Level 4 · Science S4.10

Waves

How sound, water waves, light and radio carry energy from place to place.

Start with a rope and a spring to see the two kinds of wave. Measure amplitude, wavelength, period and frequency and use v = fλ. Follow ripples, sound and ultrasound, then rays of light at mirrors, edges, fibres and lenses. End with the whole electromagnetic spectrum, how it carries calls and pages, and how its high frequency end can harm living cells.

You are ready for this book if

You can explain why a stick looks bent in water.

Written and drawn by attnom. Free to read, print and share.

S4.10.1 Chapter 1

Kinds of waves

Mei flicks one end of a long rope, and a bump runs to the other end. What actually moved along the rope?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

A wave is a disturbance that travels from one place to another and carries energy with it. A rope, a spring, the surface of a pond and the air can all carry waves.

Tie a ribbon to the middle of a long rope. Mei holds one end and flicks it up and down once. A bump runs along the rope. When the bump reaches the ribbon, the ribbon rises. Then it comes down again.

After the bump has passed, the ribbon is where it was before. The bump went on to the far end. The rope did not.

Draw it first

A bump runs past a ribbon

  1. Three pictures of a rope, one under the other, with a ribbon tied to it. A dashed line runs down through the ribbon. Row 1, labelled 0 s: a bump near the left end, labelled pulse, and the ribbon at rest, labelled ribbon. Row 2, labelled 1 s: the bump is under the ribbon and the ribbon is raised. Row 3, labelled 2 s: the bump has moved to the right and the ribbon is back at rest on the same dashed line. An arrow at the foot is labelled wave travels.0 spulseribbonwave travels

    1 At 0 s the bump is near Mei and the ribbon is at rest.

  2. Three pictures of a rope, one under the other, with a ribbon tied to it. A dashed line runs down through the ribbon. Row 1, labelled 0 s: a bump near the left end, labelled pulse, and the ribbon at rest, labelled ribbon. Row 2, labelled 1 s: the bump is under the ribbon and the ribbon is raised. Row 3, labelled 2 s: the bump has moved to the right and the ribbon is back at rest on the same dashed line. An arrow at the foot is labelled wave travels.0 s1 spulseribbonwave travels

    2 At 1 s the bump is under the ribbon, and the ribbon is at its highest.

  3. Three pictures of a rope, one under the other, with a ribbon tied to it. A dashed line runs down through the ribbon. Row 1, labelled 0 s: a bump near the left end, labelled pulse, and the ribbon at rest, labelled ribbon. Row 2, labelled 1 s: the bump is under the ribbon and the ribbon is raised. Row 3, labelled 2 s: the bump has moved to the right and the ribbon is back at rest on the same dashed line. An arrow at the foot is labelled wave travels.0 s1 s2 spulseribbonwave travels

    3 At 2 s the bump has gone on. The ribbon is back at rest, on the same dashed line. It has not moved along the rope.

The bump carried energy from Mei's hand to the far end. At the far end it can shake a hand tied there. So a wave carries energy from place to place. It does not carry the material it travels through.

Now let Mei keep flicking, up and down, again and again. A steady train of bumps and dips runs along the rope. The highest point of a wave is a crest. The lowest point is a trough. The dashed line is where the rope lay at rest.

A graph of a wave with the axes distance along the rope and sideways move. A smooth wave rises above a dashed line labelled rest line, then falls below it, then rises again. A dot at the top of the first hill is labelled crest. A dot at the bottom of the first dip is labelled trough.cresttroughsideways movedistance along the roperest line
A wave on a rope. The picture is a snapshot: each piece of rope is at its own height.

A piece of the rope moves up and down. The wave travels along the rope. These two directions are at right angles, which is 90°. A wave like this is a transverse wave.

A dot with a double headed arrow running up and down through it, labelled vibration. Below, a long arrow pointing right, labelled wave travels. The two arrows are at right angles.vibrationwave travels
In a transverse wave the vibration is across the way the wave travels.

Light is a transverse wave. So are the ripples on a pond, to a good first approximation. Strictly, each bit of water moves in a small circle. We treat ripples as transverse in this book.

Now take a long soft spring and push one end sharply towards the other end, along its length. The coils bunch up, and the bunch runs along the spring. Where coils are pressed together there is a compression. Where they are spread apart there is a rarefaction.

Two rows of upright lines, which are layers of a spring. In the top row the layers bunch together in two places and spread out in others. The bunched place is labelled compression and the spread place is labelled rarefaction. One thick layer has a short double headed arrow above it labelled vibration. In the lower row, labelled a moment later, the bunches have moved to the right. A dashed line joins a compression to its new place. The thick layer is almost where it was. An arrow at the foot is labelled wave travels.vibrationcompressionrarefactiona moment laterwave travels
A longitudinal wave on a spring. The compressions move to the right. The thick layer only shakes to and fro.

Look at the thick layer. It moves a little to the right and a little to the left, along the way the wave travels. This is a longitudinal wave. The vibration is along the direction of travel.

A dot with a double headed arrow running left and right through it, labelled vibration. Below, a long arrow pointing right, labelled wave travels. The two arrows lie on the same line.vibrationwave travels
In a longitudinal wave the vibration is along the way the wave travels.

Sound in air is a longitudinal wave. A loudspeaker pushes the air layers together and pulls them apart. The compressions and rarefactions run out through the air. In the air, as on the spring, each layer only shakes to and fro.

See it three ways

As a picture

A table with the columns Kind of wave and Vibration. Row 1: transverse, across. Row 2: longitudinal, along.Kind of waveVibrationtransverseacrosslongitudinalalong

In words

Across means at right angles to the way the wave travels. Along means on the same line as the way the wave travels.

In symbols

Transverse: vibration at 90° to travel. Longitudinal: vibration at 0° to travel, along the same line.

A table with the columns Transverse and Longitudinal. Row 1: rope, spring. Row 2: light, sound. Row 3: ripples, a pushed row.TransverseLongitudinalropespringlightsoundripplesa pushed row
Examples of the two kinds. The pushed row is the row of buttons in the second Try it.

A rope, a spring, water and air are all a medium: the material a wave travels through. Waves on a rope or a spring, ripples and sound all need a medium. Light does not. Sunlight crosses empty space to reach us, so light needs no medium.

Now think of a cork floating on a pond. Ripples pass it. The cork bobs up and down, and when the ripples have gone it is about where it was. The ripples carried energy past it. They did not carry the water or the cork along with them.

Three pictures of the surface of a pond, one under the other, with a cork on it. A dashed line runs down through the cork. Row 1, labelled 0.0 s: a ripple near the left, labelled ripple, and the cork at rest, labelled cork. Row 2, labelled 0.5 s: the ripple is under the cork and the cork is raised. Row 3, labelled 1.0 s: the ripple has moved to the right and the cork is back at rest on the same dashed line. An arrow at the foot is labelled wave travels.0.0 s0.5 s1.0 sripplecorkwave travels
A floating cork bobs as a ripple passes and stays where it is. Energy goes by, water does not.

How fast does a wave travel? The speed of a wave is the distance its pattern covers divided by the time it takes. You met this rule for moving things in the speed book: speed = distance ÷ time.

A bump that runs 10 m along a rope in 4 s has a speed of 10 m ÷ 4 s = 2.5 m/s. Waves in different materials have different speeds. The later chapters measure some.

Try it

Send a bump down a rope

You need: a long rope or string, about 5 m, a ribbon, a scrap of cloth or a piece of paper, a friend who can hold the far end, a post or a chair to tie the far end to, if you have no friend.

Be safe: Do this on flat ground, away from people walking past. Flick the rope sideways and low, not towards anyone's face.

  1. Tie the ribbon to the middle of the rope with a loose knot. Lay the rope on a flat floor, or hold it just off the ground.
  2. Hold one end. Your friend holds the other end, or ties it to the post. Pull the rope almost straight, not tight.
  3. Flick your end sideways once, quickly. Watch the bump run along the rope.
  4. Watch the ribbon as the bump reaches it. Then watch where the ribbon is after the bump has gone.
  5. Now flick your end again and again, at a steady pace. Watch how the ribbon moves.

What you will see: The ribbon moves sideways as the bump passes and then goes back to its place. It does not travel to the far end. The bump did.

If you cannot try it, this is what happens.
Three pictures of a rope with a ribbon, one under the other, with a dashed line down through the ribbon. Row start: a bump near the left end, labelled pulse, and the ribbon labelled ribbon at rest. Row under: the bump is under the ribbon and the ribbon is raised. Row past: the bump has gone on to the right and the ribbon is back at rest on the same dashed line. An arrow at the foot is labelled wave travels.startunderpastpulseribbonwave travels

Try it

A pushed row of buttons

You need: seven flat buttons of the same size, a smooth table.

Be safe: Keep the buttons away from young children, who can swallow them. Slide them along the table, not towards anyone.

  1. Put six buttons in a straight line on the table, touching each other.
  2. Slide the seventh button along the table so that it hits the end of the line, not hard.
  3. Watch which buttons move.

What you will see: The button at the far end of the line jumps away. The buttons in the middle hardly move. The push went along the row and carried energy to the last button. This is a longitudinal pulse.

If you cannot try it, this is what happens.
A table with the columns Button and What happens. Row 1: hits the end, stops. Row 2: middle ones, stay. Row 3: far end, jumps away.ButtonWhat happenshits the endstopsmiddle onesstayfar endjumps away

A wrong idea many people have

A wave carries water, or rope, or air, from one place to another.

Why it feels right: You see a bump run along a rope and ripples spread over a pond. It looks as if the stuff itself is moving.

The test that breaks it: Float a cork or tie a ribbon on. After the wave has gone by, the cork or ribbon is where it was. A wave that carried the water along would take it too.

What is true: A wave carries energy and a pattern from place to place. The material only moves a little to and fro, or up and down, and stays near where it started.

A wrong idea many people have

Sound is a wave like a rope wave, with the air moving up and down.

Why it feels right: Pictures of sound waves are drawn as a wavy line, which looks just like a rope wave.

The test that breaks it: Look at the spring. The layers move along the line of travel, not across it. Air layers do the same, so sound is not a rope type wave.

What is true: Sound is a longitudinal wave: the air shakes to and fro along the way the sound travels. The wavy line in a picture is a graph, not the path of the air.

A wrong idea many people have

Every kind of wave needs a material to travel through.

Why it feels right: Rope waves, ripples and sound all need rope, water or air, so it seems to be a rule for all waves.

The test that breaks it: The Sun is far off across empty space, and its light reaches us. If light needed a material, it could not cross the empty space between the Sun and the Earth.

What is true: Rope waves, ripples and sound need a medium. Light does not. Light crosses a vacuum, which is why we can see the Sun and the stars.

Key words

wave
a disturbance that travels and carries energy.
crest
the highest point of a wave.
trough
the lowest point of a wave.
transverse
a wave whose vibration is at right angles to the way it travels.
compression
a place in a longitudinal wave where the material is pressed together.
rarefaction
a place in a longitudinal wave where the material is spread apart.
longitudinal
a wave whose vibration is along the way it travels.

