Edexcel IGCSE Physics 3(d) Light and Sound — Notes
Written for Pearson Edexcel International GCSE Physics (4PH1), Science Double Award (4SD0) and Science Single Award (4SS0), linear and modular.
Which parts do I need?
Single Award candidates need reflection and the law of reflection, the refraction practical with blocks and prisms, total internal reflection in optical fibres and prisms, the meaning of critical angle, and the fact that light and sound can both be reflected and refracted. They skip drawing ray diagrams, \(n=\sin i/\sin r\), the refractive-index practical and \(\sin c=1/n\). Single Award wording also leaves out “transverse” for light and “longitudinal” for sound. Like Double Award, Single Award also skips the whole Physics-only sound section.
Double Award candidates skip only the Physics-only sound section: the range of human hearing, the speed-of-sound practical, the oscilloscope and its practical, and pitch and loudness.
What this page covers
| Spec | Modular no. | You need to be able to… | Route |
|---|---|---|---|
| 3.14 | 5.14 | know that light is a transverse wave that can be reflected and refracted (Single Award: reflected and refracted only) | Single · Double · Physics |
| 3.15 | 5.15 | apply the rule that the angles of incidence and reflection are equal | Single · Double · Physics |
| 3.16 | 5.16 | draw ray diagrams for reflection and refraction | Double · Physics |
| 3.17 | 5.17 | investigate how light refracts in rectangular and semicircular blocks and triangular prisms (practical) | Single · Double · Physics |
| 3.18 | 5.18 | recall and use refractive index \(n=\sin i/\sin r\) | Double · Physics |
| 3.19 | 5.19 | measure the refractive index of a glass block (practical) | Double · Physics |
| 3.20 | 5.20 | describe how total internal reflection lets optical fibres and prisms carry or redirect light | Single · Double · Physics |
| 3.21 | 5.21 | explain what the critical angle is | Single · Double · Physics |
| 3.22 | 5.22 | recall and use \(\sin c=1/n\) | Double · Physics |
| 3.23 | 5.23 | know that sound is a longitudinal wave that can be reflected and refracted (Single Award: reflected and refracted only) | Single · Double · Physics |
| 3.24P | 5.24P | know that humans hear frequencies from 20 Hz to 20 000 Hz | Physics only · 2P |
| 3.25P | 5.25P | measure the speed of sound in air (practical) | Physics only · 2P |
| 3.26P | 5.26P | understand how a microphone and oscilloscope display a sound wave | Physics only · 2P |
| 3.27P | 5.27P | find the frequency of a sound from an oscilloscope trace (practical) | Physics only · 2P |
| 3.28P | 5.28P | link pitch to the frequency of the vibrating source | Physics only · 2P |
| 3.29P | 5.29P | link loudness to the amplitude of the vibrating source | Physics only · 2P |
Modular candidates: Waves is topic 5 in the modular specifications, assessed in 4XPH1 Unit 2 (4WPH2) and 4XSD1 Unit 6 (4WSD6). The modular number is the same in both; 5.24P–5.29P are in 4XPH1 only.
In one minute
- Law of reflection: angle of incidence = angle of reflection, both measured from the normal.
- Refraction: light entering glass slows down and bends towards the normal; leaving glass it speeds up and bends away from the normal.
- Refractive index \(n=\dfrac{\sin i}{\sin r}\); critical angle \(\sin c=\dfrac{1}{n}\). Double · Physics
- Total internal reflection happens when light inside the denser material meets the boundary at more than the critical angle. It sends light along optical fibres and turns light through 90° or 180° in prisms.
- Sound can be reflected (echoes) and refracted, like light.
- Sound is a longitudinal wave. Double · Physics
- Humans hear frequencies from 20 Hz to 20 000 Hz. Physics only · 2P
- On an oscilloscope, more cycles across the screen = higher frequency = higher pitch; taller trace = larger amplitude = louder. Physics only · 2P
What kind of wave is light? Single · Double · Physics
- Light is a transverse wave: its vibrations are at right angles to the direction the light travels. It is part of the electromagnetic spectrum (3(c)). Double · Physics
- Light can be reflected (it bounces off a surface) and refracted (it changes direction when it crosses into a different material at an angle).
