Waves: A-level Physics AQA revision notes

3.3.1

Progressive and stationary waves

A progressive wave transfers energy 3.3.1.1

Definitions

Phase difference: how far out of step two points (or two waves) are, expressed as an angle or as a fraction of a full cycle.

Key results

Wave speed: $v = f λ$ ; frequency and period: $f = \frac{1}{T}$ .
Phase difference between two points a distance $Δ x$ apart on the same wave: $Δ ϕ = \frac{2 π Δ x}{λ}$ (in radians).

Notes

A progressive wave carries energy away from its source — every point along it oscillates with the same amplitude and frequency but a shifted phase, so the disturbance itself moves outward while the medium's own particles only oscillate about a fixed position.
Two snapshots of the same wave answer different questions: displacement against position (at one instant) shows the wave's shape and wavelength; displacement against time (at one fixed point) shows that point's own oscillation and period.
Points $λ /4$ apart are $90°$ ( $π /2$ ) out of phase, $λ /2$ apart are $180°$ ( $π$ ) out of phase (antiphase), and a full wavelength apart are back in phase — a direct, frequently tested consequence of $Δ ϕ = 2 π Δ x / λ$ .

Transverse, longitudinal and polarised waves 3.3.1.2

Definitions

Transverse wave: particle displacement is perpendicular to the direction of energy transfer (e.g. a wave on a string, all electromagnetic waves).
Longitudinal wave: particle displacement is parallel to the direction of energy transfer, producing alternating compressions and rarefactions (e.g. sound in air).
Plane polarisation: oscillation restricted to a single plane containing the direction of travel — only possible for a transverse wave.

Notes

Only a transverse wave can be polarised — a longitudinal wave's oscillation is already confined to the single line along which it travels, so there is no separate 'plane' left to restrict.
This makes polarisation a direct experimental test for whether a wave is transverse: if it can be polarised, it must be transverse.
A polariser only transmits the component of oscillation parallel to its own transmission axis; rotating a second polariser (an analyser) from parallel to crossed relative to the first smoothly reduces transmitted intensity from a maximum down to (ideally) zero.
All electromagnetic waves travel at $c \approx 3.00 \times 1 0^{8} m s^{- 1}$ in a vacuum and need no material medium to propagate — unlike sound, which requires one.

Superposition and stationary waves 3.3.1.3

Definitions

Principle of superposition: where two or more waves overlap, the resultant displacement at any point is the sum of the individual displacements, $y = y_{1} + y_{2}$ .
Stationary (standing) wave: formed when two progressive waves of equal amplitude and frequency travel in opposite directions and superpose — typically an incident wave and its own reflection.

Notes

Unlike a progressive wave, a stationary wave does not transfer energy along its length — the energy oscillates locally between kinetic and potential form instead.
At the instant both component waves are exactly aligned, the resultant reaches its maximum extreme profile (amplitude $2 a$ for two component waves each of amplitude $a$ ); a quarter period later, both components pass through zero simultaneously and the resultant is momentarily flat everywhere at once — the signature that distinguishes a stationary from a progressive wave pattern.
The positions where the two component waves always cancel (the nodes) stay fixed at zero displacement throughout the whole cycle, whichever instant is snapshotted.

Nodes, antinodes and phase 3.3.1.3

Definitions

Node: a fixed point of zero amplitude on a stationary wave. Antinode: a fixed point of maximum amplitude.

Key results

Distance between adjacent nodes (or adjacent antinodes): $\frac{λ}{2}$ . Distance from a node to the next antinode: $\frac{λ}{4}$ .

Notes

Every point within one loop (between adjacent nodes) oscillates in phase with a fixed amplitude that varies with position, from zero at the nodes up to $2 a$ at the antinode.
Points in adjacent loops oscillate exactly $180°$ ( $π$ rad) out of phase with each other — worth stating explicitly when asked to describe or sketch a stationary wave pattern.
No net energy transfer occurs along an ideal stationary wave — the two component travelling waves carry equal energy in opposite directions, which cancels overall.

