Notes · PhysicsUK · A-Levels

Waves

A wave carries energy without transporting matter. This chapter defines the quantities that describe progressive waves, distinguishes longitudinal from transverse waves and uses polarisation as evidence for the transverse nature of light. It develops superposition into interference, stationary waves and the diffraction grating, and closes with refraction, total internal reflection and the optical fibre - the backbone of modern communication.

5 sections~15 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

T·0333 / 13

Exam profile

AO1 · Define wave quantities and describe polarisation, superposition, stationary waves, diffraction and refractionAO2 · Calculate wave speed, phase difference, fringe spacing, harmonic frequencies, grating angles, refractive indices and critical anglesAO3 · Analyse the double-slit, grating and stationary-wave experiments and evaluate the required practicals with their uncertainties

Operators:statedescribeexplaincalculateshow thatsketch

basic level

AS-Level requires progressive and stationary waves, superposition and interference, the double-slit and grating equations, and refraction with total internal reflection.

higher level

The full A-Level treats these quantitatively and synoptically - the same superposition ideas return in the diffraction of electrons and X-rays, and optical fibres link to the electromagnetic spectrum and communication.

Depth

Text

Contents · 5 sections

Waves

Progressive waves and their properties#

FoundationAQA 7408 3.3.1.1DfE GCE Physics - progressive waves

Displacement-distance graph of a progressive wave

Fig. 1A snapshot of a wave: the amplitude is the peak displacement and the wavelength is the distance between successive points in phase.

A progressive wave transfers energy from one place to another without any net movement of the medium; each particle oscillates about a fixed point and passes energy to its neighbour. The wave is described by its amplitude

A

(the maximum displacement from equilibrium), its wavelength

\lambda

(the distance between adjacent points in phase, for example crest to crest), its frequency

f

(oscillations per second, in hertz) and its period

T

(the time for one oscillation). Frequency and period are reciprocals,

T = 1/f

The wave speed is the distance one wavefront travels per second and is linked to frequency and wavelength by the fundamental wave equation

v = f\lambda

. This follows directly from the definitions: in one period the wave advances exactly one wavelength, so

v = \lambda/T = f\lambda

. For a given medium the speed is fixed, so a higher frequency necessarily means a shorter wavelength.

Phase describes how far through its cycle a point is, measured as an angle: one whole cycle is

2\pi

radians (or

360^{\circ}

). The phase difference between two points, or two waves, tells us how their oscillations line up. Points a whole wavelength apart are in phase (phase difference

2\pi

); points half a wavelength apart are in antiphase (phase difference

\pi

). Phase difference in radians is

\dfrac{2\pi\,\Delta x}{\lambda}

for a path difference

\Delta x

Waves also reflect, refract and diffract, and superpose when they overlap - the behaviours developed through this chapter. All electromagnetic waves travel at

c = 3.00\times10^{8}\,\text{m}\,\text{s}^{-1}

in a vacuum; sound, a mechanical wave, travels at about

340\,\text{m}\,\text{s}^{-1}

in air and needs a medium.

v = f\lambda

The wave equation

Wave speed equals frequency times wavelength.

\text{phase difference} = \dfrac{2\pi\,\Delta x}{\lambda}

Phase difference from path difference

A path difference of one wavelength corresponds to a phase difference of $2\pi$ .

Worked example

Wavelength and phase difference of a sound wave

A sound wave of frequency $170\,\text{Hz}$ travels at $340\,\text{m}\,\text{s}^{-1}$ . Find its wavelength and the phase difference between two points $0.50\,\text{m}$ apart.

Wavelength
Rearrange $v = f\lambda$ to $\lambda = v/f = 340/170 = 2.0\,\text{m}$ .
Path difference in wavelengths
$0.50\,\text{m}$ is $0.50/2.0 = 0.25$ of a wavelength.
Phase difference
Multiply by $2\pi$ .
$\Delta\phi = 0.25 \times 2\pi = \tfrac{\pi}{2}\,\text{rad}\;(90^{\circ})$

Result: The wavelength is $2.0\,\text{m}$ and the two points are $\tfrac{\pi}{2}\,\text{rad}$ out of phase.

Typical mistakes

Reading wavelength from a displacement-time graph (which shows the period) instead of a displacement-distance graph.
Confusing amplitude with peak-to-peak displacement, which is twice the amplitude.