Single Award: you need to know that light can be reflected and refracted; the word “transverse” is not in your specification.
We draw light as rays: straight lines with an arrow showing the direction of travel. All angles are measured between the ray and the normal — a construction line drawn at 90° to the surface where the ray hits it.
What is the law of reflection? Single · Double · Physics
- If a question gives the angle between the ray and the mirror, subtract it from 90° to get the angle of incidence.
- A ray that hits the mirror along the normal (i = 0°) is reflected straight back along its own path.
A ray of light strikes a plane mirror. The angle between the ray and the surface of the mirror is 25°. (a) Determine the angle of reflection. (b) Determine the angle between the incident ray and the reflected ray.
(a) angle of incidence = 90° − 25° = 65°, so angle of reflection = 65°
(b) 65° + 65° = 130°
Common error: giving 25° as the angle of reflection — that angle is measured from the mirror, not from the normal.
Why does light refract? Single · Double · Physics
Light travels more slowly in glass, Perspex or water than in air. When a ray crosses a boundary at an angle, its speed changes and so does its direction.
| Ray goes from… | Speed | Bends… | Angle |
|---|---|---|---|
| air into glass (or water) | decreases | towards the normal | r is smaller than i |
| glass (or water) into air | increases | away from the normal | the angle in air is larger |
| along the normal (i = 0°) | changes | not at all | passes straight through |
How do you draw ray diagrams for reflection and refraction? Double · Physics
Not required for Single Award (4SS0).
- Use a sharp pencil and a ruler. Rays are straight lines.
- Draw the normal as a dashed line at 90° to the surface at the point where the ray hits it.
- Put an arrow on every ray to show the direction of travel.
- Reflection: measure i with a protractor and draw the reflected ray at the same angle on the other side of the normal.
- Refraction: going into a denser material, bend the ray towards the normal; coming out, bend it away. For a parallel-sided block, the emergent ray is parallel to the incident ray.
- A ray meeting a surface along the normal is drawn passing straight through, undeviated.
The marks depend on the direction of bending and whether the emerging ray is on the correct side. A ray drawn leaving a parallel-sided block at 90° to its face, or bending the wrong way, loses the marks.
Practical: how do you investigate refraction with blocks and prisms? Single · Double · Physics
You trace rays of light through a rectangular block, a semicircular block and a triangular prism to see how light refracts at each surface.
Investigating refraction using glass blocks and prisms
- Aim: find how a ray of light changes direction as it enters and leaves rectangular blocks, semicircular blocks and triangular prisms.
- Apparatus: ray box with a single slit and power supply, rectangular glass (or Perspex) block, semicircular block, triangular prism, sheet of white paper, sharp pencil, ruler, protractor.
- Independent variable: angle of incidence (and the shape of the block).
- Dependent variable: angle of refraction (and the path of the emerging ray).
- Control variables: the same block for each set of readings, the same colour of light, a narrow ray from the same slit.
Method
- Place the block on the paper and draw round it with a sharp pencil.
- Shine a narrow ray of light at one face. Mark the path of the incident ray with two crosses far apart, and the emerging ray with two more crosses.
- Remove the block. Join the crosses with a ruler to draw the incident and emergent rays, then join the entry and exit points to show the ray inside the block.
- Draw the normal at each point where the ray crosses a surface, and measure the angles of incidence and refraction with a protractor.
- Repeat for different angles of incidence. For the semicircular block, aim the ray at the centre of the flat face through the curved side, so it crosses the curved surface along a radius without bending; increase the angle to observe the critical angle and total internal reflection.
- For the prism, note that the ray bends towards the base at both faces.
Errors and improvements
- A wide ray makes it hard to mark the centre of the beam: use a narrow slit and mark the middle of the ray.
- Place the crosses as far apart as possible, so a small error in marking makes only a small error in the angle.
- Measure the angles from the normal, not from the surface of the block.
- Work in a dimmed room so the ray is easy to see.
- Safety: the ray box becomes hot — do not touch the lamp housing; keep the floor clear of bags in the darkened room.
