IGCSE Physics · Unit 3: Waves · Core + Extended

Waves CoreExtended adds refractive index, telecoms & ultrasound uses

Waves is the biggest section in the syllabus, covering general wave behaviour, light (reflection, refraction, lenses, dispersion), the electromagnetic spectrum, and sound. The unifying idea throughout: a wave transfers energy from place to place without transferring matter — everything else in this note builds on that one sentence.

Why this matters Once you’ve nailed transverse vs. longitudinal waves and v = fλ, the same handful of behaviours — reflection, refraction, diffraction — reappear again and again across light, sound, and even seismic waves. Ray diagrams for mirrors and lenses reward careful, ruler-and-protractor practice more than memorisation, so plan time to actually draw them.

1. General Properties of Waves

A wave transfers energy from one place to another without transferring matter — the particles of the medium (in a rope, spring, or water wave) oscillate about a fixed position but do not travel along with the wave itself.

FeatureMeaning
Wavefronta line/surface joining points that are in phase (e.g. all at a crest)
Wavelength (λ)the distance between two adjacent identical points on a wave (e.g. crest to crest)
Frequency (f)the number of complete waves passing a point per second
Crest / Troughthe highest / lowest point of a transverse wave
Amplitudethe maximum displacement from the undisturbed (rest) position
Wave speed (v)the speed at which the wave pattern (energy) travels
v = fλ
Worked example — using v = fλ A wave has a frequency of 5 Hz and a wavelength of 3 m. Find its speed.
v = fλ = 5 × 3 = 15 m/s

Transverse vs. Longitudinal Waves

Direction of vibrationExamples
Transverseat right angles (perpendicular) to the direction of travelelectromagnetic radiation, water waves, seismic S-waves
Longitudinalparallel to the direction of travelsound waves, seismic P-waves

Reflection, Refraction and Diffraction (Ripple Tank)

All waves can undergo: reflection at a plane surface; refraction due to a change of speed; and diffraction through a narrow gap (or at an edge). A ripple tank demonstrates each of these with water waves: reflection off a straight barrier, refraction as waves cross a boundary into shallower water (where they travel slower), and diffraction as waves spread out after passing through a gap or past an edge.

Diffraction and gap size Extended The amount of diffraction through a gap depends on the relationship between wavelength and gap size: diffraction is most noticeable when the gap is roughly the same size as the wavelength (the waves spread out strongly); if the gap is much larger than the wavelength, diffraction is barely noticeable (the waves pass through with little spreading). At an edge, a longer wavelength diffracts (bends round the edge) more than a shorter wavelength.
Try it — General Wave Properties
  1. A wave has a wavelength of 2.5 m and travels at 10 m/s. Calculate its frequency.
  2. State whether sound waves are transverse or longitudinal, and describe the direction of particle vibration relative to the direction the wave travels.
  3. Explain why radio waves (long wavelength) diffract noticeably around large obstacles like hills, while visible light (very short wavelength) does not.
Show answers
  1. f = v/λ = 10/2.5 = 4 Hz
  2. Sound waves are longitudinal — the particles of the medium vibrate parallel to (in the same direction as) the direction the wave travels.
  3. Diffraction is most significant when the wavelength is comparable in size to the obstacle or gap. Radio waves have a wavelength comparable to the size of hills and buildings, so they diffract noticeably around them; visible light has a far shorter wavelength than these obstacles, so it diffracts far too little to be noticeable, and casts sharp shadows instead.

2. Light: Reflection

The normal is an imaginary line at right angles to a surface at the point where a light ray strikes it. The angle of incidence is measured between the incoming ray and the normal; the angle of reflection is measured between the reflected ray and the normal.

angle of incidence = angle of reflection

A plane mirror forms an image that is: the same size as the object, the same distance behind the mirror as the object is in front, and virtual (it cannot be projected onto a screen, since the light rays only appear to come from behind the mirror).

Worked example — angle of reflection A ray of light strikes a plane mirror with an angle of incidence of 35°. Find the angle of reflection, and the angle between the incident ray and the mirror surface.
Angle of reflection = angle of incidence = 35°
Angle between the ray and the mirror surface = 90° − 35° = 55°

3. Light: Refraction

Refraction is the bending of light as it crosses a boundary between two different transparent materials, caused by a change in speed. Light entering a denser medium (e.g. air into glass) slows down and bends toward the normal; light leaving a denser medium speeds up and bends away from the normal.