Back to your guess

The answer is B: The bump. Each piece of rope only moved up and down a little.

A wave is a disturbance that travels. Each piece of the rope moved up and down and stayed near its place. What travelled from Mei to the far end was the bump, and with it energy. The rope itself did not travel.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A ribbon is tied to the middle of a rope. A bump runs along the rope and past the ribbon. What does the ribbon do?

    Tap your answer.
    1. Yes, that is right. The bump travels along the rope. Each piece of the rope moves up and down as the bump passes, and then comes back to rest in its place.

    2. Not this one. It is a natural thing to think, because you see the bump travel. But the bump is a disturbance. The ribbon is tied to the rope, which does not travel along. Try another.

    3. Not this one. The bump is made of the rope itself moving. As the bump passes, the piece of rope with the ribbon rises, so the ribbon rises too. Try another.

    4. Not this one. Each piece of rope moves across the rope, not along it. After the bump passes, the ribbon is back where it began. Try another.

    See the answer

    It moves up, then comes back down, in the same place along the rope.

    The bump travels along the rope. Each piece of the rope moves up and down as the bump passes, and then comes back to rest in its place.

    • If you chose B: It is a natural thing to think, because you see the bump travel. But the bump is a disturbance. The ribbon is tied to the rope, which does not travel along.
    • If you chose C: The bump is made of the rope itself moving. As the bump passes, the piece of rope with the ribbon rises, so the ribbon rises too.
    • If you chose D: Each piece of rope moves across the rope, not along it. After the bump passes, the ribbon is back where it began.
  2. The picture shows a dot that vibrates and the way a wave travels. Which kind of wave is it?

    A dot with a double headed arrow running left and right through it, labelled vibration. Below, a long arrow pointing right, labelled wave travels.vibrationwave travels
    Tap your answer.
    1. Not this one. In a transverse wave the vibration is at right angles to the way the wave travels. Here the two arrows are on one line. Try another.

    2. Not this one. The dot only shakes to and fro about its place. It does not go along with the wave. What travels is the disturbance. Try another.

    3. Yes, that is right. When the vibration is along the way the wave travels, the wave is longitudinal. Sound in air and pushes along a spring are like this.

    See the answer

    A longitudinal wave

    When the vibration is along the way the wave travels, the wave is longitudinal. Sound in air and pushes along a spring are like this.

    • If you chose A: In a transverse wave the vibration is at right angles to the way the wave travels. Here the two arrows are on one line.
    • If you chose B: The dot only shakes to and fro about its place. It does not go along with the wave. What travels is the disturbance.
  3. Which of these is a longitudinal wave?

    Tap your answer.
    1. Not this one. Light is a transverse wave. Its vibration is at right angles to the way it travels. Try another.

    2. Not this one. A sideways flick makes a transverse wave. The rope moves across the way the wave goes. Try another.

    3. Not this one. A ripple looks like a transverse wave, because the water goes up and down as it passes. We treat ripples as transverse. Try another.

    4. Yes, that is right. The air layers shake to and fro along the way the sound goes, so they make compressions and rarefactions. That is a longitudinal wave.

    See the answer

    Sound travelling through the air

    The air layers shake to and fro along the way the sound goes, so they make compressions and rarefactions. That is a longitudinal wave.

    • If you chose A: Light is a transverse wave. Its vibration is at right angles to the way it travels.
    • If you chose B: A sideways flick makes a transverse wave. The rope moves across the way the wave goes.
    • If you chose C: A ripple looks like a transverse wave, because the water goes up and down as it passes. We treat ripples as transverse.
  4. A wave travels along a rope from Mei to Ravi. What does the wave carry from Mei to Ravi?

    Tap your answer.
    1. Not this one. The pieces of rope stay near their places. If they travelled, the rope would end up in a heap at Ravi's end. Try another.

    2. Yes, that is right. A wave carries energy from place to place. The material it travels through moves only a little and stays near its place.

    3. Not this one. Only the energy travels. The rope moves up and down and stays where it was. Try another.

    See the answer

    Energy

    A wave carries energy from place to place. The material it travels through moves only a little and stays near its place.

    • If you chose A: The pieces of rope stay near their places. If they travelled, the rope would end up in a heap at Ravi's end.
    • If you chose C: Only the energy travels. The rope moves up and down and stays where it was.
  5. A cork floats on a pond. A line of ripples passes it. When the ripples have gone, where is the cork?

    Tap your answer.
    1. Not this one. Ripples carry energy, not water. A cork carried to the shore would mean the ripples had pushed the water there too. Try another.

    2. Yes, that is right. The water under the cork moves up and down as each ripple passes. The ripple moves on across the pond and the cork stays about where it was.

    3. Not this one. Nothing pulls the cork back to the start. It just bobs as the water goes up and down under it. Try another.

    See the answer

    About where it was, after bobbing up and down.

    The water under the cork moves up and down as each ripple passes. The ripple moves on across the pond and the cork stays about where it was.

    • If you chose A: Ripples carry energy, not water. A cork carried to the shore would mean the ripples had pushed the water there too.
    • If you chose C: Nothing pulls the cork back to the start. It just bobs as the water goes up and down under it.
  6. The picture shows a spring with a wave. In which part of the spring are the coils pressed closest together?

    A row of upright lines, which are layers of a spring. They bunch together in two places and spread apart in others. An arrow at the foot points right and is labelled wave travels.wave travels
    Tap your answer.
    1. Not this one. A rarefaction is where the coils are spread apart. The word means "made thinner". Try another.

    2. Not this one. Crests and troughs belong to a transverse wave. A longitudinal wave has compressions and rarefactions. Try another.

    3. Yes, that is right. Coils pressed together make a compression. Coils spread apart make a rarefaction.

    4. Not this one. The coils are pressed together where a compression is. A compression is not tied to a fixed place. It travels along the spring. Try another.

    See the answer

    In a compression

    Coils pressed together make a compression. Coils spread apart make a rarefaction.

    • If you chose A: A rarefaction is where the coils are spread apart. The word means "made thinner".
    • If you chose B: Crests and troughs belong to a transverse wave. A longitudinal wave has compressions and rarefactions.
    • If you chose D: The coils are pressed together where a compression is. A compression is not tied to a fixed place. It travels along the spring.
  7. Which kind of wave does not need a material to travel through?

    Tap your answer.
    1. Not this one. Sound needs air, water or a solid. Nothing is there to shake in a vacuum, so sound cannot cross one. Go to S3.9.6. Try another.

    2. Not this one. A rope wave is the rope moving, so it needs the rope. Try another.

    3. Not this one. Ripples are the water surface moving, so they need the water. Try another.

    4. Yes, that is right. Light can cross empty space. That is how sunlight reaches the Earth. Sound, rope waves and ripples all need a medium.

    See the answer

    Light

    Light can cross empty space. That is how sunlight reaches the Earth. Sound, rope waves and ripples all need a medium.

    • If you chose A: Sound needs air, water or a solid. Nothing is there to shake in a vacuum, so sound cannot cross one. Go to S3.9.6.
    • If you chose B: A rope wave is the rope moving, so it needs the rope.
    • If you chose C: Ripples are the water surface moving, so they need the water.
  8. A bump takes 2 s to run 6 m along a rope. How fast does it travel?

    See the answer

    3 m/s

    Speed = distance ÷ time = 6 m ÷ 2 s = 3 m/s.

Keep it

Copy this into your exercise book.

A wave carries energy, not material. Transverse: vibration at 90° to travel (rope, light). Longitudinal: vibration along travel (sound, spring). Light needs no medium.

Draw: A rope with a ribbon, a bump in three positions, and the ribbon staying on one dashed line. Beside it, a spring with one compression.

S4.10.2 Chapter 2

Measuring waves

Kofi sends a wave along a rope. Ana asks how big the wave is. Where does Kofi put his ruler?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

  • You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

To describe a wave with numbers, we need four measures. They are the amplitude, the wavelength, the period and the frequency. Each one has a unit, and each one is measured in its own way.

Start with the picture. We draw a wave on a graph. The up and down axis shows the displacement. This is how far a piece of the rope has moved from its rest place, in a unit such as cm. The dashed line through 0 is the rest line, where the rope lies when no wave is on it.

A graph of displacement against distance is a photograph of the rope at one moment. It shows the shape of the wave along the rope.

Draw it first

A rope wave on a distance graph

  1. A graph with the axes displacement (cm) and distance (m). The displacement runs from −6 to 6 cm and the distance from 0 to 2 m in steps of 0.4. Step 1: a smooth wave that rises to 2 cm at 0.4 m, falls to −2 cm at 1.2 m and rises to 2 cm at 2.0 m. Step 2: an arrow from the dashed rest line at 0 up to the first crest, labelled amplitude 2 cm. Step 3: an arrow across from the first crest to the next crest, labelled wavelength 1.6 m, with dashed lines down to the crests.−6−4−2024600.40.81.21.62displacement (cm)distance (m)

    1 Draw the wave. The dashed line through 0 is the rest line.

  2. A graph with the axes displacement (cm) and distance (m). The displacement runs from −6 to 6 cm and the distance from 0 to 2 m in steps of 0.4. Step 1: a smooth wave that rises to 2 cm at 0.4 m, falls to −2 cm at 1.2 m and rises to 2 cm at 2.0 m. Step 2: an arrow from the dashed rest line at 0 up to the first crest, labelled amplitude 2 cm. Step 3: an arrow across from the first crest to the next crest, labelled wavelength 1.6 m, with dashed lines down to the crests.−6−4−2024600.40.81.21.62amplitude 2 cmdisplacement (cm)distance (m)

    2 The amplitude is the greatest displacement. Measure it from the rest line up to a crest. Here it is 2 cm. A trough is 2 cm below the rest line, so it gives the same amplitude.

  3. A graph with the axes displacement (cm) and distance (m). The displacement runs from −6 to 6 cm and the distance from 0 to 2 m in steps of 0.4. Step 1: a smooth wave that rises to 2 cm at 0.4 m, falls to −2 cm at 1.2 m and rises to 2 cm at 2.0 m. Step 2: an arrow from the dashed rest line at 0 up to the first crest, labelled amplitude 2 cm. Step 3: an arrow across from the first crest to the next crest, labelled wavelength 1.6 m, with dashed lines down to the crests.−6−4−2024600.40.81.21.62amplitude 2 cmwavelength 1.6 mdisplacement (cm)distance (m)

    3 The wavelength is the distance between two matching points on neighbouring waves, such as crest to crest. Here it is 1.6 m. Its symbol is λ, the Greek letter lambda.

Amplitude is a distance from the rest line, so it is never negative. A bigger amplitude means more energy in the wave. For sound, a bigger amplitude makes a louder note. For a ripple, it makes a taller ripple.

Wavelength is a length along the way the wave travels, so its unit is the metre, m. You can measure it from crest to crest, trough to trough, or between any two matching points on neighbouring waves.