Harmonics on a stretched string 3.3.1.3

Key results

First harmonic (fundamental): $λ_{1} = 2 L$ , $f_{1} = \frac{1}{2 L} \frac{F}{μ}$ (tension $F$ , mass per unit length $μ$ ).
$n$ th harmonic: $f_{n} = n f_{1}$ , $λ_{n} = \frac{2 L}{n}$ .
Wave speed on the string: $v = \frac{F}{μ}$ .

Method

Required practical 1: with a vibration generator (or similar) driving a string fixed at both ends, find the resonant frequencies that produce clear standing-wave patterns, and investigate how the fundamental frequency depends on length $L$ , tension $F$ and mass per unit length $μ$ by varying one quantity at a time and keeping the others fixed.

Notes

$f \propto \frac{1}{L}$ (at fixed $F, μ$ ); $f \propto F$ (at fixed $L, μ$ ); $f \propto \frac{1}{μ}$ (at fixed $L, F$ ) — three separate proportionalities, each confirmed by varying only one quantity while holding the other two fixed.
Each successive harmonic fits one more half-wavelength into the same fixed length $L$ — a string fixed at both ends always has a node at each end, so only wavelengths satisfying $L = n \times \frac{λ}{2}$ can form a stable stationary pattern.
This is exactly why a guitar or violin string produces a discrete set of possible notes (harmonics), not a continuous range.

Stationary waves in sound and microwaves 3.3.1.3

Key results

Tube closed at one end, first harmonic: $L = \frac{λ}{4}$ — the closed end is a displacement node (and pressure antinode); the open end is a displacement antinode (and pressure node).
Microwave standing wave from an incident wave reflecting off a metal sheet: adjacent minima (or maxima) in the pattern are $\frac{λ}{2}$ apart, letting wavelength be measured directly by moving a detecting probe.

Notes

A closed end must be a displacement node (the air there cannot move) but is simultaneously a pressure antinode (pressure variation is greatest where displacement is most restricted) — the displacement and pressure stationary-wave patterns for sound are exactly out of step with each other.
An ideal conducting surface reflecting microwaves forces a displacement/electric-field node at the surface itself, exactly analogous to a mechanical wave reflecting off a fixed (rather than free) end.
Measuring the spacing between adjacent minima (or maxima) with a moving probe is a direct, practical way to find microwave wavelength, using the same $λ /2$ node-spacing relationship that applies to any stationary wave.

Worked examples

Worked example 3.3.1 · 3 marks

Two points on a progressive wave of wavelength $0.80 m$ are separated by $0.60 m$ along the direction the wave travels.

Calculate the phase difference between the two points, and state whether they oscillate in phase, in antiphase, or neither.

Show worked solution

Δ ϕ = \frac{2 π Δ x}{λ} = \frac{2 π \times 0.60}{0.80} = 1.5 π rad \approx 4.71 rad (270°)

Since $1.5 π$ is neither a multiple of $2 π$ (in phase) nor an odd multiple of $π$ (antiphase), the two points are neither in phase nor in antiphase — they are $3/4$ of a cycle out of step.

Mark scheme · 3 marks

Calculates the path-difference-to-wavelength ratio $Δ x / λ = 0.75$ 1 mark
Calculates $Δ ϕ = 1.5 π$ rad (equivalently 4.71 rad or 270°) 1 mark
States the points are neither in phase nor in antiphase, since $1.5 π$ is not a multiple of $2 π$ or an odd multiple of $π$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.3.1 · 4 marks

A student wants to determine experimentally whether an unknown wave passing through a system is transverse or longitudinal, using only a polarising filter.

Describe the test, state the two possible outcomes, and explain the physical reason the result is conclusive.

