Physics • Year 11 • Module 3 • Lesson 3
Wave Intensity and the Inverse Square Law
Build HSC Band 5–6 extended-response technique by analysing data, designing investigations, and evaluating models of intensity spreading for real physical systems.
1. Data + scenario: testing the inverse square law at the Sydney Observatory (Band 5–6)
9 marks Band 5–6
Scenario. A physics class visits Sydney Observatory and uses a small halogen lamp (treated as a point source) and a calibrated light sensor to test the inverse square law. They record light intensity at six distances along a straight track. The table below shows their results.
| Distance r (m) | 1/r² (m−²) | Measured intensity I (W/m²) | Expected from I = k/r² (k = 25 W) |
|---|---|---|---|
| 1.0 | 1.000 | 24.8 | 25.0 |
| 1.5 | 0.444 | 11.0 | 11.1 |
| 2.0 | 0.250 | 6.3 | 6.25 |
| 3.0 | 0.111 | 2.8 | 2.78 |
| 5.0 | 0.040 | 0.98 | 1.00 |
| 7.0 | 0.020 | 0.49 | 0.51 |
Illustrative data based on a standard inverse square law experiment. k = 25 W is used for predictions (this represents P/4π ≈ 25 W for a ~314 W source).
Q1. Analyse and evaluate the experimental data above to assess whether it supports the inverse square law, and discuss the physical basis of the law. In your response you must:
- Plot I vs 1/r² and describe the shape of the graph you would expect.
- Calculate the percentage deviation between the measured and expected intensity at r = 5.0 m and comment on whether the data support the inverse square law.
- Explain the geometric reason why intensity obeys the inverse square law for a point source, with reference to the surface area of a sphere.
- Identify two sources of experimental uncertainty in this investigation and suggest how each could be reduced.
- State one condition under which this inverse square model would not apply and justify your answer.
2. Experimental design — investigating the amplitude–intensity relationship (Band 5–6)
8 marks Band 5–6
Research question. A student claims that “doubling the amplitude of a sound wave doubles the intensity.” Design a scientific investigation to test whether intensity is proportional to amplitude or to amplitude squared. You must distinguish between the two models (I ∝ A and I ∝ A²) using your data.
Constraints: You have access to a signal generator, a calibrated speaker, a sound level meter (measuring intensity in W/m²), a ruler, and a laptop with data-logging software. Distance from speaker to meter is kept constant at 1.0 m throughout the investigation.
Q2. Design the investigation and present it in the format below.
- State a hypothesis that distinguishes between I ∝ A and I ∝ A², and explain why the exponent matters physically.
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including how you will control the distance and how you will determine the amplitude of the wave.
- Explain how you will analyse your data graphically to distinguish between the two models.
- State two limitations and one improvement to increase the reliability of your results.
Q1 — Sample Band 6 response (9 marks), annotated
Graph shape (1 mark): Plotting I (y-axis) vs 1/r² (x-axis) should produce a straight line through the origin, with gradient equal to k = P/(4π). This linearises the inverse square relationship and allows easy visual confirmation. [1 mark for correctly describing a straight line through the origin with explanation of why.]
Percentage deviation at r = 5.0 m (2 marks): Measured = 0.98 W/m²; expected = 1.00 W/m². % deviation = |0.98 − 1.00|/1.00 × 100 = 2.0% [1 mark for correct calculation]. This very small deviation (2%) is consistent with random measurement error and strongly supports the inverse square law; no systematic discrepancy is apparent across all six data points [1 mark for evaluative conclusion linking the deviation to support/refutation].
Geometric explanation (2 marks): For a point source, energy radiates equally in all directions, so at any distance r the total energy is spread uniformly across the surface of a sphere of radius r. The surface area of a sphere is 4πr² [1 mark for stating spherical area formula]. Since the source power P is constant (energy conservation), intensity = P/(4πr²) ∝ 1/r² — the same power is shared over four times the area when distance doubles, so intensity falls to one quarter [1 mark for linking constant power over growing area to the 1/r² dependence].
Sources of uncertainty and reductions (2 marks, 1 per source):
- The lamp may not behave as a true point source at close range (finite size); reduce by using only distances ≫ lamp diameter, or by using a smaller pinhole source [1].
