3. Waves

Exam analysis

Most tested topics

3.1 General properties of waves

A wave transfers energy from one place to another without transferring matter from the source to the destination.

The moving feature is a disturbance. In a material medium, each small part of the medium is displaced as the disturbance reaches it, then moves back towards its equilibrium position. The pattern advances, but the particles do not travel along with the wave.

A pulse sent along a rope can make a distant end move. A marked point on the rope moves briefly about its original position; it does not travel to the distant end. The pulse has carried energy through the rope without carrying that marked piece of rope with it.

Do not confuse the direction in which the wave travels with the motion of the medium. Energy follows the travelling disturbance; matter only oscillates locally when a material medium is involved.

Wave motion is the travel of a disturbance through a system while different parts of that system vibrate in sequence.

Demonstration	What is made to vibrate	What travels
flick one end of a stretched rope	each section moves across the rope's length and returns	a pulse travels along the rope
push and pull one end of a stretched spring	coils move to and fro, making compressed and spread-out regions	the disturbance travels along the spring
disturb water in a ripple tank	points on the surface move about their undisturbed positions	wavefronts travel across the surface

Watch one marked point as well as the overall pattern. The point begins moving only when the disturbance reaches it, and the next points then respond. This delayed sequence is evidence that the wave is propagating.

A travelling wave pattern does not mean the rope, spring or water is flowing from source to receiver. The examples differ in vibration direction, but each illustrates local vibration plus a travelling disturbance.

Wave features describe the size, spacing, timing and motion of a repeating disturbance.

Feature	Precise meaning
crest (peak)	highest point of a transverse wave
trough	lowest point of a transverse wave
amplitude	maximum displacement from the equilibrium position
wavelength, $\lambda$	shortest distance between two points in the same phase, such as crest to next crest
frequency, $f$	number of complete waves passing a point each second, measured in hertz (Hz)
wave speed, $v$	distance travelled by the wave per unit time, measured in $\text{m}/\text{s}$
wavefront	a line joining points on a wave that are in the same phase; adjacent crest lines are one wavelength apart

Measure amplitude from the equilibrium line to a crest or trough, not from crest to trough. Measure wavelength between matching points on consecutive cycles, not between a crest and the nearest trough.

Frequency describes how often cycles pass; speed describes how fast the disturbance moves. They are different quantities even though both affect how wave cycles are spaced.

Wave speed equals the number of complete waves produced each second multiplied by the distance occupied by one complete wave.

$v=f\lambda$

Symbol	Meaning	SI unit
$v$	wave speed	$\text{m}/\text{s}$
$f$	frequency	$\text{Hz}$
$\lambda$	wavelength	$\text{m}$

Use $v=f\lambda$ , $f=v/\lambda$ or $\lambda=v/f$ . Convert kilohertz to hertz and centimetres to metres before substituting.

For sound with $f=2.0\,\text{kHz}=2000\,\text{Hz}$ and $v=800\,\text{m}/\text{s}$ : $\lambda=v/f=800/2000=0.40\,\text{m}$ . The unit check is $(\text{m}/\text{s})/(1/\text{s})=\text{m}$ .

Use the wavelength in the same medium as the stated speed. Do not substitute $2.0$ for $2.0\,\text{kHz}$ without converting to $2000\,\text{Hz}$ .

In a transverse wave, the direction of vibration is at right angles to the direction in which the wave propagates.

Quantity	Direction in a transverse model
wave propagation and energy transfer	along the wave's travel direction
vibration	perpendicular to the travel direction

Electromagnetic radiation, water waves and seismic S-waves (secondary waves) can be modelled as transverse. For a rope pulse travelling horizontally, a marked point may vibrate vertically; those two directions are perpendicular.

A transverse wave is classified by the two directions, not by whether its drawn trace has crests and troughs. A graph can look wavy even when it represents a different quantity, so always identify what is vibrating and where the wave travels.

In a longitudinal wave, the direction of vibration is parallel to the direction in which the wave propagates.

Quantity	Direction in a longitudinal model
wave propagation and energy transfer	along the wave's travel direction
vibration	backwards and forwards along that same line

The vibration produces alternating compressions, where particles are closer together, and rarefactions, where particles are farther apart. These regions travel even though individual particles only oscillate about their equilibrium positions.

Sound waves and seismic P-waves (primary waves) can be modelled as longitudinal.

Longitudinal does not mean that particles travel all the way from source to receiver. It means their local vibration is parallel to the propagation direction.

Reflection, refraction and diffraction are different changes to a travelling wave, identified by the boundary or opening it meets.