Show worked solution

Pass the wave through a polarising filter and slowly rotate the filter about the direction of travel while observing the transmitted intensity.

If the intensity varies, falling to (near) zero at some orientation and back to a maximum a quarter turn later, the wave is transverse.

If the intensity is unaffected by rotation, the wave cannot be polarised and is longitudinal.

This is conclusive because a longitudinal wave's oscillation is already confined to the single line along which it travels, so there is no separate plane left for a filter to restrict; only a transverse wave has an oscillation direction that a polariser can select and that rotation can therefore vary.

Mark scheme · 4 marks

States the wave is passed through a polarising filter which is then rotated 1 mark
States that transmitted intensity varying (down towards zero) with rotation indicates a transverse wave 1 mark
States that transmitted intensity staying constant with rotation indicates a longitudinal wave (cannot be polarised) 1 mark
Explains a longitudinal wave's oscillation is confined to the direction of travel, leaving no plane for a polariser to restrict, unlike a transverse wave 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.3.1 · 4 marks

A string of length $1.5 m$ , fixed at both ends, is driven at a frequency that produces a stationary wave pattern with 3 antinodes visible along its length.

(a) State the harmonic number and the total number of nodes.

(b) Calculate the wavelength.

(c) State the phase relationship between the oscillation at the first antinode and the oscillation at the third antinode, with a reason.

Show worked solution

3 antinodes means this is the third harmonic ( $n = 3$ ), with 4 nodes in total once the two fixed ends are included.

λ = \frac{2 L}{n} = \frac{2 \times 1.5}{3} = 1.0 m

Adjacent antinodes (neighbouring loops) always oscillate in antiphase (180°), so the first and second antinodes are 180° out of phase, and the second and third are a further 180° out of phase.

The first and third antinodes are therefore offset by:

2 \times 180° = 360°

so they oscillate in phase with each other.

Mark scheme · 4 marks

Identifies the pattern as the third harmonic ( $n = 3$ ) with 4 nodes in total, including both fixed ends 1 mark
Uses $λ = 2 L / n$ to calculate $λ = 1.0 m$ 1 mark
States that antinodes in adjacent loops oscillate in antiphase (180°) 1 mark
Concludes the first and third antinodes, two loops apart, are offset by 360° and so oscillate in phase 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.3.1 · 6 marks

Tension F / N	Resonant frequency f₁ / Hz
1.0	28.0
2.0	39.5
3.0	48.4
4.0	55.9

In a Required Practical 1 investigation, a string of fixed length $L = 0.800 m$ is clamped horizontally and driven by a vibration generator.

The tension $F$ is varied using a hanging mass and pulley, and for each tension the frequency is tuned until the first harmonic (fundamental) stationary wave pattern is seen, giving the resonant frequency $f_{1}$ shown in the table.

Use the data to determine the mass per unit length $μ$ of the string.

Show worked solution

Rearranging:

f_{1} = \frac{1}{2 L} \frac{F}{μ}

gives:

f_{1}^{2} = \frac{F}{4 L ^{2} μ}

so a graph of $f_{1}^{2}$ against $F$ should be a straight line through the origin with gradient $\frac{1}{4 L ^{2} μ}$ .

From the data,

f_{1}^{2} = 784 Hz^{2}

at $F = 1.0 N$ and:

f_{1}^{2} = 3124.8 Hz^{2}

at $F = 4.0 N$ , giving gradient:

= \frac{3124.8 - 784}{4.0 - 1.0} \approx 780 s^{- 2} N^{- 1}

So:

μ = \frac{1}{4 L ^{2} \times gradient} = \frac{1}{4 \times 0.80 0 ^{2} \times 780} \approx 5.0 \times 1 0^{- 4} kg m^{- 1}