What is refractive index and how do you use it? Double · Physics
Not required for Single Award (4SS0).
\[ \text{refractive index}=\frac{\sin(\text{angle of incidence})}{\sin(\text{angle of refraction})}\qquad n=\frac{\sin i}{\sin r} \]- \(i\) is the angle in air, \(r\) the angle in the material. Both are measured from the normal.
- \(n\) is a ratio, so it has no unit. It is always greater than 1; a larger \(n\) means the light slows down more and bends more. Typical values: water 1.33, glass about 1.5.
- Work out the sines first, then divide. Dividing the angles themselves (50 ÷ 31) is a common error and gets no credit.
- To find \(r\): \(\sin r=\dfrac{\sin i}{n}\), then use \(\sin^{-1}\) on your calculator (in degree mode).
A ray of light enters a glass block with an angle of incidence of 50°. The angle of refraction is 31°. (a) Calculate the refractive index of the glass. (b) Another ray enters the same glass at 30°. Calculate its angle of refraction.
(a) \(n=\dfrac{\sin 50^\circ}{\sin 31^\circ}=\dfrac{0.766}{0.515}=1.487=\) 1.49 (no unit)
(b) \(\sin r=\dfrac{\sin 30^\circ}{1.487}=0.336\), so \(r=\) 20° (19.6°)
Practical: how do you measure the refractive index of glass? Double · Physics
Not required for Single Award (4SS0).
Investigating the refractive index of a glass block
- Aim: find the refractive index of glass by measuring pairs of angles of incidence and refraction.
- Apparatus: rectangular glass block, ray box with a single slit, white paper, sharp pencil, ruler, protractor.
- Independent variable: angle of incidence i (for example 20°, 30° … 70°).
- Dependent variable: angle of refraction r.
- Control variables: the same block, the same colour of light, the same face of the block.
Method
- Draw round the block. Draw a normal to the long face and mark angles of incidence from it with the protractor.
- For each angle, shine the ray along the marked line, mark the emerging ray with two crosses, remove the block and draw the ray inside it.
- Measure r from the normal inside the block. Take at least six pairs of readings.
- Calculate sin i and sin r. Plot sin i (vertical axis) against sin r (horizontal axis) and draw a straight line of best fit through the origin.
- The gradient of the line is the refractive index: \(n=\dfrac{\sin i}{\sin r}\).
Errors and improvements
- Use the gradient of a line from many readings rather than one pair of angles: this reduces the effect of random errors in measuring each angle.
- Use large angles of incidence too: at small angles a 1° error in r is a large fraction of the reading.
- Mark the rays with crosses far apart and use a narrow ray, to reduce the uncertainty in each angle.
- Safety: the ray box gets hot; handle glass blocks carefully and report chipped edges.
| i / ° | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|
| r / ° | 13 | 19 | 25 | 30 | 35 | 38 |
| sin i | 0.34 | 0.50 | 0.64 | 0.77 | 0.87 | 0.94 |
| sin r | 0.22 | 0.33 | 0.42 | 0.50 | 0.57 | 0.62 |
What is the critical angle, and what is total internal reflection? Single · Double · Physics
Single Award wording is “explain what is meant by critical angle”; Double and Physics say “explain the meaning of critical angle”. The answer is the same.
Imagine a ray inside a semicircular glass block, meeting the flat face. As the angle of incidence increases:
- i less than c: most of the light refracts out into the air, bending away from the normal; a little is reflected.
- i equal to c: the refracted ray travels along the boundary (angle of refraction = 90°).
- i greater than c: no light leaves. All of it is reflected back inside — total internal reflection (TIR), obeying the law of reflection.
How do you use \(\sin c=1/n\)? Double · Physics
Not required for Single Award (4SS0).
\[ \sin(\text{critical angle})=\frac{1}{\text{refractive index}}\qquad \sin c=\frac{1}{n} \]- A larger refractive index gives a smaller critical angle, so total internal reflection happens more easily.
- Glass (n = 1.50): \(c=\sin^{-1}(1/1.50)=41.8^\circ\). Water (n = 1.33): \(c=48.8^\circ\).