Process	Situation	What happens
reflection	wave reaches a plane surface	the wave returns into the original region in a changed direction
refraction	wave enters a region where its speed changes	wavelength changes and the wave changes direction unless it meets the boundary normally
diffraction	wave passes through a narrow gap	wavefronts spread into the region beyond the gap

The incident wave does not become a new kind of wave. These names describe what its propagation does at a surface, speed-changing boundary or gap.

A direction change alone is not enough to name refraction: it must be caused by a change of wave speed. Diffraction is spreading through a gap, not a speed change at a boundary.

A ripple tank contains shallow water. A vibrating straight bar produces regular plane wavefronts; a lamp or stroboscope makes their positions visible so the incident and resulting wavefronts can be compared.

Behaviour to show	Tank arrangement	Observation
reflection	place a straight barrier in the water	wavefronts return from the plane surface with unchanged spacing
refraction	place a flat transparent sheet under part of the water so that region is shallower; send wavefronts across the boundary at an angle	waves slow in shallow water, their spacing decreases and their direction changes
diffraction through a gap	leave a narrow opening between two barriers	wavefronts spread beyond the opening and become curved
diffraction at an edge	place one barrier so wavefronts pass its end	wavefronts curve into the region behind the edge

Keep the wave generator steady while comparing the wavefront pattern before and after each obstacle or depth boundary. Direction of travel is perpendicular to the wavefronts, and wavelength is read from their spacing.

The transparent sheet changes water depth; it is not a barrier that simply reflects the waves. For diffraction, observe spreading behind the gap or edge rather than calling every curved line refraction.

The amount of diffraction through a gap depends on wavelength compared with gap width, not on either size considered alone.

Wavelength compared with gap width	Pattern beyond the gap
wavelength similar to gap width	strong spreading; wavefronts are strongly curved
wavelength much smaller than gap width	weak spreading; the central wavefronts remain nearly straight

For a fixed gap, increasing wavelength increases diffraction. For a fixed wavelength, decreasing gap width increases diffraction. Both changes increase the ratio $\lambda/\text{gap width}$ .

Passing through the gap does not by itself change frequency, speed or wavelength. The spacing of the outgoing wavefronts stays the same as the incident spacing when the medium is unchanged; only their spread changes.

Amplitude does not determine the amount of diffraction. Compare wavelength with gap width before deciding which pattern spreads most.

When a wave passes an edge, it diffracts into the geometrical shadow region; a longer wavelength produces more spreading around the edge.

Same edge, different wavelength	Diffraction
longer wavelength	wavefronts curve farther behind the edge
shorter wavelength	less spreading; a sharper shadow region remains

Low-frequency sound has a longer wavelength than high-frequency sound when both travel at the same speed. It therefore diffracts more around a building or hill, so the low-frequency sound can be heard more clearly out of the direct line of sight.

This is diffraction around an edge, not transmission through the obstacle. Source loudness or wave amplitude can affect how detectable a signal is, but it does not set the wavelength-dependent amount of spreading.

3.2.1 Reflection of light

At the point where a light ray meets a reflecting surface, the normal is an imaginary line drawn at right angles to the surface.

Term	Meaning
incident ray	ray travelling towards the mirror
reflected ray	ray travelling away from the mirror after reflection
angle of incidence, $i$	angle between the incident ray and the normal
angle of reflection, $r$	angle between the reflected ray and the normal

Draw the normal through the exact point of incidence. Place the centre of a protractor there and measure each angle from the normal, on the appropriate side.

If the angle between an incident ray and the mirror surface is $35^\circ$ , the angle of incidence is $90^\circ-35^\circ=55^\circ$ because the normal is perpendicular to the mirror.

Angles of incidence and reflection are not measured from the mirror surface. A labelled angle touching the surface is the complement of the corresponding angle to the normal.

A plane mirror forms a virtual image behind the mirror where the reflected rays appear to come from.

Light from each point on the object reflects from the mirror and enters the observer's eye. The reflected rays do not actually meet behind the mirror, but the eye traces their straight-line paths backwards. The backward extensions meet at the corresponding image point.

Property	Plane-mirror image
size	same size as the object
position	same perpendicular distance behind the mirror as the object is in front
type	virtual: the light rays do not pass through the image position

If an object is $30\,\text{cm}$ in front of the mirror, its image is $30\,\text{cm}$ behind it, so object and image are $60\,\text{cm}$ apart.

The image is not located on the mirror surface and cannot be projected onto a screen at its apparent position. It is seen because reflected light reaches the eye, not because light travels out from behind the mirror.

For reflection at a plane mirror, the angle of incidence equals the angle of reflection when both are measured from the normal.

$i=r$

First convert any angle stated from the mirror surface into an angle from the normal. Then set $r=i$ . Keep the incident and reflected rays on opposite sides of the normal, with arrowheads showing travel towards and away from the mirror.

