Ch 3 · Electromagnetic Spectrum and Sound · IGCSE Physics 0625
Topic 3 of 3 in Waves — IGCSE Physics 0625 — 17 lessons.
Introduction
Visible light is only a tiny slice of a much larger family of electromagnetic (EM) waves that stretches from kilometre-long radio waves to gamma rays smaller than an atomic nucleus. Sound, by contrast, is a mechanical wave that needs a material to travel through. This lesson compares the two: where each part of the EM spectrum is used and what harm it can do, and how sound is produced, measured and used in echoes and ultrasound scanning.
Key ideas — The electromagnetic spectrum
1. Order of the regions. In order of increasing frequency (and decreasing wavelength):
🧑🏫 Teacher's explanation
Imagine the electromagnetic spectrum as a giant musical keyboard. On one end, you have the deep, low bass notes (radio waves) which have long, lazy waves. On the other end, you have the extremely high-pitched, frantic notes (gamma rays) which have tiny, rapid vibrations. Even though they sound different, if you were in a vacuum, every single note would travel across the room at the exact same speed: 3.0 × 10^8 m/s.
📖 Definitions
🔍 Worked example — Easy
A radio station transmits waves with a frequency of 1.0 × 10^6 Hz. Calculate the wavelength of these waves in a vacuum.
🔍 Worked example — Standard
An X-ray has a wavelength of 1.5 × 10^-10 m. Calculate its frequency.
🔍 Worked example — Exam-style
A satellite uses microwaves to transmit data. If the speed of the wave is 3.0 × 10^8 m/s and the frequency is 1.2 × 10^10 Hz, find the wavelength.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) State one typical use of infrared radiation. [1] (b) Explain why ultraviolet radiation is more dangerous to human skin than infrared radiation. [2] [3 marks]
Key ideas — Sound
6. Production and travel. Sound is produced by vibrating sources (a guitar string, a loudspeaker cone, vocal cords). The vibrations pass through the air as a longitudinal wave. Sound needs a medium — it cannot travel through a vacuum. The bell-in-a-jar experiment shows this: as air is pumped out, the ringing gets fainter although the bell can still be seen vibrating.
Practical notes
Measuring the speed of sound in air (outdoor method). Two students stand a measured distance apart, at least 100 m, measured with a trundle wheel or long tape. One makes a sharp sound that can also be seen (e.g. banging two blocks together or firing a starting pistol). The other starts a stopwatch when they see the action and stops it when they hear the sound. Light arrives almost instantly, so v = distance / time. Repeat several times and average; swap roles and directions to reduce the effect of wind and reaction time.
Worked example
Example 1 — microwave wavelength. A microwave oven uses a frequency of 2.45 GHz.
Common mistakes and examiner tips
Practice
Answers: (1) X-rays, ultraviolet, visible light, radio waves. (2) (a) gamma rays; (b) infrared; (c) microwaves; (d) ultraviolet. (3) Any two: transverse; travel through a vacuum; same speed in a vacuum (3.0 × 10^8\ m/s); obey v = fλ. (4) λ = 340/20 000 = 0.017\ m. (5) d = 340 × 3.0 = 1020\ m ≈ 1.0\ km (no halving — the sound travels one way only). (6) Plucking harder increases the amplitude, so the sound is louder; faster vibration increases the frequency, so the pitch is higher. (7) d = dfrac1540 × 1.3 × 10^-42 = 0.10\ m (10 cm). (8) Higher rate of data transmission; the signal can be regenerated without noise, so it travels further without losing quality. (9) Bluetooth uses radio waves of low power; the signal is weakened as it travels and as it passes through walls, so it becomes too weak at larger distances.