Active revision

A sound wave of frequency $256\,\text{Hz}$ travels at $340\,\text{m}\,\text{s}^{-1}$ . Calculate its wavelength and the phase difference between two points $0.33\,\text{m}$ apart.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Longitudinal, transverse waves and polarisation#

StandardAQA 7408 3.3.1.2DfE GCE Physics - polarisation

In a transverse wave the oscillations are perpendicular to the direction of energy transfer - as on a string, on water, and in all electromagnetic waves. In a longitudinal wave the oscillations are parallel to the direction of energy transfer, producing compressions and rarefactions; sound is the everyday example. This difference determines which waves can be polarised.

Polarisation restricts the oscillations of a transverse wave to a single plane. Unpolarised light contains oscillations in all planes perpendicular to its travel; a polarising filter transmits only the component in one plane, so the light that emerges is plane-polarised. Passing that light through a second filter (an analyser) reduces its intensity to zero when the analyser is turned

90^{\circ}

to the first - the two filters are then crossed.

Crucially, only transverse waves can be polarised, because only they have oscillations that can be confined to a plane; a longitudinal wave oscillates along its direction of travel and has no such plane. The fact that light can be polarised is therefore direct evidence that light is a transverse wave. Sound, being longitudinal, cannot be polarised.

Polarisation has many applications: polarising sunglasses cut glare by blocking the strongly polarised light reflected from horizontal surfaces; photographers use polarising filters to darken skies; and radio and television aerials must be aligned with the plane of polarisation of the transmitted signal to receive it strongly - a vertical transmitting aerial requires a vertical receiving aerial.

\text{unpolarised} \;\xrightarrow{\text{filter}}\; \text{plane-polarised}

Effect of a polariser

A polarising filter transmits only the oscillations in one plane.

Worked example

Reasoning about crossed filters

Light passes through a polariser and then an analyser. Describe how the transmitted intensity changes as the analyser is rotated from parallel to perpendicular, and explain why this demonstrates that light is transverse.

Parallel filters
When the analyser's transmission axis is parallel to the polariser's, the plane-polarised light passes through and the intensity is a maximum.
Rotating the analyser
As the analyser turns, only the component of the polarised light along its axis is transmitted, so the intensity falls smoothly.
Crossed filters
At $90^{\circ}$ there is no component along the analyser axis, so no light passes and the intensity is zero.

Result: The intensity varies from maximum to zero; because only a transverse wave has a plane of oscillation to block, this shows light is transverse.

Typical mistakes

Claiming that sound can be polarised - it is longitudinal, so it cannot.
Saying a single polarising filter removes half the light 'because it is dimmer', without explaining that only one plane of oscillation is transmitted.

Active revision

Explain why rotating one of two crossed polarising filters through $90^{\circ}$ changes the transmitted light from zero to a maximum, and state what this shows about the nature of light.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Superposition, interference and stationary waves#

StandardAQA 7408 3.3.1.3AQA 7408 3.3.1.4DfE GCE Physics - superposition and stationary waves

Constructive superposition of two waves

Fig. 2Two waves in phase superpose: at each point the displacements add, so the resultant (solid) has the sum of the two amplitudes.

When two waves meet, the principle of superposition states that the resultant displacement at any point is the vector sum of the individual displacements. Where two waves arrive in phase they reinforce (constructive interference, giving a larger amplitude); where they arrive in antiphase they cancel (destructive interference). For a steady, observable pattern the sources must be coherent - of the same frequency and a constant phase relationship - which is why two-source interference of light needs a single source split into two, or a laser.

The condition for interference is set by the path difference. Constructive interference occurs when the path difference is a whole number of wavelengths,

n\lambda

; destructive interference when it is an odd number of half-wavelengths,

(n + \tfrac{1}{2})\lambda

. These conditions underlie every interference calculation, from the double slit to thin films.

A stationary (standing) wave forms when two identical progressive waves travel in opposite directions and superpose - typically an incident wave and its reflection. The pattern does not move: it has nodes, points of permanently zero amplitude where the two waves always cancel, and antinodes, points of maximum amplitude, spaced half a wavelength apart. Unlike a progressive wave, a stationary wave stores energy rather than transporting it, and all points between two nodes oscillate in phase.

On a string fixed at both ends, only wavelengths that fit a whole number of half-wavelengths between the ends resonate. The lowest is the fundamental (first harmonic), with a node at each end and one antinode, so

L = \lambda/2

and

f_1 = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

, where

T

is the tension and

\mu

the mass per unit length. The higher harmonics have frequencies

2f_1, 3f_1, \dots

- the basis of stringed instruments and the stationary-waves required practical.

f_n = \dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}

Harmonics on a string

$n = 1$ is the fundamental; $T$ is tension and $\mu$ is mass per unit length.