A ray makes $35^\circ$ with the mirror surface. Its angle of incidence is $i=90^\circ-35^\circ=55^\circ$ , so its angle of reflection is also $r=55^\circ$ .

The law does not say that a ray leaves at the same angle to the mirror surface unless those surface angles are first recognised as complements of $i$ and $r$ .

Plane-mirror constructions use straight rays, a perpendicular normal and $i=r$ to locate a reflected path or virtual image accurately.

Construction job	Ordered method
complete a reflected ray	1. draw the normal at the point of incidence; 2. measure $i$ ; 3. mark the same angle $r$ on the other side; 4. draw the reflected ray with an arrow away from the mirror
locate a point image	1. draw two rays from the object to different mirror points; 2. reflect each using $i=r$ ; 3. extend the reflected rays backwards with dashed lines; 4. mark their intersection behind the mirror
find a ray seen by an eye	1. place the image the same perpendicular distance behind the mirror; 2. join image to eye; 3. where this line crosses the mirror is the reflection point; 4. join object to that point
orient a mirror for two known ray directions	from the reflection point, use the branch back towards the source and the branch towards the destination; bisect the angle between these two outward branches to obtain the normal, then draw the mirror perpendicular to it

Use a ruler for every ray and backward extension, a sharp pencil, and a protractor centred on the point of incidence. Preserve the arrow direction: object to mirror to observer.

A completed plane-mirror image must be the same perpendicular distance behind the mirror as the object is in front. Backward extensions are construction lines, not real light paths.

3.2.2 Refraction of light

At the point where a ray meets a boundary, the normal is an imaginary line drawn at right angles to the boundary.

Term	Meaning
incident ray	ray travelling towards the boundary
refracted ray	transmitted ray travelling in the second region
angle of incidence, $i$	angle between the incident ray and the normal
angle of refraction, $r$	angle between the refracted ray and the normal

Draw the normal through the exact point of incidence and place the centre of a protractor there. Measure $i$ in the first region and $r$ in the second region, both from the normal.

If the incident ray makes $50^\circ$ with the boundary, then $i=90^\circ-50^\circ=40^\circ$ .

An angle drawn against the boundary is not $i$ or $r$ ; it is the complement of the corresponding angle to the normal.

A narrow ray and a traced transparent block make the change of direction at each boundary measurable.

Step	Method
1	place a rectangular, triangular or semicircular transparent block on plain paper and trace its outline
2	direct one narrow ray from a ray box at a chosen face and mark two points on the incident path and two on the emergent path
3	remove the block, join the marked points with a ruler and reconstruct the path through the outline
4	draw a normal at every boundary crossing and measure the angles from each normal
5	repeat with a different incidence angle and with blocks of different shapes

The ray bends towards the normal when it enters a region in which it travels more slowly and away from the normal when it enters one in which it travels faster. A ray entering along the normal does not change direction. A parallel-sided block gives an emergent ray parallel to the incident ray but shifted sideways; a non-parallel face can change the final direction.

Changing the block shape changes the orientation of its boundaries; it does not by itself change the refractive behaviour of the material. Keep the ray narrow and mark points far enough apart for accurate ruler lines.

Refraction is a change in wave direction caused by a change in wave speed at a boundary between two regions.

Passage across the boundary	Ray direction	Other wave properties
into a region where light travels more slowly	bends towards the normal, so $r<i$	frequency stays constant; wavelength decreases
into a region where light travels faster	bends away from the normal, so $r>i$	frequency stays constant; wavelength increases
along the normal, $i=0^\circ$	does not bend	speed and wavelength may still change

At the first face of a parallel-sided block the ray bends towards the normal; at the second, parallel face it bends away by the matching amount. The emerging ray is parallel to the incident ray but laterally displaced.

The ray bends at the boundary, not gradually throughout a uniform material. Frequency is fixed by the source and does not change when the ray crosses the boundary.

The critical angle $c$ is the angle of incidence in the region where light travels more slowly for which the refracted ray in the faster region is at $90^\circ$ to the normal.

At $i=c$ , the refracted ray travels along the boundary. The critical angle is measured from the normal inside the slower region, not from the boundary and not in the faster region.

Incident angle in the slower region	What happens at the boundary
$i<c$	a transmitted ray refracts into the faster region, with some reflection possible
$i=c$	the transmitted ray travels along the boundary
$i>c$	no transmitted ray emerges; total internal reflection occurs

The ray at the critical angle is not an example of total internal reflection because a transmitted ray still exists along the boundary.

Internal reflection is light reflected back into its original transparent region; total internal reflection (TIR) is the limiting case in which all the light is reflected and no refracted ray crosses the boundary.

TIR requires both conditions: light approaches a boundary from a region where it travels more slowly into one where it travels faster, and its angle of incidence is greater than the critical angle.