The critical angle is the angle of incidence (in the denser medium) at which the angle of refraction is exactly 90°. Beyond the critical angle, light cannot escape the denser medium at all — instead it undergoes total internal reflection, reflecting entirely back inside the material, obeying the normal law of reflection.

Internal reflection vs. total internal reflection At any boundary, some light is partially reflected back into the medium (internal reflection) even when most of it refracts through. Once the angle of incidence exceeds the critical angle, however, all of the light is reflected back inside — none escapes. This is used in reflective road signs and prisms in binoculars/periscopes, which rely on total internal reflection rather than a silvered mirror surface.
Refractive index Extended n = sin i / sin r
n = 1 / sin c
The refractive index, n, is the ratio of the speeds of a wave in two different regions (commonly vacuum/air compared to the material).
Worked example — refractive index and critical angle Light travels from air into a glass block, with an angle of incidence of 40° and an angle of refraction of 25°. Find the refractive index of the glass, and hence its critical angle.
n = sin i / sin r = sin 40° / sin 25° = 0.643 / 0.423 ≈ 1.52
Using n = 1/sin c: sin c = 1/n = 1/1.52 ≈ 0.658
c = sin⁻¹(0.658) ≈ 41.1°
Optical fibres Extended Optical fibres use total internal reflection to carry light along a thin glass or plastic fibre, even around gentle bends, without the light escaping through the sides. This makes them extremely useful in telecommunications for carrying data as pulses of light over long distances with very little signal loss.
Try it — Refraction
  1. State whether light bends toward or away from the normal when passing from glass into air, and explain why in terms of speed.
  2. A ray of light hits a glass surface from inside the glass at an angle greater than the critical angle. State what happens to the light.
  3. A material has a critical angle of 42°. Calculate its refractive index.
Show answers
  1. Light bends away from the normal, because it speeds up as it leaves the denser glass and enters the less dense air.
  2. The light undergoes total internal reflection — none of it escapes through the surface; instead, all of it reflects back into the glass, obeying the normal law of reflection.
  3. n = 1/sin c = 1/sin 42° = 1/0.669 ≈ 1.50

4. Thin Lenses

A converging lens (convex) brings a parallel beam of light together to a single point, the principal focus (focal point), located a distance called the focal length from the lens along the principal axis. A diverging lens (concave) spreads a parallel beam of light out, as though it came from a principal focus in front of the lens.

Images are described using three pairs of terms: enlarged / same size / diminished; upright / inverted; and real / virtual. A virtual image is formed where diverging rays appear to come from when extrapolated backward — it cannot be projected onto a screen, unlike a real image.

Ray diagram rules for a converging lens — real image To locate an image formed by a converging lens, draw at least two of these three standard rays from the top of the object:
1. A ray parallel to the principal axis, which refracts through the lens and passes through the principal focus on the far side.
2. A ray passing through the centre of the lens, which continues in a straight line (undeviated).
3. A ray passing through the principal focus on the near side, which refracts through the lens and emerges parallel to the principal axis.
Where these rays cross (on the far side of the lens) locates the top of the real image.
Virtual image with a converging lens Extended When an object is placed closer to a converging lens than its focal length, the refracted rays diverge and never actually meet on the far side. Extrapolating these diverging rays backward (behind the object, on the same side as the object) shows where they appear to come from — this locates a virtual, upright, enlarged image. This is exactly how a magnifying glass works.
Correcting sight problems Extended Long-sightedness (difficulty focusing on near objects) is corrected using a converging lens, which adds extra converging power to bring the focus point forward onto the retina.
Short-sightedness (difficulty focusing on distant objects) is corrected using a diverging lens, which spreads the light out slightly before it enters the eye, moving the focus point back onto the retina.
Try it — Lenses
  1. Describe the effect of a diverging lens on a parallel beam of light.
  2. State the three describing terms needed to fully describe an image formed by a lens.
  3. Explain why a magnifying glass must be held closer to the object than the lens’s focal length to work correctly.
Show answers
  1. A diverging lens spreads a parallel beam of light out, so that the rays appear to originate from the principal focus in front of the lens.
  2. Size (enlarged/same size/diminished), orientation (upright/inverted), and type (real/virtual).
  3. Only when the object is closer to the lens than the focal length do the refracted rays diverge (rather than converge to a real image) — extrapolating these diverging rays backward produces a virtual, upright, enlarged image, which is what makes the magnifying effect possible.

