Physics • Year 11 • Module 3 • Lesson 3
Wave Intensity and the Inverse Square Law
Apply the inverse square law and the amplitude–intensity relationship to real data, scenarios, and diagrams. Practise the ratio method and avoid common exam errors.
1. Interpreting intensity data from a point source
A student places a small speaker (treated as a point source) on a bench and measures the sound intensity at several distances. The table below shows the data. 8 marks
| Distance from speaker (m) | Measured intensity (W/m²) | Intensity ratio to 1 m reading | Predicted by inverse square law? |
|---|---|---|---|
| 1 | 36.0 | 1 (reference) | |
| 2 | 9.0 | ||
| 3 | 4.0 | ||
| 6 | 1.0 | ||
| 9 | 0.44 |
1.1 Complete the “Intensity ratio” column by dividing each measured intensity by 36.0 W/m². 4 marks (1 per row, rows 2–5)
1.2 For each row, check the ratio against the inverse square prediction (I ∝ 1/r²). Write “Yes” or “No (deviation = X%)” in the final column. Show one calculation. 2 marks
1.3 The 3 m measurement deviates slightly from the inverse square prediction. Suggest one physical reason why a real measurement might not perfectly match I ∝ 1/r². 2 marks
2. Interpret graph — intensity vs distance for a torch
A student shines a torch (approximated as a point source) at a wall and measures illuminance at different distances. The graph below shows intensity vs r². 6 marks
Figure 2. Intensity of torch beam (treated as point source) plotted against 1/r². Dashed line is the best fit. Illustrative data.
2.1 Describe the shape of the graph and explain what it tells you about the relationship between I and 1/r². 2 marks
2.2 Explain why a graph of I versus r (not 1/r²) would not be a straight line for a point source. 2 marks
2.3 Use the graph’s best-fit line to estimate the source power, given that I = P/(4πr²). Show your working. 2 marks
3. Predict and justify — two scenarios
8 marks
3.1 A student moves from 5 m to 15 m from a loudspeaker. By what factor does the sound intensity change? Show your calculation using the ratio form, and state whether intensity increases or decreases. 3 marks
3.2 A sound engineer triples the amplitude of a signal from a speaker. By what factor does the intensity change? State the relevant relationship and show your reasoning. 2 marks
3.3 At a distance of 4 m from a light globe the intensity is 2.5 W/m². Predict the intensity at a distance of 1 m. Show your working using the ratio form. 3 marks
4. Compare and contrast — sound and light intensity
Complete the table below. For each feature, write a concise description that compares how sound and light intensity behave. 8 marks (1 per cell)
| Feature | Sound intensity | Light intensity |
|---|---|---|
| Wave type | ||
| Requires a medium? | ||
| Obeys inverse square law from a point source? | ||
| Factors that cause extra intensity loss in real situations |
Q1.1 — Intensity ratio column (4 marks)
Row 2 (2 m): 9.0/36.0 = 0.25. Row 3 (3 m): 4.0/36.0 = 0.111. Row 4 (6 m): 1.0/36.0 = 0.0278. Row 5 (9 m): 0.44/36.0 = 0.0122.
Marking criteria: 1 mark per correct ratio (accept 2 significant figures).
Q1.2 — Inverse square check (2 marks)
Predicted ratio from I ∝ 1/r² using reference at 1 m:
2 m: predicted = 1/4 = 0.25 → measured 0.25 — Yes. 3 m: predicted = 1/9 = 0.111 → measured 0.111 — Yes. 6 m: predicted = 1/36 = 0.0278 → measured 0.0278 — Yes. 9 m: predicted = 1/81 = 0.0123 → measured 0.0122 — Yes (within 0.8% rounding).
Sample calculation shown: At 6 m: I⊂2 = I⊂1 × (r⊂1/r⊂2)² = 36 × (1/6)² = 36 × 1/36 = 1.0 W/m² → ratio 0.0278. Yes, matches.
Marking criteria: 1 mark for correct application of the ratio formula with substituted values; 1 mark for a valid comparison statement (Yes / deviation %).
Q1.3 — Physical reason for deviation (2 marks)
Accept any valid reason: absorption/damping of sound energy by air (especially at higher frequencies) [1]; the speaker is not a perfect point source and directional effects may be present at some angles [1]; reflections from nearby surfaces (bench, walls) adding or subtracting intensity at specific locations [1]. 1 mark for each valid reason (max 2).
Q2.1 — Graph description (2 marks)
The graph shows a straight line through the origin [1]. This indicates that intensity is directly proportional to 1/r² (i.e. I = constant × 1/r²), which is the inverse square law [1].
Q2.2 — Why I vs r is not a straight line (2 marks)
Because I ∝ 1/r², the relationship between I and r is a curve (hyperbola), not a line [1]. A straight line on a graph implies direct proportionality; since r and I are inversely proportional to the square of r, the I vs r graph curves downward steeply and approaches zero asymptotically [1].
Q2.3 — Estimating source power (2 marks)
From the best-fit line: gradient ≈ 11.5 W/m² / (1.0 m−²) = 11.5 W [1]. Since the gradient = P/(4π), source power P = 11.5 × 4π ≈ 144 W [1].
Marking criteria: 1 mark for correctly reading the gradient from the best-fit line; 1 mark for P = gradient × 4π with a reasonable numerical answer consistent with the gradient read.
Q3.1 — Moving from 5 m to 15 m (3 marks)
Distance ratio: r⊂2/r⊂1 = 15/5 = 3 [1]. Apply inverse square: I⊂2/I⊂1 = (r⊂1/r⊂2)² = (5/15)² = (1/3)² = 1/9 [1]. Intensity decreases to one ninth of the original value [1].
Q3.2 — Amplitude tripled (2 marks)
Using I ∝ A²: if A⊂2 = 3A⊂1 then I⊂2/I⊂1 = (3)² = 9 [1]. Intensity increases to nine times the original [1].
Q3.3 — Light globe at 4 m and 1 m (3 marks)
Write ratio: I⊂1/I⊂2 = r⊂2²/r⊂1² [1]. Known: I⊂1 = 2.5 W/m² at r⊂1 = 4 m; find I⊂2 at r⊂2 = 1 m. Substituting: 2.5/I⊂2 = 1²/4² = 1/16 [1]. Therefore I⊂2 = 2.5 × 16 = 40 W/m² [1].
Q4 — Compare and contrast table (8 marks, 1 per cell)
Wave type: Sound — longitudinal (mechanical) / Light — transverse (electromagnetic).
Requires a medium: Sound — Yes (cannot travel in vacuum) / Light — No (travels through vacuum at c).
Obeys inverse square law from a point source: Sound — Yes (in a uniform medium without absorption) / Light — Yes (perfectly in vacuum; approximately in transparent media).
Extra intensity loss factors: Sound — absorption and damping in air (e.g. humidity, temperature gradients, wind); reflections; atmospheric turbulence / Light — absorption and scattering by particles (dust, gas); refraction at media boundaries; reflection at surfaces.