Experimental move	Observation
send a narrow ray into the curved face of a semicircular block along a radius	it reaches the flat face without bending at the curved face
increase $i$ at the flat block–air boundary	the refracted ray moves farther from the normal and the internally reflected ray remains
set $i=c$	the refracted ray runs along the flat surface
increase to $i>c$	the refracted ray disappears and only the internally reflected ray remains

A swimmer can see the water surface act like a mirror for rays striking it at sufficiently large angles. Right-angle glass prisms can also turn a ray through $90^\circ$ by TIR, as used in optical viewing devices.

A large incidence angle alone is not enough: light approaching from the faster region cannot undergo TIR at that boundary.

For a wave passing from region 1 into region 2, the relative refractive index is the ratio of its speed before the boundary to its speed after the boundary.

$n=\frac{v_1}{v_2}$

$v_1$ is the wave speed in the incident region and $v_2$ is its speed in the refracting region. For light travelling from air into a material, $v_1\approx3.0\times10^8\,\text{m s}^{-1}$ and $n\approx c/v$ .

If light travels at $3.0\times10^8\,\text{m s}^{-1}$ in air and $2.0\times10^8\,\text{m s}^{-1}$ in glass, then $n=(3.0\times10^8)/(2.0\times10^8)=1.5$ .

Refractive index has no unit because it is a ratio of two speeds with the same unit. Reversing the direction swaps the speed ratio, so identify regions 1 and 2 before substituting.

For light travelling from air into a transparent material, refractive index connects the two angles measured from the normal.

$n=\frac{\sin i}{\sin r}$

$i$ is the angle in air and $r$ is the angle in the material for this form of the equation. Check that both are measured from the normal, and put the calculator in degree mode.

For $i=46^\circ$ and $r=26^\circ$ , $n=\sin46^\circ/\sin26^\circ=1.64$ (3 s.f.). To find an unknown angle, first isolate its sine and then apply $\sin^{-1}$ .

Do not replace $\sin i/\sin r$ with $i/r$ . If the ray travels from the material into air, the angle in air still belongs in the numerator when calculating the material's refractive index.

For light at a material–air boundary, the material's refractive index $n$ and its critical angle $c$ are related because the refracted angle is $90^\circ$ at the critical condition.

$n=\frac{1}{\sin c}$

To find the critical angle, rearrange to $\sin c=1/n$ , so $c=\sin^{-1}(1/n)$ . Use degree mode and give the angle with appropriate precision.

For glass with $n=1.56$ , $c=\sin^{-1}(1/1.56)=39.9^\circ$ . At exactly this incidence angle, the refracted ray travels along the glass–air boundary.

This equation is for the critical boundary between the material and air. TIR occurs only for incidence from the material side with $i>c$ , not at $i=c$ .

An optical fibre carries information as pulses of visible or infrared light along a transparent core.

The core has a higher refractive index than the surrounding cladding. A suitably launched ray meets the core–cladding boundary from the slower core at an incidence angle greater than the critical angle, so repeated total internal reflection keeps it inside the core.

Stage	Role in telecommunications
transmitter	converts information into a timed pattern of light pulses
fibre	guides the pulses over distance by repeated TIR
detector	converts the received pulse pattern into an electrical signal that can be decoded

Optical fibres can carry large amounts of data at high rates with little signal loss, so they are used for internet, telephone and cable-television links.

The fibre does not trap every possible ray. If a ray reaches the core boundary below the required incidence angle, some light escapes into the cladding and the transmitted signal weakens.

3.2.3 Thin lenses

A thin converging lens bends a parallel beam towards the principal axis so the rays meet at the principal focus. A thin diverging lens bends the beam away from the principal axis.

Lens	Shape	Action on a parallel beam	Principal focus
converging	thicker at the centre	rays converge after the lens	real focus on the far side
diverging	thinner at the centre	rays spread out after the lens	virtual focus on the incident side, found by extending the rays backwards

A lens that changes the ray directions more strongly has a shorter focal length. A thinner converging lens is usually weaker and therefore has a longer focal length than a thicker one made from the same material.

Only rays incident parallel to the principal axis pass through, or appear to come from, the principal focus. An arbitrary ray is not forced through that point.

The principal axis is the straight line through the optical centre of the lens and perpendicular to the lens plane.

Term	Meaning for a thin converging lens
principal focus, $F$	point on the principal axis where rays initially parallel to the axis meet after refraction
focal length, $f$	distance from the optical centre of the lens to a principal focus
two principal focuses	one focus lies on each side of a thin lens, the same distance from its optical centre

To mark the focuses for a lens of focal length $f$ , measure $f$ along the principal axis from the optical centre on both sides. The focal length is a lens-to-focus distance, not an object-to-image distance.