A glass prism has refractive index 1.50. (a) Calculate the critical angle. (b) A ray inside the prism meets a face at an angle of incidence of 45°. Explain what happens to the ray.
(a) \(\sin c=\dfrac{1}{1.50}=0.667\), so \(c=41.8^\circ=\) 42°
(b) 45° is greater than the critical angle (42°) and the ray is going from glass to air, so it is totally internally reflected, at 45° to the normal.
How is total internal reflection used in optical fibres and prisms? Single · Double · Physics
Optical fibres
- An optical fibre is a very thin strand of glass. Light enters one end and meets the side of the fibre at an angle greater than the critical angle, so it is totally internally reflected.
- This repeats all the way along, even round gentle bends, so the light (and the information it carries) reaches the far end with very little loss.
- Information is sent as rapid pulses of light. Uses: telephone and internet cables; endoscopes, which carry light into the body and an image back out.
Prisms
- In a 45°–45°–90° glass prism, light that enters one short face at 90° passes straight in and meets the long face at 45°. Since 45° is greater than the critical angle (about 42°), it is totally internally reflected and turned through 90° — used in periscopes.
- Light entering the long face at 90° is totally internally reflected twice and turned through 180° — used in binoculars and bicycle reflectors.
- Prisms are better than mirrors here: total internal reflection reflects all of the light, and there is no silvering to tarnish or produce a double image.
A ray of light enters the flat end of a straight glass optical fibre, from air, with an angle of incidence of 40°. The refractive index of the glass is 1.50.
(a) Calculate the angle of refraction at the end of the fibre. (b) Show that the ray meets the side wall of the fibre at about 65° to the normal. (c) Explain whether the ray stays inside the fibre.
(a) \(\sin r=\dfrac{\sin 40^\circ}{1.50}=0.429\Rightarrow r=25.4^\circ\approx\) 25°
(b) The normal to the side wall is at 90° to the normal at the end face, so the angle of incidence at the wall = 90° − 25.4° = 64.6° ≈ 65°.
(c) Critical angle: \(\sin c=1/1.50\Rightarrow c=41.8^\circ\). 64.6° is greater than 41.8°, so the ray is totally internally reflected. Each reflection is at the same angle, so it stays inside the fibre all the way along.
What kind of wave is sound? Single · Double · Physics
- Sound is a longitudinal wave: the particles of the medium vibrate backwards and forwards along the direction the sound travels, making compressions and rarefactions. Double · Physics
- Sound needs a medium (solid, liquid or gas) — it cannot travel through a vacuum.
- Sound can be reflected: an echo is reflected sound. Hard, flat surfaces reflect sound well; soft materials such as curtains absorb it.
- Sound can be refracted: it changes speed and direction when it passes into a different medium, or through air at a different temperature. Sound travels faster in warm air than in cold air.
Single Award wording: “know that sound waves can be reflected and refracted” — you do not need “longitudinal”. Double Award wording differs only by one word (“which” for “that”); the content is identical to Physics.
What frequencies can humans hear? Physics only · 2P
- Sounds above 20 000 Hz are too high-pitched to hear; sounds below 20 Hz are too low.
Practical: how do you measure the speed of sound in air? Physics only · 2P
Investigating the speed of sound in air (echo method)
- Aim: measure the speed of sound in air by timing an echo.
- Apparatus: two wooden blocks to clap (or a starting pistol), stopwatch, long tape measure or trundle wheel, a large flat wall with open space in front of it.
- Measurements: distance from the person clapping to the wall, d; time between the clap and its echo, t.
- Control variables: the same position and the same wall for every reading; measure on the same day (the speed of sound depends on the air temperature).
Method
- Measure a distance of at least 50 m from the wall with the tape measure.
- Clap the blocks once. A second student starts the stopwatch at the sight of the clap and stops it on hearing the echo.
- Repeat several times and calculate the mean time t.
- Calculate speed = distance ÷ time \(=\dfrac{2d}{t}\).
Errors and improvements
- The time is short (about 0.3 s for 50 m), so reaction time is a large fraction of it. Improve this by clapping in a steady rhythm so each clap coincides with the echo of the last one, timing 20 claps and dividing: the distance travelled is then 20 × 2d.