Now tie a ribbon to the rope and watch it. Its displacement changes over time: up, back to the rest line, down, and up again. A graph of displacement against time shows this. It is not a photograph. It is a record of one place, as time goes by.

A graph with the axes displacement (cm) and time (s). The time runs from 0 to 2.5 s in steps of 0.5 and the displacement from −4 to 4 cm. A smooth wave rises to 2 cm at 0.5 s, falls to −2 cm at 1.5 s and rises to 2 cm again at 2.5 s. An arrow across from the first crest to the next crest is labelled period, with dashed lines down to the crests.−4−3−2−10123400.511.522.5perioddisplacement (cm)time (s)
The ribbon on the rope. The time from one crest to the next is the period.

The period T is the time for one complete wave to pass a point. It is also the time for one complete vibration. Its unit is the second, s. On the graph, it is the time between two crests.

The frequency f is the number of complete waves that pass a point each second. Its unit is the hertz, Hz. One hertz is one wave each second. A source that makes 5 waves each second has a frequency of 5 Hz. One kilohertz, kHz, is 1 000 Hz.

Period and frequency are two ways to say the same thing. If 5 waves pass each second, then one wave takes 1 ÷ 5 = 0.2 s. So f = 1 ÷ T and T = 1 ÷ f. A short period means a high frequency.

A graph with the axes displacement and time, drawn without numbers. Two waves of the same length, one over the other. The solid wave is twice as tall as the dashed wave. The key at the top says loud for the solid wave and quiet for the dashed wave.displacementtimeloudquiet
Same frequency, different amplitude. The taller wave is the louder sound.
A graph with the axes displacement and time, drawn without numbers. Two waves of the same height. The solid wave has twice as many waves in the same time as the dashed wave. The key at the top says high note for the solid wave and low note for the dashed wave.displacementtimehigh notelow note
Same amplitude, different frequency. More waves each second is a higher note.

Now put the measures together. In one period T, a wave moves forward by one wavelength λ. So its speed is the distance divided by the time: v = λ ÷ T. Since 1 ÷ T is the frequency f, this becomes v = fλ.

The wave equation is v = fλ. Speed v is in metres a second, m/s. Frequency f is in hertz, Hz. Wavelength λ is in metres, m. The units agree: m × (1 ÷ s) = m/s.

See it three ways

As a picture

A table with the columns Measure, Symbol and Unit. Row 1: speed, v, m/s. Row 2: frequency, f, Hz. Row 3: wavelength, λ, m. Row 4: period, T, s.MeasureSymbolUnitspeedvm/sfrequencyfHzwavelengthλmperiodTs

In words

The speed of a wave is its frequency times its wavelength. The frequency is one divided by the period.

In symbols

v = fλ, f = 1 ÷ T, T = 1 ÷ f. Rearranged: λ = v ÷ f and f = v ÷ λ.

A graph with the axes displacement (cm) and distance (m). The distance runs from 0 to 0.75 m in steps of 0.15 and the displacement from −6 to 6 cm. A smooth wave has crests at 0.15 m and 0.75 m. An arrow from crest to crest is labelled wavelength 0.6 m, with dashed lines down to the crests.−6−4−2024600.150.30.450.60.75wavelength 0.6 mdisplacement (cm)distance (m)
The rope wave of the next worked example. Its wavelength is 0.6 m.

Worked example

Waves on a rope have a frequency of 2.5 Hz and a wavelength of 0.6 m. Find their speed.

  1. Write the equation: v = fλ.
  2. Put in the numbers: v = 2.5 Hz × 0.6 m.
  3. v = 1.5 m/s.

Answer: 1.5 m/s

Sound in air travels at about 340 m/s, whatever the note. So a note of higher frequency must have a shorter wavelength. Rearrange the equation to find the wavelength: λ = v ÷ f.

Two rows of upright lines, which are layers of air. In the top row the layers bunch together in two places and spread out in others. The bunched places are labelled compression and the spread places are labelled rarefaction. An arrow from one compression to the next is labelled wavelength. In the lower row, labelled a moment later, the bunches have moved to the right. An arrow at the foot is labelled wave travels.wavelengthcompressionrarefactiona moment laterwave travels
A sound wave in air. Its wavelength is the distance from one compression to the next.

Worked example

Sound travels at 340 m/s in air. A note has a frequency of 0.8 kHz. What is its wavelength?

  1. Change the frequency to hertz: 0.8 kHz = 0.8 × 1 000 = 800 Hz.
  2. Rearrange v = fλ to λ = v ÷ f.
  3. Put in the numbers: λ = 340 m/s ÷ 800 Hz = 0.425 m.

Answer: 0.425 m

Both graphs together give the speed. The first graph, taken at one moment, shows the wavelength of the rope wave: 1.6 m. The second graph, taken at one place, shows the period: the time between two crests.

Worked example

The rope of the first graph has a wavelength of 1.6 m. The ribbon graph shows its period. Find the speed of the wave.

  1. From the ribbon graph, the first crest is at 0.5 s and the next crest is at 2.5 s. So T = 2.5 − 0.5 = 2 s.
  2. Find the frequency: f = 1 ÷ T = 1 ÷ 2 = 0.5 Hz.
  3. The wavelength is λ = 1.6 m.
  4. v = fλ = 0.5 Hz × 1.6 m = 0.8 m/s.

Answer: 0.8 m/s

A wave's speed is fixed by the medium it travels through. Change the frequency and the wavelength changes, but the speed does not. Raise the pitch of a note in air and the wavelength shortens. The sound still travels at about 340 m/s.

A graph with the axes displacement (cm) and distance (m). A smooth wave with a crest of 2 cm at 0.4 m and a trough of −2 cm at 1.2 m. One arrow from the dashed rest line at 0 up to the crest is labelled amplitude 2 cm. A second arrow at 1.2 m from the trough up to 2 cm is labelled 4 cm.−6−4−2024600.40.81.21.62amplitude 2 cm4 cmdisplacement (cm)distance (m)
The arrow from trough to crest is 4 cm, twice the amplitude. Measure from the rest line.
A graph with the axes displacement (cm) and distance (m). A smooth wave with crests of 2 cm at 0.4 m and 2.0 m and a trough of −2 cm at 1.2 m. A long arrow from the first crest to the next crest is labelled wavelength 1.6 m. A shorter arrow below the wave, from the first crest to the trough, is labelled 0.8 m, with dashed lines up to the crest and the trough.−6−4−2024600.40.81.21.62wavelength 1.6 m0.8 mdisplacement (cm)distance (m)
The short arrow runs from a crest to a trough. It is only half a wavelength. A wavelength is crest to crest.

Try it

The period of a swinging stone

You need: a small stone, about the size of an egg, about 1 m of string, a high hook, a nail in a beam or a branch to tie it to, a clock or phone that shows seconds.

Be safe: Hang the stone clear of faces and windows, and keep other people away from its path. Use a small stone and tie it well, so that it cannot fly off. Do not tie it where you must climb.

  1. Tie the stone firmly to the string. Tie the other end of the string up high so the stone hangs in the air.
  2. Pull the stone a short way to one side and let go. A swing to and fro and back to the start is one full swing.
  3. Start the clock when the stone passes its lowest point. Count 10 full swings and stop the clock.
  4. Work out the period: T = time for 10 swings ÷ 10.
  5. Work out the frequency: f = 1 ÷ T. Then try a shorter string, about 25 cm, and compare.

What you will see: The stone takes about 2 s for each swing on a 1 m string, so its frequency is about 0.5 Hz. A shorter string swings faster: its period is shorter and its frequency is higher. Timing 10 swings gives a better answer than timing one.

If you cannot try it, this is what happens.
A table with the columns Measure and 1 m string. Row 1: 10 swings, 20 s. Row 2: T = 20 ÷ 10, 2 s. Row 3: f = 1 ÷ T, 0.5 Hz.Measure1 m string10 swings20 sT = 20 ÷ 102 sf = 1 ÷ T0.5 Hz

A wrong idea many people have

The amplitude is the height from the trough to the crest.

Why it feels right: The whole height of a wave is the first thing you see. A ruler held beside the wave reads from the bottom to the top.

The test that breaks it: Look at the graph with the two arrows. The distance from the rest line to the crest is 2 cm. The distance from trough to crest is 4 cm. These are not the same, and the first is the amplitude.

What is true: Amplitude is measured from the rest line to a crest, or from the rest line to a trough. Crest to trough is twice the amplitude.

A wrong idea many people have

The wavelength is the distance from a crest to the next trough.

Why it feels right: A crest and a trough are the two obvious points of a wave, and they are next to each other.

The test that breaks it: Measure crest to trough on the graph: 0.8 m. Now go on to the next crest: 1.6 m in all, and the wave has repeated. A full wave needs both a crest and a trough.

What is true: Wavelength is the distance between two matching points: crest to crest or trough to trough. Crest to trough is half a wavelength.

A wrong idea many people have

A wave of higher frequency travels faster.

Why it feels right: Frequency and speed both sound like "how fast", and a high note feels quick.

The test that breaks it: Sound in air travels at about 340 m/s for every note. A high note and a low note from the same drum reach you together. If higher frequency meant a higher speed, they would arrive at different times.

What is true: Frequency says how many waves pass each second. Speed says how far one wave goes each second. The medium sets the speed. With v = fλ, a higher frequency then means a shorter wavelength.

Key words

displacement
how far a piece of the medium is from its rest place, with a direction. It has a unit such as cm.
rest line
the line on a wave graph where the medium lies when no wave is on it.
amplitude
the greatest displacement, measured from the rest line to a crest.
wavelength
the distance between two matching points on neighbouring waves, in m. Its symbol is λ.
period
the time for one complete wave to pass a point, in s. Its symbol is T.
frequency
the number of waves that pass a point each second. Its symbol is f. f = 1 ÷ T.
hertz
the unit of frequency, written Hz. One hertz is one wave each second.
wave equation
v = fλ. The speed of a wave is its frequency times its wavelength.

Back to your guess

The answer is C: From the rest line up to a crest.

The size of a wave, its amplitude, is measured from the rest line to a crest. The distance from lowest to highest is twice that. The distance from crest to crest is the wavelength, which is a different measure.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The graph shows a wave on a rope. What is its amplitude?

    A graph with the axes displacement (cm) and distance (m). The displacement runs from −4 to 4 cm and the distance from 0 to 1.5 m in steps of 0.3. A smooth wave rises to 3 cm at 0.3 m, falls to −3 cm at 0.9 m and rises again to 3 cm at 1.5 m.−4−3−2−10123400.30.60.91.21.5displacement (cm)distance (m)
    Tap your answer.
    1. Not this one. That is the height from the trough to the crest, which is twice the amplitude. Measure from the rest line to the crest. Try another.

    2. Not this one. That is a length along the other axis. It is the wavelength, not a height. Try another.

    3. Not this one. The crest is higher than 2 cm on the displacement axis. Read the crest against the numbers on that axis. Try another.

    4. Yes, that is right. The crest is 3 cm above the rest line at 0, so the amplitude is 3 cm. The trough is 3 cm below it.