A focus is a point, whereas focal length is a distance. The principal axis passes through the optical centre; it is not any convenient horizontal line in a sketch.

A converging lens forms a real image when the object is farther from the lens than one focal length. The refracted rays actually meet on the opposite side of the lens.

Start at the top of the object	Continue after the lens
ray parallel to the principal axis	through the far principal focus
ray through the optical centre	straight on without changing direction in the thin-lens model
ray through the near principal focus	parallel to the principal axis

Draw any two standard rays accurately. Their actual intersection fixes the top of the image; draw the image arrow from the principal axis to that point. Use the completed geometry to measure image position, size or focal length when the diagram is to scale.

Do not extend refracted rays backwards for a real image. The solid outgoing rays themselves must cross, and a screen placed at that crossing can receive a sharp image.

Describe a lens image with three independent comparisons: size, orientation and type.

Comparison	Allowed terms	Test
size	enlarged, same size, diminished	compare image height with object height
orientation	upright, inverted	compare which way the image points
type	real, virtual	decide whether actual rays meet and whether the image can be projected

Object position for a converging lens	Image description
beyond $2f$	diminished, inverted, real
at $2f$	same size, inverted, real
between $f$ and $2f$	enlarged, inverted, real
inside $f$	enlarged, upright, virtual

Enlarged does not imply virtual: an object between $f$ and $2f$ gives an enlarged real image. Real images made by one converging lens are inverted; the magnifying-glass image is upright and virtual.

A real image forms where light rays actually converge. It can be received as a visible projection on a screen placed at the image position.

A virtual image forms where diverging rays appear to come from when their paths are extrapolated backwards. No light rays actually pass through that apparent image position, so it cannot be projected onto a screen.

Diagram evidence	Image type
solid rays meet at the image	real
solid rays diverge but dashed backward extensions meet	virtual
a sharp image appears on a screen	real
image is seen only by looking into the outgoing rays	virtual

Virtual does not mean invisible: an eye can see a virtual image because light enters the eye along directions that appear to originate from it. The missing property is screen projection, not visibility to an observer.

Place the object between a converging lens and its near principal focus. The emerging rays diverge, so the image is virtual, upright and enlarged on the same side of the lens as the object.

Start at the top of the object	Continue after the lens
ray parallel to the principal axis	through the far focus
ray through the optical centre	straight on
ray directed towards the near focus before the lens	parallel to the principal axis

Draw two outgoing rays with solid lines. Because they spread apart, extend them backwards behind the object with dashed lines. Their backward intersection gives the top of the virtual image; draw an upright image arrow to the principal axis.

The dashed extensions show apparent origin only and are not paths travelled by light. If the object is beyond the focal point, the construction changes to an actual, inverted real-image intersection on the far side.

A magnifying glass is a single converging lens with the object placed between the lens and its principal focus. The observer looks through the lens from the opposite side.

The lens sends diverging rays into the eye. The eye traces those rays backwards to a larger upright image on the same side of the lens as the object, so the image is enlarged, upright and virtual.

Adjustment	Result
keep object distance less than $f$	maintains a virtual upright image
move lens or object while looking through it	finds a clear enlarged view
place a screen at the apparent image	no sharp projection forms because the image is virtual

An object at the principal focus gives emerging parallel rays and an image effectively at infinity; it is not the only possible magnifying-glass position. The working object position is inside one focal length.

A correcting spectacle lens changes the vergence of incoming light before it reaches the eye so that the eye lens focuses the final image on the retina.

Vision defect	Uncorrected focus for a distant or near target	Correcting lens	Action
short-sightedness	in front of the retina, especially for distant objects	diverging lens	spreads incoming rays so the eye's converging system focuses them farther back, on the retina
long-sightedness	behind the retina, especially for near objects	converging lens	pre-converges incoming rays so the eye focuses them sooner, on the retina

A diverging spectacle lens is thinner at the centre; a converging spectacle lens is thicker at the centre. In either correction, the required final condition is a sharp focus on the retina.

Do not match the lens name to the defect by shape alone. First locate the uncorrected focus: in front of the retina needs divergence; behind the retina needs extra convergence.

3.2.4 Dispersion of light

Dispersion is the separation of white light into its component colours because different frequencies are refracted by different amounts.

At the first air–glass face of a prism, each colour slows and bends towards its normal by a different amount. At the second glass–air face, each bends away from its normal. Because the faces are not parallel, the differences increase and the emerging rays spread into a spectrum.

Colour	Speed in glass	Refraction and total deviation through the prism
red	greatest among visible colours	least
violet	smallest among visible colours	greatest

A prism does not create colours or change one colour into another. White light already contains the visible frequencies; the prism separates them. A single-frequency ray refracts but does not split into a spectrum.

