The fundamental principle of waves
Waves transfer energy from one place to another without transferring matter. That single principle governs everything from the ripples spreading across a pond to the light reaching your eyes from a distant star. If you grasp this idea firmly, the rest of wave physics follows through logical chains of cause and effect. If you don't, the topic becomes a collection of disconnected facts. So start here: the medium oscillates, energy propagates, but the particles themselves stay put.
Consider a slinky spring stretched across a table. Push one end forward and a compression pulse travels to the other end, but the coils don't migrate with it. They bunch together, transfer the disturbance to their neighbours, then return to their original positions. The same logic applies to water waves, sound waves, and seismic waves. The IGCSE Physics syllabus (0625) tests this principle repeatedly, and candidates who can articulate it clearly pick up marks that others leave on the table.
Describing waves: the vocabulary that matters
Before solving any wave problem, you need precise definitions. Imprecise language costs marks in Cambridge exams, because examiners distinguish carefully between terms that students sometimes treat as interchangeable.
| Term | Definition | Unit |
|---|---|---|
| Wavelength (λ) | Distance between two consecutive points in phase (e.g. crest to crest) | metres (m) |
| Frequency (f) | Number of complete waves passing a point per second | hertz (Hz) |
| Amplitude (A) | Maximum displacement of a particle from its rest position | metres (m) |
| Wave speed (v) | Distance travelled by the wave per unit time | m/s |
| Period (T) | Time for one complete oscillation; T = 1/f | seconds (s) |
| Wavefront | A line joining points of the wave that are in phase | - |
| Crest (peak) | Point of maximum positive displacement | - |
| Trough | Point of maximum negative displacement | - |
A common exam error is measuring wavelength from crest to trough. That's only half a wavelength. Always measure from one crest to the next, or from one trough to the next, or between any two consecutive points that are doing exactly the same thing at the same time.
Transverse and longitudinal waves
All waves fall into one of two categories, and the distinction rests on a single geometric question: what is the angle between the direction of vibration and the direction the wave travels?
- Transverse waves: the vibration is perpendicular to the direction of energy transfer. Examples include electromagnetic radiation (light, radio, X-rays), water surface waves, and seismic S-waves. You can demonstrate this with a rope: flick one end sideways, and a pulse travels along the rope while the rope itself moves up and down.
- Longitudinal waves: the vibration is parallel to the direction of energy transfer. Sound waves and seismic P-waves are longitudinal. In a longitudinal wave, particles compress together (compressions) and spread apart (rarefactions) along the same axis that the wave propagates.
Why does this matter for the exam? Cambridge questions frequently ask candidates to identify wave type from a description or diagram. They also test whether students know that sound cannot be transverse and that light cannot be longitudinal. The reasoning is physical: sound requires a medium whose particles can push and pull along the travel direction, while electromagnetic waves consist of oscillating electric and magnetic fields perpendicular to travel.
The wave equation: v = fλ
The wave equation connects three quantities in one clean relationship: wave speed equals frequency multiplied by wavelength. It's deceptively simple, but exam questions probe whether you can rearrange and apply it correctly under different conditions.
Worked example 1: Calculating wave speed
A sound wave has a frequency of 440 Hz and a wavelength of 0.77 m. Calculate its speed.
- Identify the known values: f = 440 Hz, λ = 0.77 m
- Apply v = fλ = 440 x 0.77 = 338.8 m/s
- This is consistent with the speed of sound in air (approximately 340 m/s), which provides a useful sanity check.
Worked example 2: Finding wavelength from a ripple tank
In a ripple tank experiment, 20 wavefronts pass a fixed point in 10 seconds. The wave speed is measured as 0.12 m/s. Find the wavelength.
- Frequency: f = 20 waves / 10 s = 2.0 Hz
- Rearrange: λ = v / f = 0.12 / 2.0 = 0.06 m
Reflection, refraction, and diffraction
These three behaviours occur whenever a wave encounters a boundary or an obstacle. Each follows from the same underlying principle (waves interact with changes in their medium), but the outcomes differ.
Reflection
When a wave hits a plane boundary and bounces back, the angle of incidence equals the angle of reflection. Both angles are measured from the normal, the line perpendicular to the surface at the point of incidence. This law holds for all types of waves: light off a mirror, sound off a cliff face, water waves off a straight barrier in a ripple tank.
Refraction
Refraction occurs when a wave crosses a boundary between two media where its speed differs. The wave changes direction because one part of the wavefront slows down (or speeds up) before the other. A light ray passing from air into glass bends toward the normal because light travels more slowly in glass. A ray passing from glass into air bends away from the normal.
The cause-and-effect chain is precise: change in speed causes change in wavelength (frequency stays constant because it's set by the source), and if the wave hits the boundary at an angle, the speed difference causes bending. If the wave hits perpendicular to the boundary, it slows down but doesn't bend.