5. Dispersion of Light

Dispersion is the splitting of white light into a spectrum of colours, illustrated by shining white light through a glass prism. This happens because different colours (different frequencies/wavelengths) of light refract by slightly different amounts through the glass.

red, orange, yellow, green, blue, indigo, violet

This is the traditional order of the seven colours of the visible spectrum by increasing frequency (and correspondingly, decreasing wavelength) — red has the lowest frequency/longest wavelength, violet has the highest frequency/shortest wavelength.

Monochromatic light Extended is visible light of a single frequency (a single, pure colour) — a laser is a common source of monochromatic light.

Common mistakes to watch for — Section 3.2
  • Measuring the angle of incidence/reflection/refraction from the surface instead of from the normal.
  • Confusing “internal reflection” (some light reflects) with “total internal reflection” (all light reflects, only above the critical angle).
  • Forgetting that a virtual image cannot be captured on a screen, since the light rays don’t actually pass through the image location.
  • Mixing up which lens corrects which sight problem — converging lens for long-sightedness (adds focusing power), diverging lens for short-sightedness (reduces focusing power).
  • Listing the spectrum colours in the wrong order, or forgetting whether the order given is by increasing frequency or increasing wavelength (they’re opposite).

6. The Electromagnetic Spectrum

All electromagnetic waves are transverse and travel at the same high speed in a vacuum (3.0 × 10⁸ m/s, Extended), though they differ hugely in frequency and wavelength.

Region (increasing frequency →)Typical useHarmful effect of overexposure
Radio wavesradio/TV transmission, astronomy, RFID—
Microwavessatellite TV, mobile phones, microwave ovensinternal heating of body cells
Infraredelectric grills, remote controls, thermal imaging, optical fibresskin burns
Visible lightvision, photography, illumination—
Ultravioletsecurity marking, detecting fake banknotes, sterilising waterskin/eye damage, skin cancer
X-raysmedical scanning, security scannersmutation/damage to cells
Gamma rayssterilising food/equipment, cancer detection and treatmentmutation/damage to cells
Communications uses Extended Mobile phones and wireless internet use microwaves, because microwaves can penetrate some walls and only need a short aerial.
Bluetooth uses radio waves, which pass through walls but weaken as they do.
Optical fibres (visible light or infrared) are used for cable TV and broadband, since glass is transparent to these wavelengths, which can carry very high rates of data.
A digital signal has only discrete (on/off) values, while an analogue signal varies continuously. Digital signalling has the benefits of an increased rate of data transmission and increased range, since the signal can be accurately regenerated (cleaned up) at intervals, unlike an analogue signal which accumulates noise and distortion over distance.
Try it — Electromagnetic Spectrum
  1. Name the region of the electromagnetic spectrum used in a TV remote control.
  2. State one harmful effect of excessive exposure to X-rays.
  3. Explain why digital signals can be transmitted over longer distances than analogue signals without losing quality.
Show answers
  1. Infrared
  2. Mutation or damage to cells in the body.
  3. A digital signal can be accurately regenerated (restored to its original clean values) at intervals along its path, removing any noise that has been picked up. An analogue signal cannot be cleanly separated from noise in this way, so any distortion it picks up accumulates and degrades the signal over distance.

7. Sound

Sound is produced by vibrating sources and travels as a longitudinal wave. It requires a medium to travel through (it cannot travel through a vacuum), and the speed of sound in air is approximately 330–350 m/s. The human audible range is approximately 20 Hz to 20000 Hz.