Mark scheme · 6 marks

Rearranges $f_{1} = \frac{1}{2 L} F / μ$ to $f_{1}^{2} = F / (4 L^{2} μ)$ , identifying that $f_{1}^{2}$ should be plotted against $F$ 1 mark
Calculates $f_{1}^{2}$ for at least two rows of data (e.g. 784 Hz² at 1.0 N, 3124.8 Hz² at 4.0 N) 1 mark
Recognises the linear relationship between $f_{1}^{2}$ and $F$ confirms the theoretical dependence 1 mark
Calculates the gradient of $f_{1}^{2}$ against $F$ as approximately 780 s⁻² N⁻¹ 1 mark
Equates the gradient to $1/ (4 L^{2} μ)$ and rearranges for $μ$ 1 mark
Calculates $μ \approx 5.0 \times 1 0^{- 4} kg m^{- 1}$ (0.50 g m⁻¹) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.3.2

Refraction, diffraction and interference

Young's double-slit interference 3.3.2.1

Definitions

Coherent waves: waves of the same frequency with a constant phase difference.

Key results

Fringe spacing: $w = \frac{λ D}{s}$ (wavelength $λ$ , slit separation $s$ , slit-to-screen distance $D$ , valid for $D ≫ s$ ).
Constructive interference (bright fringe): path difference $Δ = mλ$ . Destructive interference (dark fringe): $Δ = (m + \frac{1}{2}) λ$ , for integer $m$ .

Method

Required practical 2: measure fringe spacing $w$ for a known slit separation $s$ and slit-to-screen distance $D$ (or use a diffraction grating and measure diffraction angles), then use $w = λ D / s$ (or the grating equation) to find the wavelength of the light — always avoiding looking directly into the laser beam or its reflection.

Notes

A single laser illuminating both slits is a simple, reliable way to guarantee the two sources are coherent, since both slits then re-radiate from the same original wavefront in a fixed phase relationship.
The fringe-spacing equation is one of the most direct ways wavelength itself is measured experimentally.
With white light instead of a laser, the fringes are only clearly distinct for the first few orders (violet nearer the centre than red, since $w \propto λ$ ) before overlapping washes them out — exactly why a monochromatic (laser) source is normally used for a clean, well-defined pattern.
Complete cancellation at a dark fringe requires the two waves to have equal amplitude as well as opposite phase, and matching polarisation — an idealisation the equation itself assumes without stating.

Diffraction at a single slit 3.3.2.2

Definitions

Diffraction: the spreading of a wave as it passes through a gap or around an obstacle, most pronounced when the gap width is comparable to the wavelength.

Notes

A narrower slit produces greater spreading of the diffracted wave; a wider slit produces less spreading, for the same wavelength — the central maximum broadens as slit width decreases.
A longer wavelength also produces a broader central diffraction maximum, for the same slit width — spreading depends on the ratio of wavelength to slit width, not on either alone.
With white light, the central maximum stays white (all wavelengths coincide there), but the side maxima split into colour, since each wavelength diffracts by a different amount — higher-order fringes increasingly overlap between colours.

Diffraction grating 3.3.2.2

Key results

Diffraction grating equation: $d sin θ = mλ$ (slit spacing $d$ , angle $θ$ to the $m$ th order maximum, $m = 0, \pm 1, \pm 2, \dots$ ).
Grating spacing from lines per unit length: $d = \frac{1}{N}$ , where $N$ is the number of lines per metre.

Notes

A diffraction grating — many closely, evenly spaced slits — produces sharp, well-separated bright maxima rather than the broad fringes of a double slit, because light from every single slit must interfere constructively at once for a maximum to appear.
Because $sin θ$ cannot exceed 1, the grating equation also sets a hard limit on how many orders a given grating and wavelength can produce — orders with $∣ m ∣ λ > d$ simply cannot form.
The zero-order ( $m = 0$ ) maximum shows no dispersion at all — with white light it stays white, since every wavelength satisfies $d sin (0°) = 0$ simultaneously; non-zero orders spread each wavelength to its own angle, with red diffracting further than violet.
Diffraction gratings are the practical tool behind spectroscopy: spreading a light source's wavelengths to distinct, measurable angles resolves it into its component spectral lines — the same principle behind identifying a distant star's composition from its spectrum.