- Room reflections from walls, ceiling and floor add stray light to the sensor reading; reduce by conducting the experiment in a dark, draped room or a long open corridor away from reflective surfaces [1].
- Accept also: sensor calibration drift; vibration; ambient light contamination.
Condition where model fails (2 marks): A collimated (directed) beam, such as a laser pointer, does not obey I ∝ 1/r² [1]. Because the light does not spread spherically, the beam cross-section remains approximately constant with distance, so intensity falls off far more slowly — often remaining nearly constant over moderate distances [1]. Accept also: an extended (non-point) source; significant absorption/scattering in a dense medium.
Marking criteria summary (9 marks): 1 = correct graph description (linear through origin); 1 = correct % deviation calculation; 1 = correct evaluative conclusion from deviation; 1 = sphere area formula stated; 1 = links constant power over area to 1/r² law; 1 = first valid source of uncertainty with reduction; 1 = second valid source of uncertainty with reduction; 1 = condition where model fails; 1 = justification for model failure (physical reasoning).
Q2 — Sample Band 6 response (8 marks), annotated
Hypothesis (1 mark): If intensity is proportional to amplitude squared (I ∝ A²), then a graph of I vs A² will be a straight line through the origin, whereas a graph of I vs A will be a non-linear curve. This matters physically because the energy of an oscillator is proportional to the square of its displacement, so the power radiated per unit area also scales with A². [1 mark for a hypothesis that explicitly distinguishes the two models graphically or mathematically, with physical justification for squaring.]
Variables (1 mark): IV = amplitude of the signal generator output (controlled by adjusting the output voltage); DV = sound intensity measured by the sound level meter in W/m²; Controlled = frequency (set to a fixed value, e.g. 1000 Hz), distance from speaker to meter (fixed at 1.0 m with a ruler), speaker orientation (unchanged). [1 mark for all three correctly identified with a clear operational definition of the IV.]
Procedure (1 mark): (1) Connect the signal generator to the speaker. Set frequency to 1000 Hz and note the baseline voltage (V⊂0}) corresponding to a known amplitude. (2) Position the sound level meter at exactly 1.0 m from the speaker centre, using a ruler, and record the initial intensity. (3) Increase the signal generator voltage in equal steps (e.g. V⊂0}, 2V⊂0}, 3V⊂0}, 4V⊂0}) and record the corresponding intensity for each step. Record the actual amplitude (or a proxy such as output voltage) for each setting. (4) Repeat each measurement three times and average to reduce random error. [1 mark for four numbered steps including fixed distance, multiple trials, and a specific means of varying amplitude.]
Graphical analysis (2 marks): Plot I vs A: if linear, supports I ∝ A; if curved, suggests higher power law [1]. Also plot I vs A²: if this is linear through the origin, the data confirm I ∝ A² [1]. A coefficient of determination (R²) close to 1 on the second graph but not the first would quantitatively confirm the squared relationship. [2 marks for correctly identifying both plots and the interpretation criterion.]
Limitations (1 mark, minimum 2 stated): The speaker’s acoustic output may not be linearly proportional to the input voltage at high amplitudes (speaker saturation or distortion), meaning the assumed amplitude may not reflect the actual amplitude at the measurement point [1]. Reflections from the lab walls, floor, and ceiling create standing wave patterns that add or subtract intensity at the fixed measurement point, introducing random error that cannot easily be controlled [1]. Accept also: the sound level meter may have a limited frequency response or accuracy range.
Improvement (1 mark): Conduct the experiment in an anechoic chamber (or a large open outdoor space) to eliminate reflections and ensure that only direct-path sound is measured, making the result much more reliable and closer to the theoretical model [1].
Marking criteria summary (8 marks): 1 = hypothesis distinguishing I∝A from I∝A² with physical reasoning; 1 = all three variables correctly identified with operational definitions; 1 = four numbered procedural steps including distance control and repeated trials; 1 = graphical plan to identify linear I vs A² plot; 1 = interpretation criterion for the graph (straight line through origin); 1 = first valid limitation; 1 = second valid limitation; 1 = specific improvement addressing one limitation.