Compression and rarefaction Extended As a sound wave travels through a medium, it creates regions of compression (particles pushed closer together, higher pressure) and rarefaction (particles spread further apart, lower pressure), alternating along the direction of travel. Sound travels faster in solids than in liquids, and faster in liquids than in gases, because particles are progressively closer together (and can transmit vibrations more quickly) in each case.
Measuring the speed of sound in air One method: stand a known, large distance from a wall and clap, or make a bang (e.g. by hitting two blocks together). Start a stopwatch at the sound of the clap and stop it when the echo is heard. The sound has travelled to the wall and back in this time, so: speed = (2 × distance) / time.
Worked example — speed of sound from an echo A person stands 85 m from a cliff and claps. The echo is heard 0.5 s later. Find the speed of sound.
Total distance travelled by sound = 2 × 85 = 170 m
speed = distance / time = 170 / 0.5 = 340 m/s

Amplitude affects loudness (larger amplitude = louder); frequency affects pitch (higher frequency = higher pitch). An echo is simply the reflection of sound waves off a surface. Ultrasound is defined as sound with a frequency higher than 20 kHz — above the range of human hearing.

Uses of ultrasound Extended Ultrasound is used in: non-destructive testing of materials (detecting internal cracks without damaging the object); medical scanning of soft tissue (e.g. imaging a foetus during pregnancy); and sonar (finding the depth of the seabed or the distance to an object by timing the reflected pulse).
For sonar, the same “distance/time” logic as the echo method applies: distance = (speed × time) / 2 (dividing by 2 because the pulse travels there and back).
Worked example — sonar depth calculation A ship sends an ultrasound pulse toward the seabed. The reflected pulse returns after 0.8 s. If the speed of sound in water is 1500 m/s, find the depth of the water.
distance = (speed × time) / 2 = (1500 × 0.8) / 2 = 1200 / 2 = 600 m
Try it — Sound
  1. Explain why sound cannot travel through a vacuum, but light can.
  2. A sound wave’s amplitude is increased while its frequency stays the same. Describe the effect on what is heard.
  3. A submarine sends an ultrasound pulse that reflects off the seabed and returns after 1.2 s. The speed of sound in water is 1500 m/s. Calculate the depth of the seabed.
Show answers
  1. Sound is a mechanical wave that relies on particles of a medium vibrating and colliding with neighbouring particles to pass the vibration along — with no particles present (a vacuum), there is nothing to vibrate or collide, so sound cannot travel. Light is an electromagnetic wave and does not need a medium, so it can travel through a vacuum.
  2. The sound becomes louder, since loudness depends on amplitude; the pitch stays the same, since pitch depends on frequency, which is unchanged.
  3. distance = (speed × time)/2 = (1500 × 1.2)/2 = 1800/2 = 900 m
Common mistakes to watch for — Sections 3.3 and 3.4
  • Forgetting to double the distance (there and back) in echo or sonar calculations before dividing by time.
  • Confusing loudness (linked to amplitude) with pitch (linked to frequency).
  • Thinking all electromagnetic waves travel at different speeds in a vacuum — they all travel at the same speed (3.0 × 10⁸ m/s) regardless of frequency.
  • Assuming ultrasound is just “very loud sound” — it is sound above the human hearing range in frequency, not necessarily loud.
  • Forgetting that sound needs a medium, while all electromagnetic waves (including light) do not.

8. Quick-Fire Challenge Round

True or False?
  1. A wave transfers matter as well as energy from one place to another.
  2. Sound waves are transverse waves.
  3. All electromagnetic waves travel at the same speed in a vacuum.
  4. Total internal reflection only occurs when the angle of incidence exceeds the critical angle.
Show answers
  1. False — a wave transfers energy without transferring matter; the particles of the medium oscillate about a fixed position rather than travelling with the wave.
  2. False — sound waves are longitudinal, with particle vibrations parallel to the direction of travel.
  3. True — all electromagnetic waves travel at 3.0 × 10⁸ m/s in a vacuum, regardless of their frequency or wavelength.
  4. True — below the critical angle, light mostly refracts out (with some partial internal reflection); only above the critical angle does total internal reflection occur, reflecting all the light back inside.
Spot the Error
  1. “A wave has a frequency of 4 Hz and a wavelength of 2 m, so its speed is 4 + 2 = 6 m/s.”
  2. “A person 100 m from a wall claps and hears the echo 0.5 s later, so the speed of sound is 100/0.5 = 200 m/s.”
  3. “A converging lens is used to correct short-sightedness, since it brings light to a sharper focus.”
Show answers
  1. Wave speed is found by multiplying, not adding: v = fλ = 4 × 2 = 8 m/s.
  2. The sound travels to the wall AND back (200 m total), not just the 100 m one-way distance. Correct: speed = (2 × 100)/0.5 = 400 m/s.
  3. Short-sightedness is corrected with a DIVERGING lens, which spreads the light out slightly before it enters the eye, moving the focus point back onto the retina. A converging lens is used to correct long-sightedness instead.