\text{constructive: } \Delta x = n\lambda, \quad \text{destructive: } \Delta x = (n + \tfrac{1}{2})\lambda

Interference conditions

Path-difference conditions for reinforcement and cancellation.

Harmonics of a stationary wave on a string

Fig. 3The first three harmonics on a string of length

L

fixed at both ends. Each fits a whole number of half-wavelengths between the fixed ends (nodes).

Worked example

Fundamental frequency of a string

A guitar string of length $0.65\,\text{m}$ has mass per unit length $4.0\times10^{-3}\,\text{kg}\,\text{m}^{-1}$ and is tuned to a tension of $70\,\text{N}$ . Find its first-harmonic frequency.

Wave speed on the string
The speed is $v = \sqrt{T/\mu} = \sqrt{70/(4.0\times10^{-3})} = \sqrt{1.75\times10^{4}} = 132\,\text{m}\,\text{s}^{-1}$ .
Fundamental wavelength
For the first harmonic $L = \lambda/2$ , so $\lambda = 2L = 1.30\,\text{m}$ .
Frequency
$f_1 = v/\lambda$ .
$f_1 = \dfrac{132}{1.30} = 102\,\text{Hz}$

Result: The first harmonic is about $1.0\times10^{2}\,\text{Hz}$ .

Typical mistakes

Thinking a stationary wave transfers energy along the string - it stores energy; nodes never move.
Forgetting that adjacent nodes (or adjacent antinodes) are half a wavelength apart, not a whole wavelength.

Active revision

A string of length $0.80\,\text{m}$ and mass per unit length $1.5\times10^{-3}\,\text{kg}\,\text{m}^{-1}$ is under a tension of $60\,\text{N}$ . Calculate the frequency of its first harmonic.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Two-source interference, double slits and diffraction gratings#

AdvancedAQA 7408 3.3.2.1AQA 7408 3.3.2.2DfE GCE Physics - interference and diffraction

Young's double-slit apparatus

Fig. 4Monochromatic light passes through two coherent slits separated by

s

; overlapping beams form fringes of spacing

w

on a screen a distance

D

away.

Young's double-slit experiment sends monochromatic light through two narrow, close slits, which act as coherent sources. The overlapping light produces a pattern of equally spaced bright and dark fringes on a distant screen. The bright fringes occur where the path difference from the two slits is a whole number of wavelengths. The fringe spacing is

w = \dfrac{\lambda D}{s}

, where

s

is the slit separation and

D

the slit-to-screen distance, so a larger wavelength or screen distance, or a smaller slit separation, widens the fringes.

The experiment is historically decisive: only a wave model can explain interference fringes, so Young's result established the wave nature of light. Using it in reverse, measuring

w

s

and

D

gives the wavelength of light - a standard required practical. With white light the central fringe is white and the others are spectra, because each colour has its own fringe spacing.

Diffraction is the spreading of a wave as it passes through a gap or around an obstacle, and it is most pronounced when the gap is comparable to the wavelength. A single slit gives a broad central maximum, twice as wide as the others and much brighter, with dimmer secondary maxima either side. Narrowing the slit widens the central maximum; using a longer wavelength does the same.

A diffraction grating has thousands of slits per millimetre and gives very sharp, widely separated maxima, so it measures wavelength far more precisely than two slits. Its maxima obey

d\sin\theta = n\lambda

, where

d

is the grating spacing (the reciprocal of the number of lines per metre) and

n

the order. Because

\sin\theta \le 1

, only a finite number of orders exists, and gratings are the heart of the spectrometer used to analyse the light from stars and flames.

w = \dfrac{\lambda D}{s}

Double-slit fringe spacing

Fringe spacing from wavelength, slit-to-screen distance and slit separation.

d\sin\theta = n\lambda

Diffraction grating

$d$ is the grating spacing and $n$ the order of the maximum.

Intensity of the double-slit fringes

Fig. 5The interference term gives equally spaced bright fringes (maxima) separated by dark fringes (minima) across the screen.

Worked example

Grating orders

Monochromatic light of wavelength $600\,\text{nm}$ is incident normally on a diffraction grating with $500\,\text{lines mm}^{-1}$ . Find the angle of the second-order maximum and the maximum order observable.

Grating spacing
$500\,\text{lines mm}^{-1} = 5.00\times10^{5}\,\text{lines m}^{-1}$ , so $d = 1/(5.00\times10^{5}) = 2.00\times10^{-6}\,\text{m}$ .
Second order angle
$\sin\theta = n\lambda/d = (2 \times 600\times10^{-9})/(2.00\times10^{-6}) = 0.600$ , so $\theta = 36.9^{\circ}$ .
Maximum order
Set $\sin\theta = 1$ : $n_{\max} = d/\lambda = 2.00\times10^{-6}/600\times10^{-9} = 3.33$ .
$n_{\max} = \left\lfloor \dfrac{d}{\lambda} \right\rfloor = 3$

Result: The second order is at $36.9^{\circ}$ ; the highest observable order is the third.