- Use a larger distance, so the time is longer compared with the reaction time.
- Alternative: place two microphones a measured distance apart (for example 1.20 m), connected to a fast electronic timer or a dual-trace oscilloscope. Make a sharp sound in line with them and measure the time delay (for example 3.5 ms): speed = 1.20 ÷ 0.0035 = 343 m/s. This removes reaction time entirely.
- Safety: do not use a starting pistol without supervision; stay well away from roads.
A student stands 80 m from a cliff and bangs two blocks together. The mean time between the bang and the echo is 0.48 s. Calculate the speed of sound.
The sound travels to the cliff and back: distance = 2 × 80 = 160 m.
speed = \(\dfrac{160}{0.48}=333=\) 330 m/s
Common error: using 80 m instead of 160 m, which gives half the correct value (167 m/s).
How do a microphone and oscilloscope display a sound wave? Physics only · 2P
- The microphone converts the pressure variations of the sound wave into an alternating electrical signal (a voltage that changes in the same pattern as the sound).
- The oscilloscope plots this voltage (vertical) against time (horizontal) as a trace on its screen.
- The timebase sets how much time each horizontal division (square) represents, for example 0.2 ms/div.
- The trace shows the pattern of the vibration over time. Although sound is longitudinal, the trace looks like a transverse wave because it is a graph of voltage against time.
Practical: how do you find the frequency of a sound using an oscilloscope? Physics only · 2P
Investigating the frequency of a sound wave using an oscilloscope
- Aim: measure the frequency of a sound (from a tuning fork, or a loudspeaker driven by a signal generator) from its oscilloscope trace.
- Apparatus: sound source, microphone, oscilloscope (with leads).
- Measurements: the number of horizontal divisions for one complete cycle; the timebase setting.
- Control variables: the same distance from source to microphone, a quiet room, the same oscilloscope settings for each reading.
Method
- Connect the microphone to the oscilloscope input and hold it a fixed distance from the source.
- Adjust the timebase until two or more complete cycles fill the screen, and adjust the vertical gain so the trace is as tall as possible without going off the screen.
- Count the number of divisions for several complete cycles and divide by the number of cycles to find the divisions per cycle.
- Period \(T\) = divisions per cycle × timebase (convert ms to s).
- Frequency \(f=\dfrac{1}{T}\). Compare with the value marked on the tuning fork or the signal generator dial.
Errors and improvements
- Measuring across several cycles reduces the uncertainty in reading the start and end of one cycle.
- Choose a timebase that spreads the cycles out, so each cycle covers several divisions.
- Background noise distorts the trace: work in a quiet room.
- Safety: keep the loudspeaker volume low to protect hearing.
The trace above is produced by a microphone placed in front of a loudspeaker. The timebase is set to 0.2 ms per division.
(a) Determine the frequency of the sound. (b) The speed of sound is 340 m/s. Calculate the wavelength of the sound in air. (c) The frequency is doubled and the loudness kept the same. Describe how the trace changes.
(a) 4 complete cycles span 10 divisions, so one cycle = 2.5 divisions. \(T=2.5\times0.2\ \text{ms}=0.50\ \text{ms}=5.0\times10^{-4}\ \text{s}\). \(f=\dfrac{1}{T}=\dfrac{1}{5.0\times10^{-4}}=\) 2000 Hz
(b) \(\lambda=\dfrac{v}{f}=\dfrac{340}{2000}=\) 0.17 m
(c) Twice as many cycles across the screen — 8 cycles, each 1.25 divisions wide — with the same height (amplitude 3 divisions).
How are pitch and loudness linked to the vibrating source? Physics only · 2P
| Property you hear | Depends on | On the oscilloscope |
|---|---|---|
| pitch (how high or low) | the frequency of vibration of the source: higher frequency → higher pitch | more cycles across the screen (shorter period) |
| loudness | the amplitude of vibration of the source: larger amplitude → louder | a taller trace (greater peak height) |
- A guitar string plucked harder vibrates with a larger amplitude: the note is louder but the pitch is unchanged.
