Light and Sound¶
Syllabus mapping
PY-11-02 — Waves. Covers: mechanical waves require a medium; sound wave energy transfer; the relationships frequency↔pitch and amplitude↔loudness; electromagnetic waves don't require a medium; the electromagnetic spectrum and its applications; the inverse square law for light intensity.
Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.
You need to know: the key difference between mechanical and electromagnetic waves, how frequency and amplitude relate to what we perceive as pitch and loudness, the structure of the EM spectrum, and how to apply the inverse square law to light intensity.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Wave Properties — wavelength, frequency, period, the wave equation
Core¶
Mechanical waves need a medium — electromagnetic waves don't¶
Mechanical waves (sound, water waves, waves on a string) transfer energy through the physical oscillation of a medium's particles — no medium, no wave. This is why sound can't travel through the vacuum of space.
Electromagnetic waves (light, radio, X-rays, and everything else in the EM spectrum) are oscillating electric and magnetic fields, and don't need a medium at all — they travel through a vacuum just as well as through air or glass (in fact, faster). The speed of light in vacuum, from your data sheet:
Sound wave energy transfer¶
Sound is a longitudinal mechanical wave: a vibrating source (a speaker cone, vocal cords) pushes air molecules together (compression) and apart (rarefaction), and this pattern propagates outward as neighbouring particles collide with and push each other in turn. Speed of sound in air, from your data sheet: \(340\ \text{m s}^{-1}\).
Frequency ↔ pitch, amplitude ↔ loudness¶
| Wave property | Perceived as |
|---|---|
| Higher frequency | Higher pitch |
| Lower frequency | Lower pitch |
| Greater amplitude | Louder |
| Smaller amplitude | Quieter |
The electromagnetic spectrum¶
🖼️ Diagram needed: labelled EM spectrum from radio to gamma, showing wavelength/frequency trend and a practical application per band. TODO — placeholder until sourced/drawn.
All EM waves travel at \(c\) in a vacuum — what distinguishes radio waves from gamma rays is purely wavelength (and correspondingly, frequency, via \(v=f\lambda\)).
| Band | Wavelength trend | A practical application |
|---|---|---|
| Radio | Longest | Broadcast communication |
| Microwave | — | Cooking, radar |
| Infrared | — | Thermal imaging, remote controls |
| Visible light | — | Human vision |
| Ultraviolet | — | Sterilisation, vitamin D synthesis |
| X-ray | — | Medical imaging |
| Gamma | Shortest | Cancer treatment, sterilisation |
Inverse square law for light intensity¶
Intensity is power spread over an area; as you move away from a point source, that same power spreads over an ever-larger area, so intensity drops off with the square of distance, not distance itself. This gives a direct way to compare intensity at two different distances from the same source, exactly as it appears on your data sheet:
Worked example
A light source has intensity \(80\ \text{W m}^{-2}\) at \(2\ \text{m}\). Find the intensity at \(4\ \text{m}\).
Doubling the distance quartered the intensity — a direct consequence of the square in the inverse square law.
Practical¶
Suggested prac: light intensity vs distance — a light sensor and a light source on an optical bench, varying distance and measuring intensity, then checking the results follow \(I \propto \dfrac{1}{r^2}\) (e.g. by graphing \(I\) against \(\dfrac{1}{r^2}\) and checking for a straight line). TODO — link this school's actual prac instructions once written.
Advanced
Where the inverse square law comes from. A point source radiates power \(P\) equally in all directions, spreading over the surface of an expanding sphere. Since a sphere's surface area is \(4\pi r^2\):
This isn't on your data sheet, but it's the reasoning behind \(I \propto \dfrac{1}{r^2}\) — worth being able to derive it, not just quote it, since it also explains why the same \(1/r^2\) pattern shows up in gravity (\(F=\dfrac{GMm}{r^2}\), already seen in Motion Relationships' data sheet formulae) and — in Year 12 — in electric fields: any quantity spreading equally in all directions from a point source picks up the same geometric \(1/r^2\) factor.
Extension
Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
The inverse square pattern isn't a coincidence specific to light — it's a consequence of living in three spatial dimensions. Any conserved quantity radiating uniformly from a point source (light intensity, gravitational field strength, electric field strength) spreads over a sphere's surface area \(4\pi r^2\), so all of them fall off as \(1/r^2\). A depth study could compare the inverse square law across two or three different contexts (light intensity, gravity, and — see the Extension box on Electrostatics once that focus area is reached — electric field strength via Coulomb's law) and demonstrate they share the same underlying geometric origin, not just a similar-looking formula.
Video/visual resources¶
- 🎥 Khan Academy — TODO: source an EM spectrum / mechanical vs EM waves explainer
- 🎥 Physics High — TODO: check for a NSW-syllabus-aligned light and sound video
- 🖥️ Interactive EM spectrum resource — TODO: source and confirm a link
Check yourself¶
-
Explain why an astronaut on the Moon could not hear an explosion happening nearby on the lunar surface.
Answer
Sound is a mechanical wave and needs a medium to travel through. The Moon has no atmosphere (no air), so there's no medium for sound to propagate through — the astronaut would see the explosion but hear nothing.
-
A speaker's sound is measured at \(64\ \text{W m}^{-2}\) at \(1\ \text{m}\). Find the intensity at \(8\ \text{m}\).
Answer
\(I_1 r_1^2 = I_2 r_2^2 \implies I_2 = \dfrac{64 \times 1^2}{8^2} = 1\ \text{W m}^{-2}\)
-
Two sounds have the same amplitude, but sound A has a higher frequency than sound B. Two other sounds have the same frequency, but sound C has a greater amplitude than sound D. Describe how each pair differs to a listener.
Answer
A and B: same loudness, but A has a higher pitch than B (frequency determines pitch).
C and D: same pitch, but C is louder than D (amplitude determines loudness).