Diffraction
Diffraction is the spreading of a wave as it passes through a gap or around an edge. The effect is most pronounced when the gap width is similar to the wavelength. A wide gap produces minimal spreading; a gap comparable to the wavelength produces significant spreading. This explains why you can hear someone talking around a corner (sound wavelengths are similar to doorway widths) but you can't see around corners (visible light wavelengths are far smaller than everyday obstacles).
Light: reflection and refraction in detail
The behaviour of light at boundaries is one of the most heavily tested areas in the waves section. Candidates need to draw accurate ray diagrams and apply the laws of reflection and refraction precisely.
For reflection in a plane mirror, three properties of the image are always tested: it is virtual (cannot be projected onto a screen), upright, laterally inverted, and the same size as the object. The image distance behind the mirror equals the object distance in front.
Snell's law and the refractive index
The refractive index n of a medium relates the speed of light in a vacuum (c) to the speed of light in that medium (v): n = c / v. For two media, Snell's law connects the angles:
n1 sin θ1 = n2 sin θ2
Extended candidates must apply this equation quantitatively. Core candidates need the qualitative understanding: light bends toward the normal when entering a denser medium, away from the normal when leaving.
Total internal reflection and the critical angle
When light travels from a denser to a less dense medium (e.g. glass to air), increasing the angle of incidence eventually reaches a point where the refracted ray runs along the boundary. This is the critical angle. Beyond it, all light reflects back into the denser medium. No refraction occurs.
The critical angle c is related to the refractive index by: sin c = 1 / n
Total internal reflection has practical applications that Cambridge examiners like to test: optical fibres use it to transmit light signals over long distances with minimal loss, and prisms in binoculars and periscopes use it to redirect light through 90 or 180 degrees.
Worked example 3: Critical angle
Glass has a refractive index of 1.50. Calculate the critical angle.
- sin c = 1 / n = 1 / 1.50 = 0.667
- c = sin-1(0.667) = 41.8 degrees
- Any ray hitting the glass-air boundary at an angle greater than 41.8 degrees from the normal will undergo total internal reflection.
Thin converging lenses
A converging (convex) lens brings parallel rays to a focus at its principal focus (focal point). The distance from the centre of the lens to the principal focus is the focal length, f. Three standard construction rays locate the image formed by a converging lens:
- A ray parallel to the principal axis refracts through the focal point on the far side.
- A ray through the centre of the lens passes straight through undeviated.
- A ray through the focal point on the near side emerges parallel to the principal axis.
The nature of the image depends on where the object sits relative to the focal point. Place the object beyond 2f: you get a real, inverted, diminished image between f and 2f on the other side. Place it at 2f: the image is real, inverted, and the same size, also at 2f. Between f and 2f: real, inverted, magnified, beyond 2f. Closer than f: the image becomes virtual, upright, and magnified. This last case is how a magnifying glass works.
Dispersion and the visible spectrum
White light is a mixture of colours, each with a different wavelength. When white light passes through a glass prism, it disperses into a spectrum because shorter wavelengths (violet) refract more than longer wavelengths (red). The sequence runs: red, orange, yellow, green, blue, indigo, violet. Red has the longest wavelength and the lowest frequency; violet has the shortest wavelength and the highest frequency.
This isn't just a pretty effect. Dispersion demonstrates that the refractive index of a material depends on wavelength, a fact that matters in lens design (chromatic aberration) and in understanding rainbows.
The electromagnetic spectrum
All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 108 m/s. They differ only in wavelength and frequency. The spectrum, from longest wavelength to shortest, runs:
| Type | Wavelength range | Key uses | Key hazards |
|---|---|---|---|
| Radio waves | km to m | Broadcasting, communication | Generally low risk |
| Microwaves | cm to mm | Cooking, satellite communication, mobile phones | Internal heating of tissue |
| Infrared | mm to 700 nm | Heating, thermal imaging, remote controls | Skin burns |
| Visible light | 700 nm to 400 nm | Vision, optical fibres, photography | Eye damage at high intensity |
| Ultraviolet | 400 nm to 10 nm | Sterilisation, fluorescent lamps, security marking | Sunburn, skin cancer, eye damage |
| X-rays | 10 nm to 0.01 nm | Medical imaging, security scanning | Cell damage, cancer |
| Gamma rays | Below 0.01 nm | Sterilisation, cancer treatment, tracing | Cell damage, cancer, mutation |
Two points that catch students out. First, all these waves are transverse. Second, the boundaries between types are not sharp; they're conventions. What matters for the exam is knowing the order, the relative wavelengths and frequencies, typical applications, and associated dangers.
Sound waves
Sound is a longitudinal wave produced by vibrating objects. A vibrating drum skin, for instance, pushes and pulls on the air molecules around it, creating alternating compressions and rarefactions that propagate outward. When these pressure variations reach your ear, the eardrum vibrates in response, and your brain interprets the signal as sound.
