Physics • Year 12 • Module 7 • Lesson 2
Properties of Electromagnetic Waves
Apply the inverse square law, interpret real intensity data, and reason about EM wave behaviour in physical scenarios.
1. Interpret experimental data, signal strength across Sydney
A radio tower in the CBD of Sydney emits 250 kW uniformly in all directions. A student records the signal intensity at five distances from the tower. The table below shows their measurements. 8 marks
| Distance from tower (r, km) | Measured intensity (W/m²) | Calculated intensity using I = P/(4πr²) (W/m²) | Agree within 10%? |
|---|---|---|---|
| 1.0 | 1.95 × 10² | ||
| 2.0 | 4.90 × 10¹ | ||
| 4.0 | 1.20 × 10¹ | ||
| 8.0 | 3.15 × 10&sup0; | ||
| 10.0 | 1.98 × 10&sup0; |
1.1 Complete the “Calculated intensity” and “Agree within 10%?” columns for all five rows. Show one full working for the 1.0 km row. 6 marks (1 per row)
1.2 The measured value at 4.0 km (1.20 × 10¹ W/m²) is slightly higher than the theoretical value. Suggest two physical reasons why a real-world measurement might exceed the ideal inverse square law prediction. 2 marks
2. Interpret a graph, intensity vs distance for a 100 W source
The graph below shows intensity (I) versus distance (r) for a 100 W point source radiating uniformly in all directions. 7 marks
Figure 2. Intensity vs distance for a 100 W point source. I = P/(4πr²). Theoretical curve.
2.1 Describe the overall shape of the graph and explain why intensity changes most rapidly at small distances. 2 marks
2.2 Use the graph to estimate the intensity at r = 2.0 m. Then calculate the exact value using the inverse square law. Comment on how well the graph matches the calculation. 3 marks
2.3 A student states: “Moving from 2 m to 4 m halves the intensity.” Use the graph and the inverse square law to explain why this statement is incorrect, and calculate the actual factor by which intensity changes. 2 marks
3. Compare EM wave interactions across five features
Complete the two-column table below. For each feature, write a concise description that contrasts reflection and refraction. 10 marks (1 per cell)
| Feature | Reflection | Refraction |
|---|---|---|
| What happens to the wave | ||
| Does wavelength change? | ||
| Does frequency change? | ||
| Governing angle/law | ||
| Real-world example |
4. Predict and justify, the Parkes Radio Telescope
The Parkes Radio Telescope (“The Dish”) in central New South Wales has a 64 m diameter dish. It was used in 1969 to relay live television from the Apollo 11 Moon landing. The transmitter on the lunar module had a power output of approximately 20 W.
5 marks
4.1 The Moon is approximately 3.8 × 10&sup8; m from Earth. Calculate the intensity of the lunar module’s radio signal at Earth’s surface. 2 marks
4.2 The 64 m diameter collecting dish intercepts a small fraction of the total power radiated. Explain why a larger dish is needed to detect such a weak signal, using the concept of intensity and collected power. Predict what would happen to the received signal if the dish diameter were halved. 3 marks
Q1.1, Calculated intensity table
Formula: I = P/(4πr²), with P = 250 × 10³ W and r in metres.
1.0 km (1000 m): I = 250000/(4π × 1000²) = 250000/12566000 ≈ 1.99 × 10¹ W/m². Measured 1.95 × 10¹ → agrees within 10%: Yes.
2.0 km (2000 m): I = 250000/(4π × 4 × 10&sup6;) ≈ 4.97 W/m². Measured 4.90 → Yes.
4.0 km: I ≈ 1.24 W/m². Measured 1.20 → Yes.
8.0 km: I ≈ 0.311 W/m². Measured 0.315 → Yes.
10.0 km: I ≈ 0.199 W/m². Measured 0.198 → Yes.
Award 1 mark per row for a correct calculated value (accept ±5%) and correct agree/disagree.
Q1.2, Reasons measurement may exceed theory (2 marks)
Any two of: (1) Reflections from buildings, terrain, or the ground add to the direct signal (multipath propagation), increasing measured intensity above the free-space prediction. (2) The transmitting antenna may not radiate uniformly in all directions; directional gain towards certain azimuths can produce higher-than-predicted intensity. (3) Atmospheric ducting under certain weather conditions can refract radio waves back toward the surface, concentrating signal. Award 1 mark each.
Q2.1, Shape description and rate of change (2 marks)
The graph is a steep, sharply decreasing curve that levels off rapidly, approaching (but never reaching) zero. [1 mark] Intensity changes most rapidly at small distances because I ∝ 1/r², a small increase in r near the source corresponds to a large change in 1/r², whereas the same increase at large distances causes a negligibly small change. [1 mark]
Q2.2, Reading and calculating at r = 2 m (3 marks)
Graph estimate: approximately 2.0 W/m² (read from graph near r = 2 m). [1 mark]
Calculation: I = 100/(4π × 2²) = 100/(4π × 4) = 100/50.27 ≈ 1.99 W/m². [1 mark]
The graph estimate of ~2.0 W/m² matches the calculated value of 1.99 W/m² very closely, confirming the inverse square law. [1 mark]
Q2.3, Correcting the “halved” misconception (2 marks)
The statement is incorrect. Using the ratio: I⊂2;/I⊂1; = (r⊂1;/r⊂2;)² = (2/4)² = 1/4. [1 mark] So moving from 2 m to 4 m reduces intensity to one-quarter (not one-half), consistent with the inverse square law where doubling the distance reduces intensity by a factor of 4. The graph confirms: ~2.0 W/m² at 2 m vs ~0.5 W/m² at 4 m. [1 mark]
Q3, Reflection vs refraction comparison
What happens: Reflection, wave bounces off a surface back into the original medium. Refraction, wave crosses into a new medium and changes direction.
Wavelength change: Reflection, no change. Refraction, yes, wavelength changes (speed changes, frequency fixed).
Frequency change: Reflection, no. Refraction, no; frequency is determined by the source and does not change.
Governing angle/law: Reflection, angle of incidence = angle of reflection (law of reflection). Refraction, the wave bends at the boundary because its speed changes; frequency stays constant while wavelength changes proportionally to the change in speed.
Example: Reflection, a mirror or metal surface reflecting visible light. Refraction, a glass lens or a prism bending visible light; atmospheric refraction of radio waves.
Q4.1, Intensity at Earth’s surface (2 marks)
I = P/(4πr²) = 20/(4π × (3.8 × 10&sup8;)²) [1 mark]
= 20/(4π × 1.444 × 1017) = 20/(1.814 × 1018) ≈ 1.1 × 10−17 W/m² [1 mark]
Q4.2, Dish size and collected power (3 marks)
Power collected by a dish = Intensity × collecting area = I × πR². Because the signal is so weak (~10−17 W/m²), only a very large collecting area can gather enough power for electronics to detect. [1 mark]
The 64 m diameter dish has area = π × 32² ≈ 3217 m², collecting about 3.5 × 10−14 W. A smaller dish collects proportionally less power. [1 mark]
If the diameter were halved (to 32 m), the area would reduce by a factor of 4 (¼ of original), so collected power would also reduce to one-quarter, making the signal 4 times harder to detect and potentially too weak to resolve. [1 mark]