9. Mixed Practice — Bringing It All Together

  1. A wave has a frequency of 50 Hz and travels at 340 m/s. Calculate its wavelength.
  2. A ray of light hits a mirror at an angle of incidence of 28°. State the angle of reflection.
  3. Explain why an optical fibre keeps light travelling along it even around gentle bends.
  4. Name the region of the electromagnetic spectrum used for satellite television.
  5. A ship’s sonar pulse returns after 2 s, and the speed of sound in water is 1500 m/s. Calculate the depth of water beneath the ship.
Show answers
  1. λ = v/f = 340/50 = 6.8 m
  2. 28° (angle of reflection = angle of incidence)
  3. Light inside the fibre strikes the internal surface at an angle greater than the critical angle, so it undergoes total internal reflection every time it reaches the edge of the fibre, keeping it travelling along the fibre’s length even around gentle curves.
  4. Microwaves
  5. distance = (speed × time)/2 = (1500 × 2)/2 = 3000/2 = 1500 m

Practice at Home

The biggest unit yet — spread this across several sittings. Parts C, D and E include Extended-only content.

Part A — General wave properties
  1. A wave has a wavelength of 4 m and a frequency of 6 Hz. Calculate its speed.
  2. State two examples of transverse waves and two examples of longitudinal waves.
  3. Describe how a ripple tank can be used to demonstrate refraction.
Part B — Reflection and lenses
  1. State the three properties of an image formed by a plane mirror.
  2. Describe the effect of a converging lens on a parallel beam of light.
  3. State the three pairs of terms used to describe an image formed by a lens.
Part C — Refraction and refractive index (Extended included)
  1. Define critical angle.
  2. Light passes from air into a material with an angle of incidence of 50° and an angle of refraction of 30°. Calculate the refractive index of the material.
  3. Calculate the critical angle of a material with a refractive index of 1.4.
Part D — Electromagnetic spectrum (Extended included)
  1. List the seven regions of the electromagnetic spectrum in order of increasing frequency.
  2. Explain why microwaves are used for mobile phone communication rather than radio waves.
  3. Explain one advantage of digital signalling over analogue signalling.
Part E — Sound and ultrasound (Extended included)
  1. State the approximate range of frequencies audible to humans.
  2. A person 150 m from a cliff claps and hears the echo 0.9 s later. Calculate the speed of sound.
  3. Describe one medical use of ultrasound.

Key Vocabulary Recap

Transverse wave — vibration at right angles to the direction of travel (e.g. light, water waves)
Longitudinal wave — vibration parallel to the direction of travel (e.g. sound)
Critical angle — the angle of incidence beyond which total internal reflection occurs
Virtual image — an image that cannot be projected onto a screen, formed where diverging rays appear to originate
Dispersion — the splitting of white light into a spectrum of colours by a prism
Ultrasound — sound with a frequency above 20 kHz, beyond human hearing

Syllabus Reference

This note covers Cambridge IGCSE Physics (0625) section 3, Waves, in full — 3.1 General properties of waves, 3.2 Light (3.2.1 Reflection, 3.2.2 Refraction, 3.2.3 Thin lenses, 3.2.4 Dispersion), 3.3 Electromagnetic spectrum, and 3.4 Sound. Core and Supplement content are both included, with Extended-only material (refractive index, optical fibres in telecommunications, digital vs. analogue signalling, sight correction with lenses, magnifying glasses, uses of ultrasound, and the speed of light in a vacuum) clearly flagged throughout.

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