Typical mistakes

Using the number of lines per millimetre directly as $d$ - convert to lines per metre and take the reciprocal for the slit spacing.
Forgetting that the central maximum in single-slit diffraction is twice as wide as the others.

Active revision

Light of wavelength $590\,\text{nm}$ falls on a grating with $300\,\text{lines mm}^{-1}$ . Calculate the angle of the first-order maximum and the highest order that can be seen.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Refraction, total internal reflection and optical fibres#

AdvancedAQA 7408 3.3.2.3DfE GCE Physics - refraction and optical fibres

Refraction at a boundary

Fig. 6Light passing from a less dense to a more dense medium bends towards the normal; the angles are measured from the normal and obey Snell's law.

When light crosses a boundary between two transparent media its speed changes, and unless it hits the boundary head-on it changes direction - it refracts. The refractive index of a medium is

n = c/c_s

, the ratio of the speed of light in a vacuum to its speed in the medium, and is always at least 1. At a boundary the directions obey Snell's law,

n_1\sin\theta_1 = n_2\sin\theta_2

, where the angles are measured from the normal. Light entering a denser (higher-

n

) medium slows and bends towards the normal.

When light travels from a denser to a less dense medium (for example glass to air) it bends away from the normal, and beyond a certain angle of incidence it cannot escape at all - it is totally internally reflected. The critical angle

\theta_c

is the angle of incidence for which the refracted ray grazes the boundary at

90^{\circ}

, given by

\sin\theta_c = n_2/n_1

(with

n_1 > n_2

). At angles greater than

\theta_c

, all the light reflects back into the denser medium.

Total internal reflection makes the step-index optical fibre possible. A thin, highly transparent core of high refractive index is surrounded by a cladding of slightly lower index; light launched into the core strikes the core-cladding boundary above the critical angle and is repeatedly totally internally reflected, guided along the fibre with very little loss over many kilometres. The cladding also protects the core and prevents light leaking between touching fibres.

Two effects broaden a light pulse and limit the data rate. Modal dispersion arises because rays taking different zig-zag paths travel different distances and arrive at slightly different times; material dispersion arises because different wavelengths travel at slightly different speeds. Both smear a sharp pulse, so real systems use monochromatic laser light and a narrow core to keep pulses distinct - the physics behind high-speed broadband and undersea communication.

n = \dfrac{c}{c_s}

Refractive index

The ratio of the speed of light in a vacuum to that in the medium.

n_1\sin\theta_1 = n_2\sin\theta_2

Snell's law

Relates the angles of incidence and refraction, measured from the normal.

\sin\theta_c = \dfrac{n_2}{n_1}\quad (n_1 > n_2)

Critical angle

Beyond this angle of incidence, light is totally internally reflected.

Worked example

Critical angle of a glass fibre core

An optical-fibre core has refractive index $1.50$ and its cladding has refractive index $1.45$ . Calculate the critical angle at the core-cladding boundary.

Identify the media
Light travels in the denser core ( $n_1 = 1.50$ ) towards the less dense cladding ( $n_2 = 1.45$ ).
Apply the critical-angle formula
$\sin\theta_c = n_2/n_1 = 1.45/1.50 = 0.9667$ .
$\theta_c = \sin^{-1}(0.9667) = 75.2^{\circ}$

Result: The critical angle is $75.2^{\circ}$ ; rays striking the boundary above this angle are guided along the fibre.

Typical mistakes

Measuring angles from the surface rather than from the normal.
Using $\sin\theta_c = n_2/n_1$ with the indices the wrong way round - the larger index must be $n_1$ , the denser medium the light starts in.

Active revision

Light travels from glass ( $n = 1.52$ ) into air. Calculate the critical angle, and find the angle of refraction when light enters the glass from air at an angle of incidence of $40^{\circ}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Waves

5 sections~15 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

Waves

Key points

Wavelength and phase difference of a sound wave

Key points

Reasoning about crossed filters

Key points

Fundamental frequency of a string

Key points

Grating orders

Key points

Critical angle of a glass fibre core

Sources

Waves

Key points

Wavelength and phase difference of a sound wave

Key points

Reasoning about crossed filters

Key points

Fundamental frequency of a string

Key points

Grating orders

Key points

Critical angle of a glass fibre core

Sources