Intensity, amplitude and inverse-square spreading
Key idea: H2 Physics lessons on progressive-wave models, standing waves, interference, diffraction and resolution.
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The core idea
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Big question: How does wave amplitude control energy flow?
Intensity is power per unit area. For a wave in the same medium, intensity is proportional to amplitude squared, so doubling amplitude gives four times the intensity. For an isotropic point source with no absorption, the same power spreads over area 4πr² and intensity follows an inverse-square law.
Separate power from intensity
Power is the rate at which a source transfers energy. Intensity I is power per area normal to the direction of travel: I = P/A. Its unit is W m⁻². The same power can therefore give different intensities when spread over different areas.
For waves of the same type in the same medium, transported energy is proportional to amplitude squared, so I ∝ A². Amplitude here is the maximum value of the oscillating quantity, not peak-to-peak size.
Check your understanding: Amplitude falls to 30% of its original value. What happens to intensity?
It falls to 0.30² = 0.090, or 9.0%, of its original value.
State the assumptions behind inverse-square spreading
An isotropic point source distributes power over a sphere of area 4πr², giving I = P/(4πr²). Hence I ∝ 1/r² only when source power is constant and absorption, reflection and directional focusing are negligible.
The inverse-square law is geometric spreading, not a universal rule for every wave. A collimated beam, a cylindrical wave or a lossy medium needs a different model.
Check your understanding: Why may laser intensity not follow 1/r² over a short distance?
The beam is strongly directional and approximately collimated, so its cross-sectional area does not grow like 4πr².
Key ideas to keep
- Amplitude and intensity are not directly proportional.
- Inverse-square spreading assumes the source power is constant and propagation is unobstructed.
- Use distance from the source, not distance travelled between two points.
See the reasoning
Worked example
Combine spreading and amplitude information
Question: An isotropic source emits 18 W. Find intensity at 3.0 m. At another point the measured amplitude is half the amplitude there; find its intensity if the medium is unchanged.
Step 1: Choose the area
Why: An isotropic point source spreads over a sphere.
Working: A = 4πr² = 4π(3.0)².
Step 2: Calculate intensity
Why: Intensity is power divided by perpendicular area.
Working: I = 18/[4π(3.0)²] = 0.159 W m⁻².
Step 3: Apply the amplitude square
Why: Intensity, not amplitude, measures energy flow.
Working: I′ = (1/2)²I = 0.0398 W m⁻².
Answer: The intensities are 0.159 W m⁻² and 3.98 × 10⁻² W m⁻².
Check: A halved amplitude must produce a quarter intensity, not half.
Another worked model
Question
An isotropic source has intensity 0.318 W m⁻² at 5.00 m. Find its power and the amplitude factor at 10.0 m.
Check the worked solution
P = I4πr² = 0.318(4π)(5.00²) ≈ 100 W. At double distance, intensity is one quarter. Since amplitude is proportional to √I, amplitude is one half.
Use a hint if needed
Practise with support
Try this
A point source gives 6.0 × 10⁻³ W m⁻² at 2.0 m. Find intensity at 5.0 m and the amplitude ratio A₅/A₂.
Hint: Use I₂/I₁ = (r₁/r₂)², then amplitude ∝ √I.
Check your answer
I₅ = 6.0 × 10⁻³(2.0/5.0)² = 9.6 × 10⁻⁴ W m⁻². A₅/A₂ = √(I₅/I₂) = 2.0/5.0 = 0.40.
Now work without the hint
Practise independently
Your turn
Without absorption, intensity falls from 0.80 to 0.20 W m⁻². State the distance and amplitude factors.
Check your answer
The intensity factor is 1/4. From I ∝ 1/r², distance doubles. From I ∝ A², amplitude halves.
Avoid these traps
Common mistakes
Common mistake
Halving amplitude halves intensity.
What is wrong with this reasoning?
Show better thinking
For a progressive wave, I ∝ A², so halving amplitude reduces intensity to one quarter.
Common mistake
Every measured intensity obeys an inverse-square law.
What is wrong with this reasoning?
Show better thinking
The model requires point-source spreading without energy loss. Absorption, directional emission or nearby boundaries break the simple relation.
Write for the examiner
Exam guidance
State whether you are comparing amplitudes or distances before forming an intensity ratio.
Exam-style practice [6 marks]
A detector reads 2.4 W m⁻² at distance r from a source. Predict the reading at 3r, then explain two reasons a real measurement may differ from the prediction.
Plan before you answer
- Apply the radius-squared ratio.
- State the model assumptions.
- Name physical departures, not measurement vagueness.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
I(3r) = 2.4(r/3r)² = 0.27 W m⁻². This assumes a constant-power isotropic point source with no absorption or reflection. A directional source does not spread over a full sphere, while an absorbing medium removes power before it reaches the detector, so either can change the reading.
Come back in three days
Check what stayed with you
Recall question 1
Define intensity.
Check the answer
Power transferred per unit area normal to propagation.
Recall question 2
How does intensity depend on amplitude?
Check the answer
I ∝ A² for the same wave type and medium.
Recall question 3
What geometric area gives inverse-square spreading?
Check the answer
The spherical area 4πr².
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 10 states no explicit exclusions. Topic 11(j) does not require knowledge of spectrometer structure or use. Graph interpretations distinguish time traces at one position from spatial profiles at one instant, and all small-angle equations are used only with their stated geometry.
- GCE A-Level H2 PhysicsTopic 10(g) / Topic 10(h) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027