AS Physics: Waves — Revision Notes

Condensed for the final weeks. For the full explanation, use the Waves study guide.

Wave quantities

Quantity	Symbol	Meaning
Displacement	x	Distance of a point from equilibrium
Amplitude	A	Maximum displacement
Wavelength	λ	Distance between adjacent points in phase
Period	T	Time for one complete oscillation
Frequency	f	Oscillations per second (Hz)
Wave speed	v	Distance travelled per second

v = f λ            f = 1 / T

Intensity ∝ amplitude² — doubling the amplitude quadruples the intensity. And intensity ∝ 1/r² for an ideal isotropic point source radiating equally in all directions with negligible absorption, since the same total power is spread over the surface of an ever-larger sphere as distance from the source increases — doubling the distance from such a source cuts the intensity to a quarter, not a half. A directional source, a source near a reflecting boundary, or significant absorption in the medium can all make the real fall-off deviate from a simple inverse-square law.

Transverse vs longitudinal

	Transverse	Longitudinal
Oscillation	Perpendicular to energy transfer	Parallel to energy transfer
Structure	Crests and troughs	Compressions and rarefactions
Polarisable?	Yes	No
Examples	Light, all EM waves, water waves	Sound, ultrasound

Only transverse waves can be polarised — the standard reason sound cannot be, since a longitudinal oscillation has only one direction (parallel to travel) to begin with, so there is no second plane for a filter to restrict it to.

Polarisation and Malus’s law

Unpolarised light has oscillations in all planes perpendicular to travel. A polariser transmits one plane only, halving the intensity.

I = I0 cos^2(theta)

Worked example: plane-polarised light of intensity 8.0 W/m2 passes through
a filter at 30 degrees to the plane of polarisation.
I = 8.0 x cos^2(30) = 8.0 x 0.75 = 6.0 W/m2

This formula applies to light that is already plane-polarised — the case tested at this level, not the separate question of what happens when unpolarised light first meets a polariser. θ is measured between the filter’s own transmission axis and the plane in which the incoming light is already polarised, not any other reference direction.

Two polarisers at 90° (“crossed”) transmit zero intensity.

The electromagnetic spectrum

radio -> microwave -> infrared -> VISIBLE -> ultraviolet -> X-ray -> gamma
LONGEST wavelength, LOWEST frequency ------> SHORTEST, HIGHEST

Visible light: roughly 400 nm (violet) to 700 nm (red). All EM waves are transverse, travel at 3.00 × 10⁸ m/s in a vacuum, and are progressive transfers of energy — carrying energy from source to receiver without transferring matter, exactly like every other wave in this topic.

The Doppler effect

For a source moving relative to an observer:

f_observed = f_source x v / (v +/- v_s)

approaching -> use MINUS in the denominator -> frequency INCREASES
receding    -> use PLUS                     -> frequency DECREASES

The wavelength is compressed ahead of the source and stretched behind it. The source frequency itself does not change.

Worked example: a train's horn emits sound at 500 Hz and approaches a
platform at 20 m/s. Speed of sound = 340 m/s.
f_observed = 500 x 340 / (340 - 20) = 500 x 340 / 320 = 531 Hz

Only the stationary observer, moving source case is required at this level — the more general case, where the observer also moves, isn’t examined at AS. See the Waves practice questions for the full worked-answer versions of both calculations above.

Exam traps

• Intensity ∝ amplitude squared, not amplitude.
• Sound cannot be polarised — it is longitudinal.
• Phase difference in radians (2π per cycle) or degrees (360° per cycle); state which.
• In the Doppler equation, approaching gives a smaller denominator and therefore a higher frequency.
• Wave speed depends on the medium; frequency is set by the source and does not change on refraction — the wavelength does.
• Malus’s law (I = I₀cos²θ) only applies to light that is already plane-polarised — this specification does not require the separate case of unpolarised light meeting a first filter.
• Crossed polarisers (90° apart) give zero, not a small non-zero, transmitted intensity — cos²(90°) = 0 exactly.

Self-test

1. State the wave equation and the relationship between f and T.
2. Why can light be polarised but sound cannot?
3. Amplitude is doubled. What happens to intensity?
4. Order the EM spectrum from longest to shortest wavelength.
5. A siren approaches you. Does the observed frequency rise or fall, and why?
6. Plane-polarised light of intensity 12 W/m² passes through a filter at 60° to the plane of polarisation. Find the transmitted intensity.
7. Two polarisers are crossed at 90°. What intensity is transmitted, and why?

Answers: 1. v = fλ; f = 1/T. 2. Light is transverse, so oscillations occur in many planes perpendicular to travel and one plane can be selected; sound is longitudinal, oscillating parallel to travel, so there is no plane to filter. 3. It quadruples (I ∝ A²). 4. Radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. 5. It rises — the wavefronts ahead of the source are compressed, shortening the observed wavelength and raising the observed frequency, although the source frequency is unchanged. 6. I = 12 × cos²(60°) = 12 × 0.25 = 3.0 W/m². 7. Zero — cos²(90°) = 0, so no light with its plane of polarisation at 90° to the filter’s transmission axis can pass through.