Year 12 Physics Module 7 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 2 of 14

Properties of Electromagnetic Waves

In 1887, Albert Michelson and Edward Morley used an interferometer to test for the expected effect of Earth's motion through a hypothetical light-carrying aether. The predicted fringe shift was about 0.4 fringe, but the observed shift was below 0.01 fringe. This null result was inconsistent with the expected aether wind and became important evidence in the development of a theory of light that needs no material medium.

Today's hook: Michelson and Morley expected an aether wind to produce a fringe shift of about 0.4 fringe; they observed less than 0.01. The null result did not by itself prove Einstein's later postulates, but it was inconsistent with the expected stationary-aether model. Now consider: if an ideal mobile-phone transmitter radiated in all directions, would moving twice as far away halve the intensity, or change it by a different factor?
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Orient and predict

Predict how a radio signal weakens with distance, then set up the vocabulary for EM wave properties.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

A mobile phone tower emits radio waves uniformly in all directions. A person stands 100 m from the tower and measures a certain signal strength.

Before reading on, answer:

  1. If the person moves to 200 m from the tower, what happens to the signal strength? Is it halved, quartered, or something else?
  2. Why does the signal spread out and weaken with distance?
  3. A flashlight beam appears to travel in a straight line and does not spread noticeably. Why does a radio tower's signal spread in all directions?

Warm-up, electromagnetic waves require which medium to travel through space?

Learning Intentions
goals

Know, EM Wave Production

  • Accelerating charges produce electromagnetic waves
  • Transverse nature: E and B fields perpendicular to each other and to propagation
  • All EM waves travel at $c$ in vacuum

Understand, Intensity & Inverse Square Law

  • Intensity $I = P/(4\pi r^2)$ for a point source
  • Intensity decreases with the square of distance
  • Energy transport by EM waves

Can Do, Analyse Wave Behaviour

  • Calculate intensity at different distances
  • Explain reflection, refraction and absorption
  • Relate wave properties to real applications
Scan these before reading
vocab
Transverse waveA wave in which oscillations are perpendicular to the direction of energy transfer.
Intensity ($I$)Power per unit area; $I = P/A$ measured in W/m².
Inverse square lawIntensity is inversely proportional to the square of distance from a point source.
Cross-lesson links: L01 described what EM waves are. L02 examines their key properties, transverse polarisation, constant speed $c$, and the absence of a medium for propagation. The Michelson-Morley result is historically crucial: within the experiment's sensitivity it removed the observational support for a stationary aether, and set the stage for Einstein's special relativity in L11.

Core Content

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Explain how EM waves are produced

Link accelerating charges to self-sustaining perpendicular E and B fields, and match each spectrum band to its source.

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How Electromagnetic Waves Are Produced
+5 XP

Accelerating charges create ripples in the electromagnetic field

Switch on a mobile phone and an alternating current drives electrons back and forth along the antenna at hundreds of millions of times per second. Each time those electrons reverse direction they accelerate, and that acceleration sends electromagnetic waves rippling outward through empty space at $c = 3.00 \times 10^8$ m/s, reaching a base station kilometres away without any physical medium connecting them. Electromagnetic waves are produced whenever electric charges accelerate: the oscillating electron creates a changing electric field, which in turn creates a changing magnetic field, forming a self-sustaining wave.

The key features of an electromagnetic wave are:

  • The electric field ($\vec{E}$) and magnetic field ($\vec{B}$) oscillate perpendicular to each other.
  • Both fields are perpendicular to the direction of propagation making EM waves transverse.
  • The ratio of field amplitudes is fixed: $E/B = c$.
E field B field (dots = out of page) Direction of propagation

Figure 1, Electromagnetic wave: E field (orange) oscillates vertically, B field (blue) oscillates perpendicular to it, wave propagates horizontally

Different regions of the spectrum are produced by different mechanisms:

  • Radio waves: Accelerated electrons in antenna circuits.
  • Microwaves: Electron transitions in molecules (rotational) and special vacuum tubes (magnetrons).
  • Infrared: Molecular vibrations and thermal motion of atoms.
  • Visible light and UV: Electron transitions between energy levels in atoms.
  • X-rays: Rapid deceleration of high-energy electrons (bremsstrahlung) or electron transitions in inner atomic shells.
  • Gamma rays: Nuclear decay and transitions within the nucleus.
Stop & Check

Explain why a stationary charge does not emit electromagnetic waves, while an oscillating charge does. What is the key difference?