    See the answer

    3 cm

    The crest is 3 cm above the rest line at 0, so the amplitude is 3 cm. The trough is 3 cm below it.

    • If you chose A: That is the height from the trough to the crest, which is twice the amplitude. Measure from the rest line to the crest.
    • If you chose B: That is a length along the other axis. It is the wavelength, not a height.
    • If you chose C: The crest is higher than 2 cm on the displacement axis. Read the crest against the numbers on that axis.
  2. The graph shows how a ribbon on a rope moves as time goes by. What is the frequency of the wave?

    A graph with the axes displacement (cm) and time (s). The time runs from 0 to 1 s in steps of 0.2 and the displacement from −3 to 3 cm. A smooth wave has a crest of 2 cm at 0.2 s and the next crest at 1 s.−3−2−1012300.20.40.60.81displacement (cm)time (s)
    Tap your answer.
    1. Yes, that is right. The crests are at 0.2 s and 1.0 s, so the period is T = 0.8 s. The frequency is f = 1 ÷ 0.8 = 1.25 Hz.

    2. Not this one. You used 0.2 s, the time of the first crest. The period is the time between two crests. Try another.

    3. Not this one. 0.8 s is the period. The frequency is 1 divided by the period, and its unit is Hz. Try another.

    See the answer

    1.25 Hz

    The crests are at 0.2 s and 1.0 s, so the period is T = 0.8 s. The frequency is f = 1 ÷ 0.8 = 1.25 Hz.

    • If you chose B: You used 0.2 s, the time of the first crest. The period is the time between two crests.
    • If you chose C: 0.8 s is the period. The frequency is 1 divided by the period, and its unit is Hz.
  3. The graph is a photograph of a wave on a rope. What is its wavelength?

    A graph with the axes displacement (cm) and distance (m). The distance runs from 0 to 3 m in steps of 0.6 and the displacement from −3 to 3 cm. A smooth wave has a crest of 2 cm at 0.6 m, a trough of −2 cm at 1.8 m and the next crest at 3 m.−3−2−1012300.61.21.82.43displacement (cm)distance (m)
    Tap your answer.
    1. Not this one. That is the whole length of the picture, which holds more than one wave. Measure from crest to the next crest. Try another.

    2. Yes, that is right. The crests are at 0.6 m and 3.0 m. The distance from one crest to the next is 3.0 − 0.6 = 2.4 m.

    3. Not this one. That is the distance from crest to trough, which is only half a wavelength. Try another.

    4. Not this one. 2 cm is a height, the amplitude. A wavelength is a length along the rope. Try another.

    See the answer

    2.4 m

    The crests are at 0.6 m and 3.0 m. The distance from one crest to the next is 3.0 − 0.6 = 2.4 m.

    • If you chose A: That is the whole length of the picture, which holds more than one wave. Measure from crest to the next crest.
    • If you chose C: That is the distance from crest to trough, which is only half a wavelength.
    • If you chose D: 2 cm is a height, the amplitude. A wavelength is a length along the rope.
  4. A wave has a frequency of 20 Hz and a wavelength of 0.5 m. What is its speed?

    Tap your answer.
    1. Yes, that is right. v = fλ = 20 Hz × 0.5 m = 10 m/s.

    2. Not this one. You divided the frequency by the wavelength. The wave equation is a product: v = fλ. Try another.

    3. Not this one. You added 20 and 0.5. The speed is the frequency times the wavelength. Try another.

    4. Not this one. You divided the wavelength by the frequency. Multiply the two instead. Try another.

    See the answer

    10 m/s

    v = fλ = 20 Hz × 0.5 m = 10 m/s.

    • If you chose B: You divided the frequency by the wavelength. The wave equation is a product: v = fλ.
    • If you chose C: You added 20 and 0.5. The speed is the frequency times the wavelength.
    • If you chose D: You divided the wavelength by the frequency. Multiply the two instead.
  5. Ravi shakes a rope faster, so its waves have a higher frequency. The rope and its tension stay the same. What happens?

    Tap your answer.
    1. Not this one. The speed is set by the rope and its tension, not by how fast you shake it. A faster shake only makes more waves each second. Try another.

    2. Not this one. With v fixed, v = fλ means a bigger f needs a smaller λ. So the wavelength shortens. Try another.

    3. Yes, that is right. The speed stays the same because the rope is the same. Since v = fλ, a higher f must go with a shorter λ.

    See the answer

    The waves travel at the same speed and the wavelength gets shorter.

    The speed stays the same because the rope is the same. Since v = fλ, a higher f must go with a shorter λ.

    • If you chose A: The speed is set by the rope and its tension, not by how fast you shake it. A faster shake only makes more waves each second.
    • If you chose B: With v fixed, v = fλ means a bigger f needs a smaller λ. So the wavelength shortens.
  6. A guitar string vibrates 250 times each second. What is its period?

    Tap your answer.
    1. Not this one. 250 is the frequency in hertz. The period is 1 divided by it. Try another.

    2. Not this one. A period of 0.25 s would be a frequency of 4 Hz. Work out 1 ÷ 250 again. Try another.

    3. Not this one. You divided 1 000 by 250. The period is 1 ÷ 250, much shorter than a second. Try another.

    4. Yes, that is right. T = 1 ÷ f = 1 ÷ 250 = 0.004 s. A vibration this fast takes a very short time.

    See the answer

    0.004 s

    T = 1 ÷ f = 1 ÷ 250 = 0.004 s. A vibration this fast takes a very short time.

    • If you chose A: 250 is the frequency in hertz. The period is 1 divided by it.
    • If you chose B: A period of 0.25 s would be a frequency of 4 Hz. Work out 1 ÷ 250 again.
    • If you chose C: You divided 1 000 by 250. The period is 1 ÷ 250, much shorter than a second.
  7. A drum is hit harder, so the note is louder, but it stays the same note. What changes in the wave?

    Tap your answer.
    1. Not this one. A bigger frequency is a higher note, not a louder one. Pitch depends on frequency. Go to S3.9.8. Try another.

    2. Yes, that is right. Loudness depends on the amplitude. The note, which is the frequency, does not change, and the wavelength is v ÷ f, so it does not change either.

    3. Not this one. The note is the same, so the frequency is the same. The speed of sound in air is fixed, so the wavelength is the same too. Try another.

    See the answer

    The amplitude gets bigger. The frequency and wavelength stay the same.

    Loudness depends on the amplitude. The note, which is the frequency, does not change, and the wavelength is v ÷ f, so it does not change either.

    • If you chose A: A bigger frequency is a higher note, not a louder one. Pitch depends on frequency. Go to S3.9.8.
    • If you chose C: The note is the same, so the frequency is the same. The speed of sound in air is fixed, so the wavelength is the same too.
  8. From an earlier chapter

    In which kind of wave does the vibration run along the same line as the way the wave travels?

    Tap your answer.
    1. Not this one. In a transverse wave the vibration is at right angles to the way the wave travels, as in a rope. Go to S4.10.1. Try another.

    2. Not this one. Waves carry energy, not matter. The material only shakes about its place. Go to S4.10.1. Try another.

    3. Yes, that is right. In a longitudinal wave, such as sound or a push along a spring, the vibration is along the direction of travel.

    See the answer

    A longitudinal wave

    In a longitudinal wave, such as sound or a push along a spring, the vibration is along the direction of travel.

    • If you chose A: In a transverse wave the vibration is at right angles to the way the wave travels, as in a rope. Go to S4.10.1.
    • If you chose B: Waves carry energy, not matter. The material only shakes about its place. Go to S4.10.1.
  9. A wave has a frequency of 4 Hz and a wavelength of 0.75 m. Find its speed.

    See the answer

    3 m/s

    v = fλ = 4 Hz × 0.75 m = 3 m/s.

  10. A wave has a frequency of 40 Hz. What is its period?

    See the answer

    0.025 s

    T = 1 ÷ f = 1 ÷ 40 = 0.025 s.

Keep it

Copy this into your exercise book.

Amplitude: rest line to crest. Wavelength λ: crest to crest. Period T and frequency f = 1 ÷ T. Wave equation v = fλ. The medium sets v.

Draw: A wave on axes with the rest line. Add an arrow for the amplitude and an arrow for the wavelength. Write v = fλ beside it.

S4.10.3 Chapter 3

Waves on water

Priya's ripples run from deep water onto a shallow part of the tray. What happens to them?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

  • You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Drop a pebble into a pond and rings spread out from the place it fell. Each ring is a line of crests. Scientists study water waves in a shallow tray, seen from above, called a ripple tank.

Seen from above, a line that joins the crests is a wavefront. The wave travels at right angles to its fronts. A tap on the water makes circular wavefronts. A straight bar moved up and down makes straight ones.

Four circles round a dot, seen from above, each one the same distance outside the last. An arrow runs along a radius from the first circle to the second and is labelled wavelength.wavelength
Ripples from one tap. The rings are equally far apart. That gap is the wavelength.

The gap between two neighbouring rings, or two neighbouring straight fronts, is one wavelength. To measure it well, measure across several waves and divide. Do not measure a single gap, because the reading error is a bigger share of a small length.

A tray seen from above with five straight wavefronts, each the same distance from the next. An arrow above the tray runs from the first front to the last and is labelled 4 wavelengths. A label wavefront is under the first front. An arrow below the tray is labelled wave travels. The tray is labelled.tray4 wavelengthswavefrontwave travels
Five fronts have four gaps between them. Four gaps are four wavelengths.

Count the gaps, not the fronts. Five fronts leave four gaps, so the distance from the first front to the last is four wavelengths. A common slip is to divide by five.

To find the speed, use the wave equation from the last chapter: v = fλ. The frequency f is how many taps you make each second. The wavelength λ is the length you measure. Keep λ and v in the same length unit, such as cm and cm/s.

Worked example

Taps at 4 Hz make five fronts with 28 cm from the first to the last. Find the wavelength and the speed.

  1. Five fronts have four gaps. So four wavelengths together are 28 cm.
  2. One wavelength is λ = 28 cm ÷ 4 = 7 cm.
  3. The speed is v = fλ = 4 Hz × 7 cm = 28 cm/s.

Answer: 7 cm and 28 cm/s

Try it

Find the speed of ripples in a tray

You need: a flat tray or a wide dish, with water about 1.5 cm deep, a ruler, a pencil or a stick, a torch, a clock or phone that shows seconds, a phone camera to take a photo from above, if you have one.

Be safe: Keep the tray and the water away from sockets, wires and anything that runs on electricity. Wipe up spills at once, because a wet floor is slippery. Do not shine the torch into anyone's eyes.

  1. Put the tray on a table or the floor, on a dark cloth. Pour in water about 1.5 cm deep. Wait until the water is still.
  2. Lay the ruler flat on the edge of the tray where it shows along the water. Shine the torch low from the side, so the crests show as bright and dark lines.
  3. Tap the water with the tip of the pencil at one end, at a steady beat. Count your taps for 10 s and divide by 10. This is the frequency f in Hz. Try for about 4 taps each second.
  4. Look from straight above, or take a photo. Measure the distance across 5 fronts, which is 4 gaps. Divide by 4. This is the wavelength λ.
  5. Work out v = fλ.