- A shorter or tighter string vibrates at a higher frequency: the note has a higher pitch.
- Compare traces only when the oscilloscope settings are the same.
Exam tips for 3(d)
The angles of incidence, reflection and refraction are all measured from the normal. In ray-tracing questions, the most common lost mark is an angle measured from the surface of the block.
\(n=\sin i/\sin r\) means find each sine, then divide. Dividing the two angles gives a wrong answer even when the equation has been written correctly — use the equation you wrote down.
To explain total internal reflection, give both conditions: light travelling from glass (denser) towards air, and angle of incidence greater than the critical angle. Quote numbers if you have them (“45° > 42°”).
In echo calculations the sound covers twice the distance to the wall. Write “distance = 2 × …” as your first line so it is clear you have accounted for it.
Count divisions for several cycles, multiply by the timebase, convert ms to s, then use \(f=1/T\). When asked to draw a trace, get both the number of cycles and the height right — each earns its own mark.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Light bends away from the normal when it enters glass. | It slows down in glass. | Entering a denser material, the ray bends towards the normal. |
| Total internal reflection can happen when light goes from air into glass. | There is always a refracted ray going into the denser material. | TIR only happens from the denser to the less dense material, above the critical angle. |
| Refractive index has the unit degrees. | It is a ratio of two sines. | \(n\) has no unit. |
| Louder sounds have a higher pitch. | Loudness and pitch depend on different properties. | Loudness ↔ amplitude; pitch ↔ frequency. |
| The loudspeaker should be connected to the oscilloscope to display the sound. | The oscilloscope must detect the sound wave itself. | The microphone converts the sound into an electrical signal for the oscilloscope. |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| angle of incidence = angle of reflection | the direction of a reflected ray | All | no — learn it |
| refractive index = sin(angle of incidence) ÷ sin(angle of refraction), \(n=\dfrac{\sin i}{\sin r}\) | n, or an angle of refraction | Double · Physics | yes — all papers |
| \(\sin c=\dfrac{1}{n}\) | the critical angle, or n from c | Double · Physics | yes — all papers |
| average speed = distance moved ÷ time taken (with distance = 2d for an echo) | the speed of sound | Physics only · 2P | yes — all papers |
| \(f=\dfrac{1}{T}\), with T = divisions × timebase | the frequency from an oscilloscope trace | Physics only · 2P | yes — all papers |
Pearson includes the Equation List with the question papers, but statements 3.18 and 3.22 say “know and use”: you may be asked to state the equation for a mark, so learn them.
Check your understanding
1. State the law of reflection. All
The angle of incidence equals the angle of reflection, both measured from the normal.
2. Explain what is meant by the critical angle. All
The angle of incidence in the denser material (e.g. glass) for which the angle of refraction in air is 90°. Above it, total internal reflection occurs.
3. Describe how an optical fibre carries a signal from one end to the other. All
Light enters the fibre and hits the side at an angle greater than the critical angle, so it is totally internally reflected. This repeats along the fibre, so the light pulses (the signal) reach the far end.
4. Light enters a Perspex block with i = 35° and r = 22°. Calculate the refractive index of the Perspex. Double · Physics
\(n=\dfrac{\sin35^\circ}{\sin22^\circ}=\dfrac{0.574}{0.375}=\) 1.53
5. The refractive index of water is 1.33. Calculate the critical angle for water. Double · Physics
\(\sin c=\dfrac{1}{1.33}=0.752\Rightarrow c=48.8^\circ\approx\) 49°
6. A ray of light travels into a rectangular glass block at an angle and out of the opposite face. Describe the path of the emerging ray. All
It bends towards the normal on entering, away from the normal on leaving, and emerges parallel to the incident ray, displaced sideways.
7. (a) State the frequency range of human hearing. (b) Calculate the wavelengths in air of the lowest and highest frequencies. (speed of sound = 340 m/s) Physics only · 2P
(a) 20 Hz to 20 000 Hz. (b) 340 ÷ 20 = 17 m; 340 ÷ 20 000 = 0.017 m (1.7 cm).
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