Electromagnetic waves are produced by accelerating (oscillating) charges. The electric field $\vec{E}$ and magnetic field $\vec{B}$ oscillate perpendicular to each other and to the direction of propagation, making EM waves transverse. Each changing field sustains the other ($E/B = c$), so no medium is required.

Pause, copy the highlighted definition into your book before moving on.

Which of the following produces infrared radiation?

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Model intensity and boundary behaviour

Derive the inverse square law from spherical spreading, then compare reflection, refraction and absorption.

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Intensity and the Inverse Square Law
+5 XP

Why signals fade with distance

We just saw that EM waves are produced by accelerating charges and require no medium. That raises a question: as those waves travel outward in all directions, how does their strength change with distance? This card answers it → the inverse square law $I = P/(4\pi r^2)$.

As an electromagnetic wave spreads out from a point source, its energy is distributed over an increasingly large area. The intensity $I$ at distance $r$ from a source of power $P$ is:

$$I = \dfrac{P}{4\pi r^2}$$

This is the inverse square law: intensity is inversely proportional to the square of the distance from the source. Doubling the distance reduces intensity to one-quarter. Tripling the distance reduces it to one-ninth.

Intensity Equations

$I = P/A$  ,   Intensity = power / area (W/m²)

$I = \dfrac{P}{4\pi r^2}$  ,   Inverse square law for point source

$\dfrac{I_1}{I_2} = \left(\dfrac{r_2}{r_1}\right)^2$  ,   Ratio of intensities at two distances

Reflection occurs when an EM wave strikes a boundary and bounces back. The angle of incidence equals the angle of reflection. Smooth surfaces (mirrors) produce specular reflection; rough surfaces produce diffuse reflection.

Refraction occurs when an EM wave crosses a boundary between media with different refractive indices, changing direction and speed. The frequency stays constant; wavelength and speed change.

Absorption occurs when the wave's energy is transferred to the medium. Different materials absorb different wavelengths, this is why glass is transparent to visible light but opaque to UV, and why skin absorbs UV and gets sunburned.

Stop & Check

A 100 W light bulb emits visible light uniformly in all directions. Calculate the intensity at 2.0 m from the bulb. A person moves from 2.0 m to 4.0 m away. By what factor does the intensity change?

Intensity from a point source: $I = P/(4\pi r^2)$ (W/m²). Doubling the distance reduces intensity to $1/4$, the inverse square law. Ratio form: $I_1/I_2 = (r_2/r_1)^2$. At boundaries: reflection (angle in = angle out), refraction (frequency unchanged, speed and wavelength change), absorption (energy transfers to medium).

Pause, write the highlighted inverse square law formula and key points into your book before the check below.

If the distance from a point source is tripled, the intensity of radiation:

4

Investigate the inverse square law

Drive the simulator, check each reading by hand and find the distance where intensity falls to 1%.

Activity 1, Inverse Square Law Simulator
ApplyBand 4

Use the interactive to explore how intensity changes with distance

P = 100 W
r = 3.0 m
I = 0.88 W/m²
Compared to 1 m: 11.1%
  1. Set power to 100 W and distance to 1 m. Record the intensity. Then move to 2 m. Verify that the new intensity is exactly $1/4$ of the original.
  2. At what distance does the intensity drop to 1% of the value at 1 m?
  3. A phone tower emits 500 W. Calculate the intensity at 5.0 km. Convert this to a percentage of the intensity at 1.0 km.
  4. Explain why radio telescopes must be extremely large (like the 64 m Parkes dish) to detect weak signals from distant space.
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Solve intensity problems

Work the three standard question types: direct substitution, the ratio method and rearranging for distance.

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Worked Example, Inverse Square Law
+5 XP

Calculating intensity and safe distances

We just saw that intensity follows $I = P/(4\pi r^2)$. That raises a question: how do you apply this formula in multi-step problems, finding intensity at a given distance, solving for distance, or finding a safe distance? This card answers it → worked examples using direct substitution and the ratio method.