The traditional visible-spectrum sequence from lowest frequency to highest frequency is red, orange, yellow, green, blue, indigo, violet.

Required order	Seven colours
increasing frequency	red → orange → yellow → green → blue → indigo → violet
increasing wavelength	violet → indigo → blue → green → yellow → orange → red

In a vacuum, and approximately in air, every visible colour has the same speed. Since wave speed equals frequency multiplied by wavelength, higher frequency then means shorter wavelength, so the two colour orders run in opposite directions.

State the direction requested before writing the list. Red has the longest visible wavelength and lowest visible frequency; violet has the shortest visible wavelength and highest visible frequency.

Monochromatic visible light has a single frequency.

In one specified medium, a single frequency also corresponds to a single wavelength. Frequency is the safest defining quantity because it remains unchanged when light crosses a boundary, while wavelength changes with speed.

Light	Frequency content	Passage through a glass prism
monochromatic light	one frequency	refracts as one ray; no colour spectrum is formed
white light	a range of visible frequencies	refracts by different amounts and disperses into a spectrum

Monochromatic does not mean one amplitude, one direction or no refraction. It specifies frequency content only; a monochromatic ray can still reflect and refract.

3.3 Electromagnetic spectrum

All electromagnetic waves are transverse waves. They form one continuous spectrum divided into named regions by frequency and wavelength.

Direction	Regions
increasing frequency	radio waves → microwaves → infrared → visible light → ultraviolet → X-rays → gamma rays
increasing wavelength	gamma rays → X-rays → ultraviolet → visible light → infrared → microwaves → radio waves

In a vacuum, wave speed is common to every region and equals frequency multiplied by wavelength. Therefore higher frequency means shorter wavelength.

The spectrum order concerns frequency or wavelength, not wave speed. Radio waves have the longest wavelengths and lowest frequencies; gamma rays have the shortest wavelengths and highest frequencies.

Every region of the electromagnetic spectrum travels at the same high speed in a vacuum.

Radio waves, infrared, visible light, ultraviolet, X-rays and gamma rays sent together through the same vacuum cover the same distance in the same time, even though their frequencies and wavelengths differ.

Property across EM regions in vacuum	Same or different?
speed	same
frequency	different
wavelength	different and inversely ordered to frequency

Do not transfer this rule to sound or ultrasound: mechanical waves require a medium and travel far more slowly. EM waves can travel through empty space.

A region is chosen because its interaction with matter, transmission through materials or atmosphere, detectability and energy suit the task.

Region	Typical uses required here
radio waves	radio and television transmission; astronomy; RFID
microwaves	satellite television; mobile phones; microwave ovens
infrared	electric grills; television remotes; intruder alarms; thermal imaging; optical fibres
visible light	vision; photography; illumination
ultraviolet	security marking; detecting fake banknotes; sterilising water
X-rays	medical scanning; security scanners
gamma rays	sterilising food and medical equipment; detecting and treating cancer

Use the exact region–application pair before explaining it: for example, infrared reveals temperature patterns in thermal imaging, X-rays penetrate soft tissue more than bone, and gamma radiation can destroy living cells and microorganisms.

Do not swap satellite television to radio waves or cancer treatment to X-rays in this syllabus mapping: satellite links use microwaves, while gamma rays are specified for cancer detection/treatment and sterilisation.

Harm depends on absorbed energy and dose: greater intensity, longer exposure or repeated exposure increases risk.

Radiation	Harm from excessive exposure
microwaves	internal heating of body cells
infrared	skin burns
ultraviolet	damage to surface cells and eyes; skin cancer and eye conditions
X-rays and gamma rays	mutation or damage to cells

Risk is reduced by limiting exposure time, increasing distance where appropriate, using shielding and avoiding unnecessary exposure. Medical use balances a controlled dose against its benefit.

A useful application does not make radiation harmless. The specified hazard follows the radiation and absorbed dose, not whether the source is labelled medical, domestic or industrial.

Communication with artificial satellites is mainly by microwaves, which can pass through the atmosphere to and from an aerial or dish.

Satellite arrangement	Syllabus communication example	Key feature
low orbit	some satellite phones	satellite is relatively close but moves across the sky; a network or handover can maintain coverage
geostationary orbit	some satellite phones and direct-broadcast satellite television	satellite stays above the same point on the equator, so a fixed dish can keep pointing at it

A ground transmitter sends a microwave uplink to the satellite; the satellite relays a downlink towards the receiving region. The journey still takes time because the signal covers a large distance.

Geostationary does not mean stationary in space: the satellite orbits once per day in the same direction as Earth rotates, so it appears fixed from the ground.

The speed of every electromagnetic wave in a vacuum is 3.0 × 10^8 m/s. Its speed in air is approximately the same.