What you will see: The ripples are a few centimetres apart and travel several tens of centimetres each second. Your numbers will differ with the depth of the water and how fast you tap. Measuring four gaps together is far better than measuring one.

If you cannot try it, this is what happens.
A table with the columns Measure and Value. Row 1: 40 taps in 10 s, f = 4 Hz. Row 2: 4 gaps, 32 cm. Row 3: wavelength, 8 cm. Row 4: speed 4 × 8, 32 cm/s.MeasureValue40 taps in 10 sf = 4 Hz4 gaps32 cmwavelength8 cmspeed 4 × 832 cm/s

Now change the depth. Once the water is shallower than about half a wavelength, water waves travel more slowly where it is shallower. Ripples in a tray 1.5 cm deep are in this range. This is also why ocean waves slow, shorten and rise as they reach a beach.

A tray of water seen from the side. The water is 1.5 cm deep at the left, marked with an arrow. At the right a plate lies on the bottom, shown with hatching and labelled plate, and the water over it is 0.5 cm deep, marked with an arrow. The water and the tray are labelled.1.5 cm0.5 cmwaterplatetray
A flat plate in the tray makes part of the water shallow.

Send straight ripples across the tray from the deep end to the plate. Here is what happens at the edge of the shallow part. The frequency does not change. The source makes the same number of waves each second, and the same number cross the edge each second.

The speed falls in the shallow water. The wave equation v = fλ then tells us the wavelength must fall too, because f stays the same. So the fronts bunch up in the shallow water. The new wavelength is λ = v ÷ f.

Next, let the ripples meet the edge at a slant. Each part of a front slows down as it reaches the shallow part. The part that arrives first slows first, so the front turns. It is like a cart whose wheel meets sand first and swings round.

To describe the turn, draw the normal: the line at 90° to the edge. The ripples travel along arrows at right angles to the fronts. In the slower, shallow water, the ripples turn towards the normal. This bending is refraction.

A square seen from above, split across the middle by an edge. The top half is deep water, labelled deep. The bottom half is shallow water, with dots, labelled shallow. Straight wavefronts slope across both halves. The fronts are further apart in the deep half and closer in the shallow half, and they meet the edge in the same places. A dashed line at right angles to the edge is labelled normal. Two thick arrows show the way the waves travel, one in each half. The one in the shallow half is closer to the normal. A short arrow between two fronts in each half is labelled λ.λλnormaldeepshallow
Ripples cross from deep to shallow water. The fronts are closer together, and the way they travel turns towards the normal. It is drawn with the speed ratio 0.6 of the worked example below. In a real tray the bend is often smaller, because the edge of the plate is not sharp.

Look at the edge. Every front that reaches the edge carries on into the shallow water without a break. So the fronts meet the edge equally far apart on both sides. That joining is why the direction must turn when the wavelength changes.

If the waves run from shallow water into deep water, they speed up and the wavelength grows. They then turn away from the normal. Along the normal itself, waves do not turn. Their wavelength and speed still change.

Worked example

In the deep part of a tray the wavelength is 5 cm. Ripples with a frequency of 4 Hz then cross to the shallow part, where the wavelength is 3 cm. Find the speed in each part.

  1. In the deep part v = fλ = 4 Hz × 5 cm = 20 cm/s.
  2. The frequency stays 4 Hz in the shallow part, because the source does not change.
  3. In the shallow part v = fλ = 4 Hz × 3 cm = 12 cm/s.

Answer: 20 cm/s and 12 cm/s

The ratio of the two speeds is 12 ÷ 20 = 0.6, which is the same as the ratio of the wavelengths, 3 ÷ 5. This works because the frequency is the same in both parts. The picture above is drawn with this ratio of 0.6.

Going further: the sea shows the same slowing on a huge scale. For a wave much longer than the water is deep, the speed is v = √(g × h). Here h is the depth in m, and g is 9.8 m/s². This rule is for tides and tsunamis. It is not for small ripples in a tray.

A table with the columns Sea depth and Speed. Row 1: 4 000 m, 198 m/s. Row 2: 100 m, 31 m/s. Row 3: 10 m, 10 m/s.Sea depthSpeed4 000 m198 m/s100 m31 m/s10 m10 m/s
The speed of a long wave in the sea. 198 m/s is about 710 km/h, nearly as fast as a jet airliner. The shallow sea near a coast slows it to about 36 km/h.

Worked example

A long wave crosses a sea that is 90 m deep. Find its speed.

  1. The rule for a long wave is v = √(g × h).
  2. Put in g = 9.8 m/s² and h = 90 m: g × h = 9.8 × 90 = 882.
  3. The square root is √882 = 29.7.
  4. So the speed is about 30 m/s.

Answer: About 30 m/s

As such a wave nears a coast, the water gets shallower and the wave slows. The energy the wave carries is squeezed into less water, so the wave grows taller. This is why a tsunami is small far out at sea and tall near the shore.

Try it

Ripples over a plate

You need: the tray and the water from before, about 1.5 cm deep, a flat tile or flat piece of board about 1 cm thick, with its top just under the surface, a pencil, a ruler and a torch.

Be safe: Keep the tray away from electrical things, and wipe up spills. Set the tile down gently, because it can crack the tray.

  1. Lay the tile flat at one end of the tray, so that the water over it is about 0.5 cm deep.
  2. Tap the water at the other end at a steady beat. Watch the ripples cross onto the tile.
  3. Look at how far apart the crests are over the tile compared with the deep part. Measure both, across four gaps.
  4. Try to tap so that the ripples meet the edge of the tile at a slant. Watch the way they turn.

What you will see: The crests are closer together over the tile. That is a shorter wavelength at the same frequency, so the waves are slower there. At a slant the ripples turn towards the normal on the tile. The effect is small, so look carefully, or take a photo from above.

If you cannot try it, this is what happens.
A table with the columns Part and Four gaps. Row 1: deep part, 24 cm. Row 2: over the tile, 17 cm.PartFour gapsdeep part24 cmover the tile17 cm
A tray seen from above with five straight wavefronts, each the same distance from the next. An arrow above the tray runs from the first front to the last and is labelled 20 cm. The tray is labelled, and an arrow below it is labelled wave travels.tray20 cmwave travels
Five fronts, with 20 cm across them. Count the gaps between them: one, two, three, four.

A wrong idea many people have

To find the wavelength, divide the distance across the fronts by the number of fronts.

Why it feels right: We count things to share a total. Five fronts look like five waves, so it seems right to divide by five.

The test that breaks it: Look at the picture: 20 cm across four gaps, not five. Divide 20 cm by 5 and you get 4 cm. The real gaps are 20 ÷ 4 = 5 cm.

What is true: Divide by the number of gaps, which is one less than the number of fronts. Six fronts have five gaps.

A wrong idea many people have

The frequency of ripples changes when they pass into shallow water.

Why it feels right: The ripples look different in the shallow part, with the crests closer together. It seems that the number of waves must have changed.

The test that breaks it: Count crests that pass a point in the deep part in 10 s, then in the shallow part in 10 s. The counts are the same. If one more arrived than left, the water at the edge would pile up forever.

What is true: The source sets the frequency, and it does not change when the wave crosses an edge. The speed and the wavelength change. The wavelength is v ÷ f.

A wrong idea many people have

Water waves speed up in shallow water.

Why it feels right: A wave on a beach seems to rush up the sand, and it looks fast as it breaks.

The test that breaks it: The ripples over the tile have crests closer together at the same frequency. With v = fλ, a smaller λ and the same f give a smaller v.

What is true: Waves slow down as the water gets shallower. A breaking wave looks quick because it is rising and tipping over, not because its speed has gone up.

Key words

ripple tank
a shallow tray of water, seen from above, for studying water waves.
wavefront
a line joining the crests of a wave, seen from above. The wave travels at right angles to it.
normal
a line at right angles (90°) to an edge, used to measure angles.
refraction
the turning of a wave when its speed changes at a slanted edge.

Back to your guess

The answer is A: Ripples slow down and bunch up as they run onto shallow water.

Water waves travel more slowly in shallower water. Their frequency stays the same, so the wavelength gets shorter and the crests bunch up. If they meet the shallow part at a slant, they also bend.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The picture shows six wavefronts in a tray, with the distance across them. What is the wavelength?

    A tray seen from above with six straight wavefronts, each the same distance from the next. An arrow above the tray runs from the first front to the last and is labelled 30 cm. The tray is labelled.tray30 cm
    Tap your answer.
    1. Not this one. You divided 30 by 6, the number of fronts. Six fronts have only five gaps, and each gap is one wavelength. Try another.

    2. Yes, that is right. Six fronts have five gaps. Five wavelengths are 30 cm, so one wavelength is 30 ÷ 5 = 6 cm.

    3. Not this one. 30 cm is the distance across all five gaps. One wavelength is one gap. Try another.

    See the answer

    6 cm

    Six fronts have five gaps. Five wavelengths are 30 cm, so one wavelength is 30 ÷ 5 = 6 cm.

    • If you chose A: You divided 30 by 6, the number of fronts. Six fronts have only five gaps, and each gap is one wavelength.
    • If you chose C: 30 cm is the distance across all five gaps. One wavelength is one gap.
  2. Taps on water make 6 waves each second. The wavelength is 3 cm. How fast do the ripples travel?

    Tap your answer.
    1. Not this one. You added 6 and 3. The speed is the frequency times the wavelength. Try another.

    2. Not this one. You divided 6 by 3. The wave equation multiplies: v = fλ. Try another.

    3. Yes, that is right. v = fλ = 6 Hz × 3 cm = 18 cm/s.

    4. Not this one. You divided 3 by 6. Multiply the frequency and the wavelength instead. Try another.

    See the answer

    18 cm/s

    v = fλ = 6 Hz × 3 cm = 18 cm/s.

    • If you chose A: You added 6 and 3. The speed is the frequency times the wavelength.
    • If you chose B: You divided 6 by 3. The wave equation multiplies: v = fλ.
    • If you chose D: You divided 3 by 6. Multiply the frequency and the wavelength instead.
  3. Ripples run from deep water onto a plate where the water is shallow. What happens to their frequency?

    Tap your answer.
    1. Yes, that is right. The source sets the frequency. The same number of crests cross the edge every second, so the frequency is the same on both sides.

    2. Not this one. Closer crests mean a shorter wavelength, not a higher frequency. The same number of waves cross the edge each second. Try another.

    3. Not this one. The speed falls, but the frequency is set by the source. The same number of waves cross the edge each second. Try another.

    See the answer

    It stays the same.

    The source sets the frequency. The same number of crests cross the edge every second, so the frequency is the same on both sides.

    • If you chose B: Closer crests mean a shorter wavelength, not a higher frequency. The same number of waves cross the edge each second.
    • If you chose C: The speed falls, but the frequency is set by the source. The same number of waves cross the edge each second.
  4. Which statement describes ripples after they cross from deep water into shallow water?