Refraction 3.3.2.3

Definitions

Refractive index: $n = \frac{c}{v}$ , comparing the speed of light in a vacuum to its speed in the medium.

Key results

Snell's law: $n_{1} sin i = n_{2} sin r$ .

Notes

Refraction occurs because a wave changes speed when it crosses into a medium of different optical density, bending its direction unless it travels exactly along the normal.
Entering a higher-index medium: speed decreases, wavelength decreases (since $v = f λ$ and frequency cannot change), and the ray bends towards the normal — frequency alone stays fixed, set entirely by the source.
Refractive index can also be found from $n = \frac{sin i}{sin r}$ directly from Snell's law when one medium is air/vacuum ( $n \approx 1$ ) — a frequently used simplification in practical measurements.

Total internal reflection 3.3.2.3

Definitions

Critical angle, $θ_{c}$ : the angle of incidence (in the denser medium) at which the refracted ray grazes exactly along the boundary ( $r = 90°$ ).

Key results

$sin θ_{c} = \frac{n _{2}}{n _{1}}$ (for light travelling from denser medium $n_{1}$ into less dense medium $n_{2} < n_{1}$ ).

Notes

Below $θ_{c}$ : the ray both refracts and partially reflects. At exactly $θ_{c}$ : the refracted ray runs along the boundary itself ( $r = 90°$ ). Above $θ_{c}$ : no refracted ray exists at all, and all the light is reflected back into the denser medium — total internal reflection.
Total internal reflection requires travelling from a denser into a less dense medium ( $n_{1} > n_{2}$ ) — it can never happen the other way round, whatever the angle of incidence.

Optical fibres 3.3.2.3

Definitions

Step-index fibre: an optical fibre with a core of higher refractive index than its surrounding cladding, so light is guided along the core by total internal reflection at the core-cladding boundary.

Notes

The cladding's role is to guarantee a reliably lower refractive index at the core boundary everywhere along the fibre, regardless of what the fibre's outer surface happens to touch — without it, total internal reflection would depend unpredictably on the external environment.
Modal dispersion: different guided rays take different zigzag paths (and so different path lengths) along the fibre, arriving at slightly different times and broadening an input pulse.
Material dispersion: different wavelengths within the source's own spectrum travel at slightly different speeds in the fibre material, which also broadens a pulse over distance.
Pulse broadening limits how closely spaced successive digital pulses can be sent before they start to overlap and become indistinguishable — the practical limit on a fibre's maximum data rate over a given length.
Both dispersion effects are reduced by using a single-mode fibre (removing modal dispersion, since only one path is possible) together with a source of narrow wavelength range (reducing material dispersion).

Worked examples

Worked example 3.3.2 · 5 marks

D / m	w / mm
1.0	1.3
2.0	2.6
3.0	3.9
4.0	5.2

In a Required Practical 2 investigation, a laser illuminates two slits of separation $s = 0.50 mm$ , and the fringe spacing $w$ on a screen is measured for several slit-to-screen distances $D$ , as shown in the table.

Use the data to determine the wavelength of the laser light, and identify the likely colour of the laser.

Show worked solution

Since:

w = \frac{λ D}{s}

$w$ is proportional to $D$ with gradient $λ / s$ .

From the data, gradient:

= \frac{( 5.2 - 1.3 ) \times 1 0 ^{- 3}}{4.0 - 1.0} = 1.3 \times 1 0^{- 3}

(converting mm to m).

So:

λ = gradient \times s = 1.3 \times 1 0^{- 3} \times 5.0 \times 1 0^{- 4} = 6.5 \times 1 0^{- 7} m = 650 nm

which corresponds to red light.