Equivalent form	Value
metres per second	300 000 000 m/s
kilometres per second	300 000 km/s

For one-way travel, distance = speed × time. For a reflected signal that returns to its starting point, the measured time covers twice the one-way distance, so distance to the reflector = speed × total time ÷ 2.

Do not use 340 m/s: that is approximately the speed of sound in air. Also keep distance units consistent with m/s before multiplying or dividing.

System	Radiation	Why it suits the system
mobile phones and wireless internet	microwaves	penetrate some walls and require only short aerials for transmission and reception
Bluetooth	radio waves	pass through walls, although the signal weakens as it travels
optical-fibre cable television and high-speed broadband	visible light or short-wavelength infrared	glass transmits these waves and they can carry high data rates

A communication system encodes information at a transmitter, sends it on an electromagnetic carrier through air or glass, and detects and decodes it at the receiver.

Match all three parts of an explanation: named system, correct EM region, and the specific transmission or aerial property that makes it useful.

Optical fibre does not use microwaves in this syllabus, and Bluetooth is assigned to radio waves. Signal weakening through walls explains Bluetooth's limited useful range; it does not mean radio waves cannot pass through walls.

An analogue signal varies continuously and can take any value within a range.

A digital signal uses discrete levels; in the binary case it has two allowed states, commonly labelled 0 and 1 or low and high.

Feature	Analogue	Digital
allowed values	continuous range	two discrete binary states
typical trace	smoothly varying level	sequence of high and low levels
small added disturbance	directly changes the represented level	can often be rejected if each level remains recognisable

Digital does not mean the physical voltage changes instantaneously or that the carrier is always on/off. It means the information is represented by discrete states rather than a continuous range.

A microphone converts changing air pressure from a sound into a changing electrical signal.

Route	Representation of the sound
analogue transmission	the continuously varying signal follows the sound waveform and is carried or modulated for transmission
digital transmission	the analogue signal is sampled, each sample is encoded as numbers/binary data, and the bit sequence is transmitted

At the receiver, an analogue signal can drive a loudspeaker after processing. Digital data is decoded and converted back to an analogue electrical waveform before the loudspeaker recreates the sound.

The original sound wave in air is not itself digital. Digital describes how information about the sound is represented and transmitted after conversion and encoding.

Digital signalling supports a high rate of data transmission and a long useful range because discrete states can be regenerated accurately.

Stage	What happens	Benefit
encode and transmit bits	information is carried as sequences of discrete states	large data streams can be processed, compressed and sent at high rates
noise and attenuation affect the link	received pulses become weaker or distorted	states can still be identified while they remain above decision margins
repeater/regenerator decides 0 or 1	outputs clean standard pulses rather than simply amplifying the distorted waveform	noise does not accumulate in the same way, increasing reliable range

An analogue amplifier boosts both the wanted waveform and superposed noise. A digital regenerator reconstructs the intended discrete levels, provided the corruption has not made them ambiguous.

Digital transmission is not immune to noise and does not guarantee perfect reception. Its range advantage depends on sufficiently accurate detection and regeneration before errors become too large.

3.4 Sound

A sound starts when a source vibrates: it moves repeatedly backwards and forwards about an equilibrium position.

Stage	What happens
source	an object such as a loudspeaker cone, tuning fork or vocal cord vibrates
nearby medium	the source pushes and pulls neighbouring particles
outward transfer	each disturbed region affects the next, so the sound disturbance travels away

Touching a sounding tuning fork lightly to water can reveal its vibration by making splashes. Stopping the fork from vibrating stops the sound it produces.

The source must vibrate, but the source itself does not travel to the listener. It transfers a disturbance into the surrounding medium.

A sound wave is longitudinal: particles of the medium vibrate parallel to the direction in which the wave and its energy travel.

Direction	In a sound wave
wave travel	from the source towards the listener
particle vibration	backwards and forwards along the same line

A particle oscillates about its own equilibrium position as the disturbance passes. It transfers energy to neighbouring particles rather than being carried from source to listener.

A graph of sound may be drawn with peaks and troughs, but that drawing does not make sound transverse. Classify the wave by the direction of particle vibration relative to wave travel.

A healthy human with normal hearing can hear approximately 20 Hz to 20 000 Hz. The upper value is also written as 20 kHz.

Frequency	Relative to the approximate human range
below 20 Hz	below the audible range
20 Hz to 20 000 Hz	audible range
above 20 000 Hz	above the audible range

Convert before comparing: 1 kHz = 1000 Hz, so 15 kHz = 15 000 Hz and lies inside the approximate range.

The limits are approximate and vary between people and with age. A sound outside the human range may still be detected by another animal or by equipment.

Sound is a mechanical wave, so it needs a material medium to transmit its disturbance.