    Tap your answer.
    1. Not this one. The wavelength does change, and the frequency does not. The source sets the frequency. Try another.

    2. Not this one. Shallow water slows waves. A longer wavelength would need a bigger speed at the same frequency. Try another.

    3. Not this one. With the same frequency, a shorter wavelength means a smaller speed. The speed does change. Try another.

    4. Yes, that is right. In shallow water the speed falls. The frequency is fixed by the source, so v = fλ gives a shorter wavelength.

    See the answer

    They go slower, with a shorter wavelength and the same frequency.

    In shallow water the speed falls. The frequency is fixed by the source, so v = fλ gives a shorter wavelength.

    • If you chose A: The wavelength does change, and the frequency does not. The source sets the frequency.
    • If you chose B: Shallow water slows waves. A longer wavelength would need a bigger speed at the same frequency.
    • If you chose C: With the same frequency, a shorter wavelength means a smaller speed. The speed does change.
  5. The picture shows a tray with a plate. Where do ripples travel slowest?

    A tray of water seen from the side. The water is 1.5 cm deep at the left, marked with an arrow. At the right a plate lies on the bottom, shown with hatching and labelled plate, and the water over it is 0.5 cm deep, marked with an arrow. The tray is labelled.1.5 cm0.5 cmplatetray
    Tap your answer.
    1. Not this one. There is more water to move, but that does not slow the wave. Waves travel faster in deeper water. Try another.

    2. Yes, that is right. Water waves slow down as the water gets shallower, so the ripples travel slowest over the plate.

    3. Not this one. The depth changes the speed. A wave over the plate is in less water and is slower. Try another.

    See the answer

    Over the plate, where the water is shallow.

    Water waves slow down as the water gets shallower, so the ripples travel slowest over the plate.

    • If you chose A: There is more water to move, but that does not slow the wave. Waves travel faster in deeper water.
    • If you chose C: The depth changes the speed. A wave over the plate is in less water and is slower.
  6. A long wave in the sea reaches the shore, where the water is shallow. What happens to its speed?

    Tap your answer.
    1. Yes, that is right. The speed of a long wave is √(g × h). As the depth h falls, the speed falls.

    2. Not this one. Nothing pushes the wave faster. Its speed follows the depth, and the depth falls near the shore. Try another.

    3. Not this one. The speed depends on the depth where the wave is now, not on where it began. Try another.

    4. Not this one. The source sets the frequency, and it does not change. The speed falls with the depth, and the wavelength gets shorter. Try another.

    See the answer

    It slows down.

    The speed of a long wave is √(g × h). As the depth h falls, the speed falls.

    • If you chose B: Nothing pushes the wave faster. Its speed follows the depth, and the depth falls near the shore.
    • If you chose C: The speed depends on the depth where the wave is now, not on where it began.
    • If you chose D: The source sets the frequency, and it does not change. The speed falls with the depth, and the wavelength gets shorter.
  7. Priya wants the speed of ripples in her tray. Which two measurements does she need?

    Tap your answer.
    1. Not this one. The amplitude, the height of the ripples, does not enter the wave equation. The speed needs the frequency and the wavelength. Go to S4.10.2. Try another.

    2. Yes, that is right. The wave equation v = fλ needs the frequency and the wavelength. She counts the taps to get f, and she measures across four gaps to get λ.

    3. Not this one. The amplitude does not enter the wave equation v = fλ. The wavelength is needed too. Go to S4.10.2. Try another.

    4. Not this one. The depth changes the speed, but it does not give the speed in this method. Use v = fλ. Go to S4.10.2. Try another.

    See the answer

    The frequency and the wavelength

    The wave equation v = fλ needs the frequency and the wavelength. She counts the taps to get f, and she measures across four gaps to get λ.

    • If you chose A: The amplitude, the height of the ripples, does not enter the wave equation. The speed needs the frequency and the wavelength. Go to S4.10.2.
    • If you chose C: The amplitude does not enter the wave equation v = fλ. The wavelength is needed too. Go to S4.10.2.
    • If you chose D: The depth changes the speed, but it does not give the speed in this method. Use v = fλ. Go to S4.10.2.
  8. From an earlier chapter

    A wave has a period of 0.5 s. What is its frequency?

    Tap your answer.
    1. Not this one. 0.5 is the period in seconds. The frequency is 1 divided by it. Go to S4.10.2. Try another.

    2. Not this one. You may have used 1 ÷ 0.05. The frequency is 1 ÷ 0.5. Go to S4.10.2. Try another.

    3. Yes, that is right. f = 1 ÷ T = 1 ÷ 0.5 = 2 Hz. Two waves pass each second.

    See the answer

    2 Hz

    f = 1 ÷ T = 1 ÷ 0.5 = 2 Hz. Two waves pass each second.

    • If you chose A: 0.5 is the period in seconds. The frequency is 1 divided by it. Go to S4.10.2.
    • If you chose B: You may have used 1 ÷ 0.05. The frequency is 1 ÷ 0.5. Go to S4.10.2.
  9. A long wave crosses a sea that is 600 m deep. Find its speed to the nearest whole m/s. Use g = 9.8 m/s².

    See the answer

    About 77 m/s

    v = √(g × h) = √(9.8 × 600) = √5 880 = 76.7, which is about 77 m/s.

  10. Ripples travel at 27 cm/s and their frequency is 3 Hz. What is the wavelength?

    See the answer

    9 cm

    λ = v ÷ f = 27 cm/s ÷ 3 Hz = 9 cm.

Keep it

Copy this into your exercise book.

v = fλ. Measure λ across several gaps: fronts − 1. Shallow water slows waves. f stays the same, so λ shortens and the wave bends towards the normal.

Draw: Straight wavefronts crossing an edge at a slant, closer together in the shallow part, with the normal and two arrows.

S4.10.4 Chapter 4

Sound and ultrasound

Tariq claps in front of a tall wall 85 m away and hears an echo. How long after the clap does the echo reach him?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Sound is a longitudinal wave. Layers of the medium are pushed together and pulled apart, and the pattern of compressions and rarefactions runs outwards from the source.

Sound needs a medium to travel through. It can travel through a gas, a liquid or a solid. It is usually slowest in a gas, faster in a liquid and fastest in a solid. In a solid the particles are close and spring back quickly.

A table with the columns Medium and Speed of sound. Row 1: air, 340 m/s. Row 2: water, 1 500 m/s. Row 3: steel, 5 000 m/s.MediumSpeedair340 m/swater1 500 m/ssteel5 000 m/s
Sound is about 340 m/s in air at room temperature. It is a little slower when the air is colder and a little faster when it is warmer.

In a vacuum there are no particles, so there is nothing to push together. Sound cannot travel in a vacuum. Scientists show this with a bell in a jar. They cannot do it at home, because it needs a pump that takes the air out.

A glass jar with a bell hanging in it on a thread. The jar is joined by a pipe to a pump. The jar holds air, shown as dots. Curved lines outside the jar, labelled sound, show that the ringing is heard. The bell, the jar and the pump are labelled.belljarpumpsound
With air in the jar you hear the bell.
The same glass jar and bell, joined to a pump. This time the jar has no dots, because the air has been pumped out. There are no curved lines outside the jar. The bell, the jar and the pump are labelled.belljarpump
With the air pumped out, the bell still moves, but you hear almost nothing.

As the air is pumped out, the ringing gets fainter. When almost no air is left, it is almost silent. The bell hangs on a thin thread, so that the sound cannot run along something solid to the glass. You can still see the bell move, because light does cross a vacuum.

Light is far faster than sound. In a vacuum, light travels at 3 × 10⁸ m/s. That is nearly a million times the speed of sound in air. This is why a flash of lightning is seen long before the thunder is heard.

Now use an echo to measure the speed of sound. Stand in front of a large wall and clap. The sound travels to the wall and bounces back. You hear the echo a short time later.

Sound goes to the wall and comes back. So the distance it travels is twice the distance d to the wall. The speed is distance divided by time: v = 2d ÷ t.

A person on the ground on the left, labelled you, and a hatched wall on the right, labelled wall. A long arrow from the person to the wall is labelled sound goes. A second arrow of the same length from the wall back to the person is labelled echo. A mark under the ground between the person and the wall is labelled 85 m.sound goesecho85 myouwall
The sound covers the 85 m twice, there and back.

Worked example

A person stands 85 m from a wall. An echo comes back 0.5 s after a clap. Find the speed of sound.

  1. The sound goes to the wall and back, so it travels 2 × 85 m = 170 m.
  2. The speed is v = distance ÷ time = 170 m ÷ 0.5 s = 340 m/s.

Answer: 340 m/s

A hand held stopwatch cannot time 0.5 s well, because your reaction time is about 0.2 s. The trick is to clap again and again, so that each clap comes with the echo of the one before. Then the time between claps is the echo time. Count many claps and divide.

Try it

Measure the speed of sound with an echo

You need: a large flat wall, such as the side of a tall building, at least 50 m away across open ground, a long string or tape, or a known length of pace, to find the distance, a clock or phone that shows seconds, two flat pieces of wood to clap together.

Be safe: Stay well away from roads and from the edges of cliffs, roofs or deep water. Do not clap close to anyone's ears. Use a wall that belongs to nobody who would mind the noise.

  1. Find the distance d from where you stand to the wall. Walk it and count paces, after you have measured the length of one pace on a string.
  2. Clap the two pieces together. Listen for the echo.
  3. Clap again so that your clap comes just as the echo of the last clap arrives. Keep a steady rhythm.
  4. When the rhythm is steady, count 20 claps and time them. Divide the time by 20. This is the echo time t.
  5. Work out v = 2d ÷ t.

What you will see: Your answer will be close to 340 m/s, but not exactly. Timing 20 claps instead of one clap makes the answer much better. The speed of sound also changes a little with the temperature of the air.

If you cannot try it, this is what happens.
A table with the columns Measure and Value. Row 1: distance to wall, 102 m. Row 2: 20 claps take, 11.8 s. Row 3: t = 11.8 ÷ 20, 0.59 s. Row 4: v = 2 × 102 ÷ 0.59, 346 m/s.MeasureValuedistance to wall102 m20 claps take11.8 st = 11.8 ÷ 200.59 sv = 2 × 102 ÷ 0.59346 m/s

The same idea gives the distance to a reflector. If you know the speed of sound and the echo time, then v × t is the whole way there and back. Half of that is the distance.

A person on the ground on the left, labelled you, and a hatched wall on the right, labelled cliff. A long arrow from the person to the cliff is labelled sound goes. A second arrow of the same length from the cliff back to the person is labelled echo. A mark under the ground between the person and the cliff is labelled ? m.sound goesecho? myoucliff
An echo gives the distance to a cliff when the speed and the time are known.

Worked example

A shout echoes from a cliff and the echo returns 1.2 s later. Take the speed of sound as 340 m/s. How far away is the cliff?

  1. The sound goes there and back in 1.2 s.
  2. The whole way is v × t = 340 m/s × 1.2 s = 408 m.
  3. The cliff is half of the whole way: 408 ÷ 2 = 204 m.

