Lesson 12: Electromagnetic Waves and the Spectrum

Radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays are one family: electromagnetic waves, oscillating electric and magnetic fields travelling at c = 3.00 × 10⁸ m/s in a vacuum, needing no medium. This lesson covers the spectrum as a single continuous story of frequency, energy and use — plus the inverse-square law that governs how intensity fades with distance.

The spectrum and photon energy

All EM waves obey v = fλ, and each carries energy in photons of E = hf. The families are defined by how they are produced and detected rather than by sharp boundaries: radio (λ from kilometres down to ~0.1 m), microwaves (0.1 m to 1 mm), infrared, visible (a narrow band, 700 nm red to 400 nm violet), ultraviolet, X-rays (down to ~10⁻¹¹ m) and gamma rays. As frequency rises, photon energy rises and waves become more penetrating and more biologically damaging: radio photons (~10⁻⁷ eV) cannot break chemical bonds, ultraviolet (~5 eV) can — hence sunburn — and gamma photons can ionise atoms deep inside tissue.

Because waves spread out, intensity follows the inverse-square law: for a source radiating power P equally in all directions, I = P ÷ 4πd². Double the distance and intensity falls to a quarter. This law applies to light, sound intensity (approximately), gravity and radiation exposure alike — it is pure geometry: the same energy spread over the surface of a growing sphere.

Worked example 1: the microwave oven

A microwave oven operates at 2.45 GHz. Find the wavelength of its radiation, and hence explain the hot spots in heated food. λ = c ÷ f = 3.00 × 10⁸ ÷ 2.45 × 10⁹ = 0.122 m (12.2 cm). Inside the oven, reflections set up standing waves with antinodes every λ ÷ 2 ≈ 6.1 cm — matching the spacing of hot and cold patches in food that is not rotated, which is exactly why ovens have turntables.

Worked example 2: sunlight and the inverse-square law

The Sun radiates P = 3.83 × 10²⁶ W. Earth is 1.50 × 10¹¹ m away. Find the intensity of sunlight arriving at Earth. I = P ÷ 4πd² = 3.83 × 10²⁶ ÷ (4π × (1.50 × 10¹¹)²) = 3.83 × 10²⁶ ÷ 2.83 × 10²³ = 1,354 W/m² — the famous "solar constant" of about 1.4 kW on every sun-facing square metre at the top of the atmosphere. For a simpler sense of scale, a 100 W isotropic bulb gives I = 100 ÷ (4π × 2.0²) = 2.0 W/m² at 2.0 m, and only 0.50 W/m² at 4.0 m — the quartering in action.

In the lab: testing the inverse-square law

A phone light sensor measures the illuminance from a small lamp at several distances (background light subtracted):

Test the law by computing I × d²: 250 × 0.040 = 10; 111 × 0.090 = 10; 63 × 0.16 = 10; 40 × 0.25 = 10. The product is constant, confirming I ∝ 1 ÷ d². Uncertainties: distance should be measured to the lamp's filament, not its glass (an offset error that matters most at small d — one reason the closest point often deviates), the sensor must face the lamp squarely, and ambient light must be subtracted or it inflates distant readings. Plotting I against 1 ÷ d² and checking for a straight line through the origin is the rigorous version of the same test.

Physics in the real world

Each band earns its keep: radio for broadcasting and radar; microwaves for cooking (2.45 GHz couples strongly to water molecules), Wi-Fi and satellite links; infrared for thermal imaging, remote controls and the James Webb Space Telescope's view of cold, distant objects; ultraviolet for sterilisation and fluorescence; X-rays for imaging bones and airport security; gamma rays for radiotherapy and industrial weld inspection. The same physics sets the hazards: dose falls with the inverse-square law, so distance is the cheapest radiation protection in medicine and industry, alongside time limits and shielding.

Common mistakes

• Thinking EM waves need a medium — they are self-propagating field oscillations; that is how sunlight crosses the vacuum of space.
• Confusing radio waves with sound waves; a radio is a receiver of light-speed EM, not of sound.
• Believing brighter light means more energetic photons — brightness is photon *rate*; photon energy is set by frequency alone.
• Mis-ordering the spectrum: remember frequency rises radio → gamma, so wavelength falls.
• Applying the inverse-square law to a laser beam or a light pipe, where the energy is deliberately *not* spreading spherically.

Exam Experience

1. Which list orders electromagnetic waves by increasing frequency? A) radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. B) gamma, X-ray, ultraviolet, visible, infrared, microwave, radio. C) microwave, radio, infrared, visible, X-ray, ultraviolet, gamma. D) infrared, radio, visible, microwave, gamma, X-ray, ultraviolet. Answer: A. Frequency increases steadily across the spectrum from radio (lowest) to gamma rays (highest); wavelength decreases in the same sequence.