Region	Can sound be transmitted?	Reason
solid, liquid or gas	yes	particles can interact and pass on the vibration
vacuum	no	there are no particles to pass on the vibration

An astronaut outside a spacecraft cannot hear machinery through the vacuum between them, even if the machinery is loud and its frequency is in the human audible range.

A vacuum does not merely make sound quieter: it prevents sound transmission. Electromagnetic waves such as light can cross a vacuum, but sound cannot.

The speed of sound in air is approximately 330–350 m/s; 340 m/s is a useful typical value.

Statement	Meaning
340 m/s	sound travels about 340 m through air in 1 s
0.34 km/s	the same typical speed expressed in kilometres per second

For one-way travel, use speed = distance ÷ time. A sound that travels 680 m through air in 2.0 s has speed 680 ÷ 2.0 = 340 m/s.

Do not use 3.0 × 10^8 m/s: that is the speed of electromagnetic waves in a vacuum. The sound value is approximate, so a value within 330–350 m/s may be appropriate.

Choose a long, measured distance between a sound source and one observer. A starting pistol is useful because its puff of smoke and bang are produced together.

Step	Measurement action
1	measure the source-to-observer distance with a long tape or trundle wheel
2	the observer starts the stopwatch on seeing the smoke
3	the same observer stops the stopwatch on hearing the bang
4	calculate speed = measured distance ÷ measured time

Use a long distance so the sound delay is much larger than the timing resolution. Repeat the measurement and average the times; keep the source and observer positions fixed.

The light-travel time is negligible over this distance, so the smoke gives the start time. Do not double the measured distance in this direct one-way method; doubling applies to an echo that travels out and back.

Amplitude controls loudness, while frequency controls pitch.

Wave change	Heard effect
greater amplitude	louder sound
smaller amplitude	quieter sound
greater frequency	higher pitch
smaller frequency	lower pitch

The two properties can change independently: a note can become louder at the same pitch, or higher-pitched at the same loudness.

Do not use amplitude to rank pitch or frequency to rank loudness. A high-pitched sound is not necessarily loud, and a loud sound is not necessarily high-pitched.

An echo is a sound heard after a sound wave reflects from a surface and returns to the listener or detector.

Stage	Path
outward	source → reflecting surface
return	reflecting surface → receiver
measured delay	time for both outward and return journeys

If the receiver is beside the source, distance to the reflector = speed × echo time ÷ 2. At 330 m/s, an echo after 0.60 s comes from a surface about 99 m away.

Reflection changes the direction of the sound path. The measured echo time is round-trip time, so using speed × time without dividing by 2 gives twice the distance to the surface.

Ultrasound is sound with a frequency higher than 20 kHz, which is higher than 20 000 Hz.

Sound	Frequency condition	Human with normal hearing
audible sound	approximately 20 Hz to 20 kHz	normally audible
ultrasound	greater than 20 kHz	not audible

Ultrasound is still a sound wave: it is longitudinal and needs a material medium. Its name describes frequency, not a different type of wave motion.

The definition is frequency greater than 20 kHz, not great loudness, small amplitude or high speed. A value exactly at 20 kHz is the approximate upper hearing boundary, not higher than it.

A longitudinal sound wave consists of alternating compressions and rarefactions in the medium.

Region	Particle spacing	Density and pressure
compression	particles closer together than normal	higher density and pressure
rarefaction	particles farther apart than normal	lower density and pressure

One wavelength is the distance from the centre of one compression to the centre of the next compression, or from one rarefaction centre to the next.

Compressions are not individual particles moving all the way through the medium. The pattern travels while each particle vibrates backwards and forwards about its own equilibrium position.

In general, sound travels fastest in solids, more slowly in liquids and slowest in gases.

State	General relative speed	Typical scale
gas	slowest	air about 340 m/s
liquid	faster	water about 1500 m/s
solid	fastest	many solids: several thousand m/s

For equal path lengths, the sound travelling through the solid generally arrives first, followed by the liquid path and then the gas path.

This is a general order, not a claim that every solid has one fixed speed or that density alone determines speed. Always use a supplied material value when a calculation provides one.

Ultrasound pulses can travel through a material and reflect at a boundary. The returning echo reveals where a boundary, flaw or object is located.

Use	How ultrasound provides information
non-destructive testing	echoes from cracks, bubbles or internal boundaries locate defects without cutting open the material
medical scanning of soft tissue	different tissue boundaries return echoes that are processed into an image
sonar	echoes from the seabed or an underwater object give depth or distance

For a pulse that returns to its source, depth or distance = wave speed × total echo time ÷ 2. In seawater at 1500 m/s, a return time of 0.080 s gives 1500 × 0.080 ÷ 2 = 60 m.

Use the wave speed in the actual medium and convert time to seconds. The factor of 2 is required because the measured time includes the outward and return journeys.