Answer: 204 m

Ordinary sound is used in this way, but sound that we can hear is a nuisance for it. For many jobs, scientists use sound that is too high for us to hear. Sound with a frequency above 20 000 Hz is called ultrasound.

Humans hear from about 20 Hz to about 20 000 Hz. The top of the range falls as people grow older. Sound below 20 Hz is called infrasound. Elephants use it. Many bats and dolphins use ultrasound.

A bar on a scale from 1 to 100 000 hertz, with each ten times further along the scale. The bar has three parts. The first part, hatched, is labelled infrasound. The middle part, the thick one, is labelled human hearing, with 20 Hz at its left end and 20 000 Hz at its right end. The last part, cross hatched, is labelled ultrasound. The scale numbers are 10, 1 000 and 100 000. The axis is labelled frequency (Hz).human hearing20 Hz20 000 Hzinfrasoundultrasound101 000100 000frequency (Hz)
The scale goes up in tens: each tick is ten times the one before. The top of the human range is 1 000 times the bottom.

Ultrasound is the same kind of wave as ordinary sound. It is longitudinal, and in a given medium it travels at the same speed as ordinary sound. It differs only in having a higher frequency, and so a shorter wavelength.

A ship uses ultrasound to find the depth of the sea. This is sonar. The ship sends a short pulse of ultrasound down. It reflects from the sea floor, or from a shoal of fish, and comes back up. The ship times the echo.

A ship on the sea surface, labelled ship, with the water shown by dots and the sea floor hatched, labelled sea floor. A downward arrow, labelled pulse, and an upward arrow of the same length, labelled echo. A mark at the right from the surface to the sea floor is labelled ? m.? mpulseechoshipsea floor
Sonar. The pulse goes down and the echo comes up, so the ship halves the whole way.

Sound in sea water travels at about 1 500 m/s. The speed changes a little with temperature, salt and depth, so the ship uses a value from measurements. The depth is half the distance there and back: d = v × t ÷ 2.

Worked example

A ship's sonar pulse returns 0.8 s after it was sent. Sound travels at 1 500 m/s in the sea. Find the depth.

  1. The echo sounder times the pulse as it goes down and comes back.
  2. Going down alone takes half the time: 0.8 s ÷ 2 = 0.4 s.
  3. The depth is speed × time = 1 500 m/s × 0.4 s = 600 m.
  4. In kilometres: 600 m ÷ 1 000 = 0.6 km.

Answer: 600 m, which is 0.6 km

Doctors use ultrasound to look inside the body without cutting it. The scanner has a probe that sends in pulses of ultrasound, with frequencies of several million hertz. A clear gel on the skin keeps air from lying between the probe and the skin. Air reflects almost all the ultrasound.

Echoes come back from every boundary where one kind of tissue meets another: skin, fluid, the baby and bone. The scanner times each echo, and notes how strong it is. A computer turns these into a picture. Sound in soft tissue travels at about 1 540 m/s, so the echoes come back in about a tenth of a millisecond.

A cut through the body seen from the side. A probe sits on the skin at the top, with a thick line under it labelled gel. A pulse arrow, labelled pulse, goes down towards a dotted oval labelled baby, and an echo arrow, labelled echo, comes back up. A mark at the right from the skin to the top of the baby is labelled 8 cm.8 cmpulseechoprobebabygel
A scan. The echo from the baby comes back to the probe, and the time says how deep it is.

Ultrasound does not ionise, which means it does not knock electrons off atoms. At the low powers used for scans, no harm to the mother or the baby has been found. This is why it is used to look at unborn babies, in place of X-rays. Scans are still only done when they are useful.

A wrong idea many people have

The echo time is the time the sound takes to reach the wall.

Why it feels right: We think of the wall as the place the sound is going to, and the echo as the end of the journey.

The test that breaks it: Take a wall 85 m away and the speed 340 m/s. Reaching the wall alone takes 85 ÷ 340 = 0.25 s. The echo reaches you after 0.5 s, twice as long, because it must come back.

What is true: The echo time covers the way there and the way back. The distance to the wall is half of v × t.

A wrong idea many people have

Light and sound travel at about the same speed.

Why it feels right: A clap seems to be seen and heard together. A flash and a bang from a nearby firework seem to come at once.

The test that breaks it: Watch a firework across a wide field or a far storm. The flash comes first, and the bang follows seconds later. At 1 km, light takes about 3 millionths of a second, and sound takes about 3 s.

What is true: Light is nearly a million times faster than sound. Sound is slow, about 340 m/s in air, and it can never be faster than light.

A wrong idea many people have

Ultrasound travels faster than ordinary sound, because it is higher.

Why it feels right: Higher frequency sounds like "more" and "quicker", and we call the sound "ultra".

The test that breaks it: In air at 340 m/s, a note of 400 Hz and an ultrasound of 40 000 Hz both travel at 340 m/s. The ultrasound has 100 times the frequency, so with v = fλ it has one hundredth of the wavelength.

What is true: The speed is set by the medium. Ultrasound has the same speed as ordinary sound in the same medium. Only its frequency and wavelength differ.

Key words

vacuum
a space with no particles in it. Sound cannot cross it, and light can.
echo
sound that reflects from a surface and returns. It travels the distance there and back.
ultrasound
sound with a frequency above 20 000 Hz, too high for people to hear.
infrasound
sound with a frequency below 20 Hz, too low for people to hear.
sonar
a way to find depths and objects in water by timing the echo of a pulse of sound.

Back to your guess

The answer is B: About 0.5 s, because the sound goes to the wall and back.

The sound has to go to the wall and come back, a total of 170 m. At about 340 m/s that takes 0.5 s. An echo always covers the distance twice.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A clap is echoed from a wall, and the echo comes back 0.8 s later. The speed of sound is 340 m/s. How far away is the wall?

    A person on the ground on the left, labelled you, and a hatched wall on the right, labelled wall. A long arrow from the person to the wall is labelled sound goes. A second arrow of the same length from the wall back to the person is labelled echo. A mark under the ground between the person and the wall is labelled ? m.sound goesecho? myouwall
    Tap your answer.
    1. Not this one. That is the whole way there and back, 340 × 0.8. The wall is half that far away. Try another.

    2. Not this one. You divided 340 by 0.8. Distance is speed times time, and the sound went there and back. Try another.

    3. Yes, that is right. The sound travels 340 m/s × 0.8 s = 272 m there and back. The wall is half of that: 272 ÷ 2 = 136 m.

    4. Not this one. You halved twice. Halve the whole way once, and no more. Try another.

    See the answer

    136 m

    The sound travels 340 m/s × 0.8 s = 272 m there and back. The wall is half of that: 272 ÷ 2 = 136 m.

    • If you chose A: That is the whole way there and back, 340 × 0.8. The wall is half that far away.
    • If you chose B: You divided 340 by 0.8. Distance is speed times time, and the sound went there and back.
    • If you chose D: You halved twice. Halve the whole way once, and no more.
  2. A ship's sonar pulse returns after 0.4 s. Sound travels at 1 500 m/s in sea water. How deep is the sea?

    Tap your answer.
    1. Not this one. That is the whole way down and up. The depth is half of it. Try another.

    2. Not this one. You divided 1 500 by 0.4. Multiply the speed by the time, then halve. Try another.

    3. Not this one. You halved twice. The time is already the there and back time, so halve once. Try another.

    4. Yes, that is right. The pulse travels 1 500 m/s × 0.4 s = 600 m down and up. The depth is half of that: 600 ÷ 2 = 300 m.

    See the answer

    300 m

    The pulse travels 1 500 m/s × 0.4 s = 600 m down and up. The depth is half of that: 600 ÷ 2 = 300 m.

    • If you chose A: That is the whole way down and up. The depth is half of it.
    • If you chose B: You divided 1 500 by 0.4. Multiply the speed by the time, then halve.
    • If you chose C: You halved twice. The time is already the there and back time, so halve once.
  3. Why does a doctor put gel on the skin before an ultrasound scan?

    Tap your answer.
    1. Not this one. Ultrasound cannot be heard at all. Its frequency is above the range of human hearing. Try another.

    2. Yes, that is right. Air would reflect almost all the ultrasound at the skin, so little would get into the body. The gel fills the gap so the ultrasound enters the body.

    3. Not this one. The speed depends on the tissue, not on the gel. The gel helps the ultrasound get in. Try another.

    See the answer

    The gel keeps air out between the probe and the skin, because air reflects nearly all the ultrasound.

    Air would reflect almost all the ultrasound at the skin, so little would get into the body. The gel fills the gap so the ultrasound enters the body.

    • If you chose A: Ultrasound cannot be heard at all. Its frequency is above the range of human hearing.
    • If you chose C: The speed depends on the tissue, not on the gel. The gel helps the ultrasound get in.
  4. What is ultrasound?

    Tap your answer.
    1. Yes, that is right. Ultrasound has a frequency above 20 000 Hz, which is higher than people can hear. It is the same kind of wave as ordinary sound.

    2. Not this one. In the same medium, ultrasound travels at the same speed as ordinary sound. It has a higher frequency. Try another.

    3. Not this one. That is infrasound. Ultrasound has a frequency above 20 000 Hz, above the top of human hearing. Try another.

    4. Not this one. Loudness depends on the amplitude. Ultrasound is defined by its frequency, and people cannot hear it however large it is. Try another.

    See the answer

    Sound with a frequency above 20 000 Hz

    Ultrasound has a frequency above 20 000 Hz, which is higher than people can hear. It is the same kind of wave as ordinary sound.

    • If you chose B: In the same medium, ultrasound travels at the same speed as ordinary sound. It has a higher frequency.
    • If you chose C: That is infrasound. Ultrasound has a frequency above 20 000 Hz, above the top of human hearing.
    • If you chose D: Loudness depends on the amplitude. Ultrasound is defined by its frequency, and people cannot hear it however large it is.
  5. Lightning flashes, and the thunder is heard 6 s later. What does this show?

    Tap your answer.
    1. Not this one. It is the other way round. The flash reaches you first because light is much faster than sound. Try another.

    2. Not this one. The flash and the thunder begin together. Light arrives at once and sound is slow, which makes the gap. Go to S3.9.7. Try another.

    3. Yes, that is right. The flash and the bang start together. Light reaches you almost at once, and sound takes about 3 s for each kilometre, so you hear it later.

    See the answer

    Light travels much faster than sound, so the flash arrives first.

    The flash and the bang start together. Light reaches you almost at once, and sound takes about 3 s for each kilometre, so you hear it later.

    • If you chose A: It is the other way round. The flash reaches you first because light is much faster than sound.
    • If you chose B: The flash and the thunder begin together. Light arrives at once and sound is slow, which makes the gap. Go to S3.9.7.
  6. A bell rings in a jar, and a pump takes the air out. What do you hear, and why?

    Tap your answer.
    1. Not this one. Sound needs the air. Without air, nothing carries the vibrations to the glass and to your ears. Try another.

    2. Yes, that is right. Sound needs a medium. As the air goes, the sound fades. Light still crosses the empty jar, so you can see the bell move.