Problem

A microwave transmitter emits 200 W of power uniformly in all directions.

  1. Calculate the intensity at 4.0 m from the transmitter.
  2. A person stands 2.0 m from the transmitter and measures an intensity of 4.0 W/m². At what distance would the intensity drop to 0.50 W/m²?
  3. Safety guidelines recommend maximum exposure of 10 W/m² for microwaves. What is the minimum safe distance from this transmitter?
Step 1, Intensity at 4.0 m

$I = P/(4\pi r^2) = 200 / (4\pi \times 4.0^2) = 200 / 201.1 \approx$ 1.0 W/m²

Step 2, Distance for 0.50 W/m²

$I_1/I_2 = (r_2/r_1)^2$

$4.0/0.50 = (r_2/2.0)^2 \Rightarrow 8.0 = (r_2/2.0)^2$

$r_2 = 2.0 \times \sqrt{8.0} = 2.0 \times 2.83 \approx$ 5.7 m

Step 3, Minimum safe distance

$r = \sqrt{P/(4\pi I_{max})} = \sqrt{200/(4\pi \times 10)} = \sqrt{1.59} \approx$ 1.3 m

Stop & Check

The Sun emits approximately $3.8\times10^{26}$ W. Calculate the intensity of solar radiation at Earth's orbit ($r = 1.5\times10^{11}$ m). This value is known as the solar constant.

Three-step inverse square law method: (1) direct calculation $I = P/(4\pi r^2)$; (2) ratio form $I_1/I_2 = (r_2/r_1)^2$ when power is not given; (3) rearrange for distance $r = \sqrt{P/(4\pi I)}$. HSC trap: square the distance ratio and then invert it to get the intensity ratio.

Add the highlighted method steps to your notes before the check below.

A 60 W light bulb emits uniformly in all directions. What is the intensity at 3.0 m from the bulb?

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Apply and consolidate

Avoid the standard exam traps, apply the ratio form yourself and pull the three big ideas together.

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HSC Tip: Inverse Square Law Traps & Spherical Spreading
+5 XP

Avoid the most common exam errors

We just saw how to apply $I = P/(4\pi r^2)$ in worked examples. That raises a question: what are the most common mistakes students make in exam questions about the inverse square law? This card answers it → the key conceptual traps and a reminder that only point sources obey this law.

The most common error with the inverse square law is confusing the factor by which distance changes with the factor by which intensity changes.

  • If distance doubles ($\times 2$), intensity becomes $1/2^2 = 1/4$.
  • If distance triples ($\times 3$), intensity becomes $1/9$.
  • Square the distance ratio, then invert.

Also, the inverse square law only applies to point sources radiating uniformly in all directions. Lasers and focused beams do not obey this law because their energy is collimated.

Source r r I I/4 Area ∝ r² At 2r: area = 4× larger Same power spread over 4× area ∴ I drops to 1/4

Figure 2, Spherical spreading: doubling distance quadruples the area, quartering the intensity

ApplyBand 5

Activity 2: A person measures an intensity of 8.0 W/m² at 2.0 m from a point source. What is the intensity at 8.0 m? Show your working using the ratio form of the inverse square law.

The inverse square law $I \propto 1/r^2$ applies only to point sources radiating uniformly in all directions (area $= 4\pi r^2$). Lasers and collimated beams do NOT obey it. Common trap: doubling distance reduces intensity by factor 4, not 2, always square the distance ratio and invert.

Pause, write the highlighted principle into your book before the check below.

Doubling the distance from a point source reduces the intensity to one-quarter.

A laser beam obeys the inverse square law because it is an electromagnetic wave.

The electric and magnetic fields in an EM wave are perpendicular to each other and to the direction of propagation.

Synthesis, connect the ideas

In this lesson, three ideas lock together:

  • EM waves are produced by accelerating charges, both $\vec{E}$ and $\vec{B}$ oscillate perpendicular to propagation.
  • Intensity from a point source follows $I = P/(4\pi r^2)$, it falls as $1/r^2$, not $1/r$.
  • Different behaviours at boundaries, reflection (angle in = angle out), refraction (frequency preserved, speed and $\lambda$ change), absorption (energy transferred to medium).