Properties of sound
- Speed: sound travels at approximately 340 m/s in air at room temperature. It travels faster in liquids (around 1500 m/s in water) and faster still in solids (around 5000 m/s in steel). The reason is straightforward: particles in solids are closer together and more tightly bonded, so they transmit vibrations more quickly.
- Pitch: determined by frequency. A higher frequency means a higher pitch. The human ear detects sounds from roughly 20 Hz to 20 000 Hz.
- Loudness: determined by amplitude. A larger amplitude means a louder sound.
- Sound requires a medium: place an electric bell inside a vacuum jar and pump out the air. You'll see the hammer striking the bell but hear nothing. Sound cannot travel through a vacuum because there are no particles to compress and rarefy.
Echoes and reflection of sound
Sound reflects off hard, flat surfaces. An echo is simply a reflected sound wave that reaches the listener with enough delay to be heard separately from the original. The speed of sound can be measured using echoes: stand a known distance d from a wall, clap, time the interval t for the echo to return, and calculate v = 2d / t.
Worked example 4: Speed of sound from an echo
A student stands 170 m from a cliff and claps. The echo returns 1.0 s later. Calculate the speed of sound.
- Total distance = 2 x 170 = 340 m (there and back)
- v = distance / time = 340 / 1.0 = 340 m/s
Ultrasound
Sound above 20 000 Hz is called ultrasound. Humans can't hear it, but it has practical applications that IGCSE candidates are expected to know. Medical ultrasound imaging sends pulses into the body; reflections from boundaries between different tissues return at different times, and a computer builds an image from the timing data. Sonar works on the same principle: a ship emits an ultrasound pulse downward, measures the time for the echo from the seabed, and calculates the depth using v = 2d / t. Industrial quality control uses ultrasound to detect internal cracks in metal components.
Common mistakes and how to avoid them
| Mistake | Why it happens | How to fix it |
|---|---|---|
| Measuring wavelength as crest to trough | Confusing half a cycle with a full cycle | Wavelength is crest-to-crest or trough-to-trough. If given a diagram, count the full cycles and divide the total distance. |
| Saying sound is transverse | Water waves (which are visible and familiar) are transverse, leading to a false generalisation | Sound is longitudinal. Particles vibrate parallel to the direction of energy transfer. |
| Forgetting to double the distance in echo calculations | The sound travels to the reflector and back | Always use 2d in the equation, not d. |
| Confusing frequency and wavelength with speed | Assuming higher frequency means faster wave | Wave speed depends on the medium, not on the frequency. Changing frequency changes wavelength, but v stays the same in a given medium. |
| Drawing the normal incorrectly in ray diagrams | Drawing the normal along the surface instead of perpendicular to it | The normal is always at 90 degrees to the reflecting or refracting surface at the point of incidence. |
| Stating that EM waves need a medium | Confusion with sound waves | Electromagnetic waves travel through a vacuum. Sound does not. This is a fundamental distinction. |
Self-check questions
Attempt each question fully before reading the answers.
- A wave has a frequency of 500 Hz and travels at 1500 m/s through water. Calculate its wavelength.
- Explain why sound cannot travel through a vacuum but light can.
- A ray of light travels from water (n = 1.33) into air. Calculate the critical angle for total internal reflection.
- A ship sends an ultrasound pulse toward the seabed. The pulse returns after 0.4 s. The speed of sound in seawater is 1500 m/s. Calculate the depth of the water.
- State two differences between transverse and longitudinal waves, giving one example of each.
Exam strategy for waves
Waves questions appear across multiple papers in the IGCSE Physics exam, and the topic rewards candidates who think systematically. For calculation questions, always write the formula first, substitute values with units, and check whether your answer is physically reasonable. A wavelength of 300 m for visible light should immediately raise a red flag; visible wavelengths are measured in hundreds of nanometres.
Ray diagrams demand precision. Use a sharp pencil, a ruler, and label every angle, normal line, and medium. Sloppy diagrams lose marks even when the physics behind them is correct. For refraction questions, mark the normal first, then draw the rays relative to it.
Qualitative questions on the electromagnetic spectrum test recall, but they also test application. Knowing that microwaves are used in satellite communication isn't enough: you should be able to explain why (they can pass through the atmosphere and can be focused into narrow beams). Similarly, knowing that X-rays are used in medical imaging isn't enough: you should understand that they pass through soft tissue but are absorbed by bone, creating a shadow image.
For sound questions, the echo method and ultrasound applications follow the same mathematical pattern. Master the v = 2d / t calculation once, and you can apply it to cliffs, seabeds, and medical scanners alike. The physics is identical; only the context changes. Recognising this kind of structural similarity across different problems is what separates strong candidates from average ones.
A rigorous breakdown of wave physics for Cambridge IGCSE (0625), covering transverse and longitudinal waves, the wave equation, reflection, refraction, total internal reflection, the electromagnetic spectrum, and sound, with worked examples, common pitfalls, and exam-focused problem-solving strategies.
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