubled → intensity becomes $1/2^2 = 1/4$ (quarter).
2. Distance halved → intensity becomes $1/(0.5)^2 = 4$ times (quadruple).
3. Amplitude doubled → intensity becomes $2^2 = 4$ times (quadruple).
4. Distance tripled gives $1/9$; amplitude tripled gives $9$; combined: $(1/9) \times 9 = 1$. Intensity stays the same as original.
1. D — doubling distance gives one quarter intensity.
2. B — moving from 3 m to 6 m doubles distance, so intensity becomes one quarter.
3. C — the same energy is spread over an area that grows with $r^2$.
4. A — doubling amplitude quadruples intensity.
5. D — halving distance from 4 m to 2 m makes intensity 4 times as large.
6. B — tripling amplitude makes intensity 9 times as large.
Q7 (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. As a result, the energy received per unit area each second becomes smaller.
Q8 (3 marks): Use $\dfrac{I_1}{I_2} = \dfrac{r_2^2}{r_1^2}$. So $\dfrac{36}{I_2} = \dfrac{6^2}{2^2} = \dfrac{36}{4} = 9$. Therefore $36/I_2 = 9$, so $I_2 = 4$ units.
Q9 (4 marks): Since the listener is the same distance from both speakers, distance does not change the comparison. Intensity depends on amplitude squared, so $I \propto A^2$. If speaker B has twice the amplitude of speaker A, then $I_B/I_A = 2^2 = 4$. Therefore speaker B produces 4 times the intensity of speaker A at the listener.
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