2. A source radiates equally in all directions. At 3.0 m the intensity is 12 W/m². At 6.0 m it will be: A) 6.0 W/m². B) 3.0 W/m². C) 24 W/m². D) 1.3 W/m². Answer: B. Doubling distance quarters the intensity: 12 ÷ 4 = 3.0 W/m².

3. Electromagnetic waves in a vacuum all: A) have the same wavelength. B) have the same frequency. C) travel at 3.00 × 10⁸ m/s. D) carry the same photon energy. Answer: C. The speed is universal in vacuum; wavelength, frequency and photon energy are linked but vary across the spectrum.

Constructed response (4 marks). An FM radio station broadcasts at 98.5 MHz. (a) Calculate the wavelength of its waves. (b) Calculate the energy of one of its photons, in eV, and comment on why radio waves do not cause sunburn. (1 eV = 1.602 × 10⁻¹⁹ J.) Scoring guide: (a) 1 mark for λ = c ÷ f = 3.00 × 10⁸ ÷ 9.85 × 10⁷ = 3.0 m. (b) 1 mark for E = hf = 6.63 × 10⁻³⁴ × 9.85 × 10⁷ = 6.5 × 10⁻²⁶ J; 1 mark for converting: 6.5 × 10⁻²⁶ ÷ 1.602 × 10⁻¹⁹ = 4.1 × 10⁻⁷ eV; 1 mark for the comment that this is millions of times smaller than the few eV needed to disrupt molecules/ionise, so no chemical damage occurs regardless of intensity at broadcast levels. Common error: computing photon energy in joules and comparing directly with an eV threshold without converting.

Going further: counting photons, and phoning via satellite

Two quick calculations show the quantum graininess of light in everyday settings. A green laser pointer emits 1.0 mW (1.0 × 10⁻³ J per second) at 532 nm. Each photon carries E = hc ÷ λ = 3.74 × 10⁻¹⁹ J, so the beam contains 1.0 × 10⁻³ ÷ 3.74 × 10⁻¹⁹ = 2.7 × 10¹⁵ photons per second. Even a feeble beam is a torrent of quanta — which is why light usually *looks* continuous, and why detecting single photons needs cryogenically cooled laboratory sensors. Sunlight, at about 1,400 W/m² with typical photon energies near 2 eV, delivers roughly 4 × 10²¹ photons to every sunlit square metre each second. For communications, distance becomes delay. A geostationary satellite sits 3.58 × 10⁷ m up; a signal making the round trip (up to the satellite and back down, twice for a two-way call) travels about 4 × 3.58 × 10⁷ m, taking 1.43 × 10⁸ ÷ 3.00 × 10⁸ ≈ 0.48 s for a full exchange — the noticeable pause on satellite phone calls and older international broadcasts. Fibre signals travel at about 2.0 × 10⁸ m/s (c ÷ n for glass), so a 5,600 km transatlantic cable imposes roughly 28 ms one-way — invisible in conversation, but enough to matter enormously to financial trading networks, which pay premium prices for straighter cable routes.

In focus: why thermal cameras see in the infrared

Apply Wien's law to a person: skin temperature is about 310 K (37 °C), so the peak emission wavelength is λ_max = 2.898 × 10⁻³ ÷ 310 = 9.3 × 10⁻⁶ m = 9.3 μm — deep in the infrared, far beyond the red end of vision. Humans glow brightly; we simply lack the sensors. Thermal cameras are built around this number: their detectors respond to the 8–14 μm band precisely because everything at everyday temperatures peaks there, and because the atmosphere is fairly transparent in that window. Firefighters use them to find people in smoke (smoke particles scatter visible light but pass long-wave infrared), building inspectors find missing insulation as warm patches, and border and wildlife sensors work at night when the visible spectrum has nothing to offer. The inverse-square law rides along: a thermal camera does not measure a surface's temperature directly, it measures arriving infrared flux, so distance and the surface's emissivity (how efficiently it radiates compared with a perfect black body) both enter the calibration. Polished metal, an excellent reflector and poor emitter, can fool a thermal camera into reporting the temperature of whatever it reflects — another practical case where the naive reading and the physics part company.

Formula card

Exam technique: spectrum questions reward ordering confidence — anchor the sequence with visible light in the middle and one fact at each end (radio: metres to kilometres; gamma: nuclear origin). For inverse-square problems, work with ratios: new intensity = old × (old distance ÷ new distance)², which avoids computing 4π at all.