Mark scheme · 5 marks

Recognises $w = λ D / s$ predicts $w$ proportional to $D$ , with gradient equal to $λ / s$ 1 mark
Calculates the gradient of $w$ (in m) against $D$ from the data, $\approx 1.3 \times 1 0^{- 3}$ 1 mark
Substitutes $s = 5.0 \times 1 0^{- 4} m$ to write $λ = gradient \times s$ 1 mark
Calculates $λ \approx 6.5 \times 1 0^{- 7} m$ (650 nm) 1 mark
States this wavelength corresponds to red light 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.3.2 · 4 marks

A diffraction grating has 300 lines per mm.

Monochromatic light of wavelength 500 nm is incident normally on the grating.

Calculate the highest order of diffraction maximum that can be observed.

Show worked solution

d = \frac{1}{300 \times 1000} = 3.33 \times 1 0^{- 6} m

An order exists only while $sin θ \leq 1$ , i.e.

m \leq \frac{d}{λ} = \frac{3.33 \times 1 0 ^{- 6}}{500 \times 1 0 ^{- 9}} \approx 6.67

Since $m$ must be an integer, the highest observable order is $m = 6$ .

Mark scheme · 4 marks

Calculates the grating spacing $d = \frac{1}{300 \times 1000} = 3.33 \times 1 0^{- 6} m$ 1 mark
States the condition for an order to exist is $sin θ \leq 1$ , i.e. $m \leq d / λ$ 1 mark
Calculates $d / λ \approx 6.67$ 1 mark
States the highest observable order is $m = 6$ , since $m$ must be an integer 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.3.2 · 3 marks

A ray of light travelling in air is incident on a glass block of refractive index $n = 1.52$ at an angle of incidence of $35°$ to the normal.

Calculate the angle of refraction inside the glass.

Show worked solution

$n_{1} sin i = n_{2} sin r$ with $n_{1} = 1.00$ (air) and $n_{2} = 1.52$ :

sin r = \frac{sin 35°}{1.52} = \frac{0.574}{1.52} \approx 0.377

so $r \approx 22.2°$ .

Mark scheme · 3 marks

Writes Snell's law $n_{1} sin i = n_{2} sin r$ with $n_{1} = 1.00$ and $n_{2} = 1.52$ 1 mark
Calculates $sin r = sin 35°/1.52 \approx 0.377$ 1 mark
Calculates $r \approx 22.2°$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.3.2 · 4 marks

An optical fibre core has refractive index $n_{1} = 1.48$ ; the surrounding cladding has refractive index $n_{2} = 1.46$ .

(a) Calculate the critical angle at the core-cladding boundary.

(b) A ray inside the core strikes the boundary at $79°$ to the normal — state, with a reason, whether it is totally internally reflected.

(c) Explain why the fibre has a cladding layer rather than relying on total internal reflection at a bare core-air boundary.

Show worked solution

(a):

sin θ_{c} = \frac{n _{2}}{n _{1}} = \frac{1.46}{1.48} \approx 0.986

so $θ_{c} \approx 80.6°$ .

(b) Since $79° < θ_{c}$ , the ray is not totally internally reflected — it partially refracts out through the boundary and partially reflects.

(c) The cladding guarantees a reliably lower refractive index at the core boundary all along the fibre, regardless of what the fibre's outer surface happens to touch (dirt, fingers, scratches); a bare core-air interface would fail total internal reflection unpredictably wherever the surface contacted a higher-index material.

Mark scheme · 4 marks

Calculates $sin θ_{c} = n_{2} / n_{1} = 1.46/1.48 \approx 0.986$ 1 mark
Calculates $θ_{c} \approx 80.6°$ 1 mark
States the ray is not totally internally reflected because $79° < θ_{c}$ 1 mark
Explains the cladding guarantees a reliably lower refractive index at the core boundary independent of the fibre's outer environment, unlike a bare core-air interface 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.