    3. Not this one. Sound is a wave of the air. With the air taken out, nothing is left to pass on the sound. Go to S4.10.1. Try another.

    See the answer

    The ringing gets fainter, because there is less air to carry the sound.

    Sound needs a medium. As the air goes, the sound fades. Light still crosses the empty jar, so you can see the bell move.

    • If you chose A: Sound needs the air. Without air, nothing carries the vibrations to the glass and to your ears.
    • If you chose C: Sound is a wave of the air. With the air taken out, nothing is left to pass on the sound. Go to S4.10.1.
  7. An ultrasound echo comes back from a boundary 0.12 ms after the pulse was sent. The speed of sound in the body is 1 540 m/s. How deep is the boundary, to one decimal place? Note: 1 ms = 0.001 s.

    Tap your answer.
    1. Yes, that is right. First change the time: 0.12 ms = 0.00012 s. The whole way is 1 540 × 0.00012 = 0.185 m. Half of it is 0.0924 m, which is 9.2 cm.

    2. Not this one. That is the whole way there and back. The boundary is half that deep. Try another.

    3. Not this one. You may have used 0.12 s instead of 0.12 ms. That gives 1 540 × 0.12 ÷ 2 = 92.4 m. 0.12 ms is only 0.00012 s. Try another.

    See the answer

    9.2 cm

    First change the time: 0.12 ms = 0.00012 s. The whole way is 1 540 × 0.00012 = 0.185 m. Half of it is 0.0924 m, which is 9.2 cm.

    • If you chose B: That is the whole way there and back. The boundary is half that deep.
    • If you chose C: You may have used 0.12 s instead of 0.12 ms. That gives 1 540 × 0.12 ÷ 2 = 92.4 m. 0.12 ms is only 0.00012 s.
  8. From an earlier chapter

    Sound of frequency 800 Hz travels at 340 m/s in air. What is its wavelength?

    Tap your answer.
    1. Not this one. You multiplied 340 by 800. For the wavelength divide the speed by the frequency: λ = v ÷ f. Go to S4.10.2. Try another.

    2. Not this one. You divided 800 by 340. The wavelength is the speed divided by the frequency, not the other way round. Go to S4.10.2. Try another.

    3. Yes, that is right. λ = v ÷ f = 340 m/s ÷ 800 Hz = 0.425 m.

    See the answer

    0.425 m

    λ = v ÷ f = 340 m/s ÷ 800 Hz = 0.425 m.

    • If you chose A: You multiplied 340 by 800. For the wavelength divide the speed by the frequency: λ = v ÷ f. Go to S4.10.2.
    • If you chose B: You divided 800 by 340. The wavelength is the speed divided by the frequency, not the other way round. Go to S4.10.2.
  9. A sonar pulse returns after 0.5 s. Sound travels at 1 500 m/s in the sea. How deep is the sea?

    See the answer

    375 m

    The whole way is 1 500 × 0.5 = 750 m. The depth is half of it: 375 m.

  10. You clap, and the echo from a cliff comes back 1.5 s later. The speed of sound is 340 m/s. How far away is the cliff?

    See the answer

    255 m

    The whole way is 340 × 1.5 = 510 m. The cliff is half of it: 255 m.

Keep it

Copy this into your exercise book.

Sound needs a medium and cannot cross a vacuum. Echo: v = 2d ÷ t. Depth = v × t ÷ 2. Ultrasound is above 20 000 Hz. Light is far faster than sound.

Draw: A ship, the sea and the sea floor. Draw a pulse arrow going down and an echo arrow coming up. Mark the depth.

S4.10.5 Chapter 5

Reflection and refraction

A ray of light meets a mirror and makes an angle of 20° with its surface. At what angle was the ray arriving, in the way scientists measure it?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Light travels in straight lines called rays. A ray is drawn as a straight line with an arrow that shows the way the light goes. This chapter follows rays when they meet a mirror, and when they meet the edge of glass or water.

Every angle is measured from the normal. The normal is a dashed line at 90° to the surface, drawn at the point where the ray meets it. The ray that arrives is the incident ray. The angle of incidence is the angle between the incident ray and the normal. The ray that leaves a mirror is the reflected ray. The angle of reflection is the angle between that ray and the normal.

A flat mirror drawn as a thick line, with hatching behind it, labelled mirror. A dashed line at right angles to the mirror is labelled normal. A ray labelled incident ray arrives from the upper left and meets the mirror at the foot of the normal. A ray labelled reflected ray leaves towards the upper right. The angle between each ray and the normal is marked 40°.mirrornormalincident rayreflected ray40°40°
The law of reflection: the angle of incidence equals the angle of reflection. Both are measured from the normal.

The law of reflection says two things. The angle of incidence equals the angle of reflection. And the incident ray, the normal and the reflected ray all lie in one flat plane. On a diagram, that plane is the page.

A very common slip is to measure the angle from the mirror, not from the normal. If a question says a ray makes 30° with the mirror, the angle of incidence is not 30°. It is 90° − 30° = 60°. Always draw the normal first.

The law holds for every ray on every surface. A rough surface, such as paper, is made of tiny patches that face different ways. Each ray obeys the law at its own patch, but the patches point in many directions, so the light scatters. A smooth mirror keeps parallel rays parallel, and you see an image.

Now look at your own image in a flat mirror. Rays from your face reflect from the mirror and reach your eye. Your eye follows each ray back in a straight line, to a point behind the mirror, where the image seems to be.

A flat mirror seen from the side, labelled mirror. An upright arrow in front of it is labelled object. A dashed arrow of the same height, the same distance behind the mirror, is labelled image. Two rays go from the tip and the foot of the object to the mirror and reflect to an eye on the left, labelled eye. Dotted lines go back behind the mirror from the mirror to the image. Marks under the picture show the distance from the object to the mirror and from the mirror to the image, both labelled 4 cm.objectimagemirroreye4 cm4 cm
The image is as far behind the mirror as the object is in front. The rays only seem to come from it.

The image in a flat mirror is the same size as the object. It is as far behind the mirror as the object is in front. It is upright. It is left and right reversed. It is a virtual image: light does not really pass through it, so you cannot catch it on a screen.

Now turn to refraction. Light travels at 3 × 10⁸ m/s in a vacuum and almost as fast in air. In glass and in water it is slower. The refractive index n of a material is the speed of light in a vacuum divided by the speed in the material.

A table with the columns Material, n and Speed. Row 1: air, 1.00, 3.0 × 10⁸. Row 2: water, 1.33, 2.26 × 10⁸. Row 3: glass, 1.5, 2.0 × 10⁸. Row 4: diamond, 2.42, 1.24 × 10⁸. The speeds are in metres a second.MaterialnSpeedair1.003.0 × 10⁸water1.332.26 × 10⁸glass1.52.0 × 10⁸diamond2.421.24 × 10⁸
Refractive index and the speed of light in m/s. Ordinary glass is about 1.5. Other kinds of glass differ a little.

When a ray goes into a material where light is slower, it bends towards the normal. When it goes into a material where light is faster, it bends away from the normal. A ray that travels along the normal does not bend, but it still changes speed.

A ray going from air into glass. The edge is a horizontal line, with the glass below it, shaded with dots and labelled glass, and the air above labelled air. A dashed line at right angles to the edge is labelled normal. The ray arrives from the upper left at 40° to the normal and bends towards the normal inside the glass, leaving at 25° to the normal.airglassnormal40°25°
Air to glass: the light slows down and bends towards the normal.
A ray going from glass into air. The edge is a horizontal line, with the glass below it, shaded with dots and labelled glass, and the air above labelled air. A dashed line at right angles to the edge is labelled normal. The ray rises from the lower left at 30° to the normal and bends away from the normal in the air, leaving at 49° to the normal.glassairnormal30°49°
Glass to air: the light speeds up and bends away from the normal.

How much does a ray bend? Snell's law gives the answer. For light going from air into a material of index n, sin i ÷ sin r = n. Here i is the angle of incidence and r is the angle of refraction, both from the normal. The sine of an angle is the ratio of two sides of a right angled triangle.

In general, light goes from a material of index n₁ into one of index n₂. Then n₁ × sin i = n₂ × sin r. In every case the ray bends towards the normal when n goes up, and away when n goes down.

A ray going from air into water. The edge is a horizontal line, with the water below it, shaded with dots and labelled water, and the air above labelled air. A dashed line at right angles to the edge is labelled normal. The ray arrives from the upper left and bends towards the normal in the water.airwaternormal
A ray enters water at 45° to the normal.

Worked example

A ray goes from air into water, whose refractive index is 1.33. The angle of incidence is 45°. Find the angle of refraction.

  1. Use sin i ÷ sin r = n, so sin r = sin i ÷ n.
  2. sin 45° = 0.7071, so sin r = 0.7071 ÷ 1.33 = 0.5317.
  3. The angle whose sine is 0.5317 is r = 32.1°.

Answer: 32.1°

A rectangular block of glass shows both bends. The ray bends towards the normal as it goes in, and away from the normal as it comes out. The two bends are equal and opposite, so the ray that leaves is parallel to the ray that went in. It is moved sideways.

A ray crossing a rectangular block of glass, labelled glass. The ray arrives from the upper left at 40° to the normal, marked 40°, and bends towards the normal inside the block at 25°, marked 25°. It leaves at the bottom parallel to the first ray but moved to one side. A dotted line shows the straight on path, labelled straight on. Two dashed normals are labelled normal.glassnormalstraight on40°25°
Through a glass block, the ray leaves parallel to the way it came in. It is at 40° to the normal again, but shifted sideways.

Now go the other way. A ray inside glass or water meets the edge with air. It bends away from the normal. As the angle of incidence grows, the bent ray swings closer to the edge. At one angle it runs along the edge. That angle is the critical angle, c.

At the critical angle, the angle of refraction is 90°. So sin c = 1 ÷ n for light going from a material of index n into air. For glass of index 1.5, c is about 42°.

Past the critical angle, no light can leave. All of it is reflected back inside, by the law of reflection. This is total internal reflection.

Three rays of light inside glass, all meeting the top edge at the same point, with air above. The glass is below the edge, shaded with dots and labelled glass. The air is labelled air. A dashed normal is labelled normal. The ray at the smallest angle bends away from the normal and leaves into the air, labelled bends out. The ray at the middle angle, the critical angle, runs along the edge, labelled along edge. The ray at the largest angle reflects back into the glass, labelled reflected.glassairnormalbends outalong edgereflected
Three rays in glass. Under the critical angle the light goes out. At it, the light runs along the edge. Beyond it, all the light is reflected.

Total internal reflection has two conditions. The light must go from a higher n to a lower n, such as glass to air. And the angle of incidence must be greater than the critical angle. Light going from air into glass can never be totally reflected.

Even when light does cross an edge, a little of it is also reflected. The diagrams leave that weak reflection out. Total internal reflection is different: there, all the light is reflected.

A ray inside water meeting the edge with air above. The water is below the edge, shaded with dots and labelled water. The air is labelled air. A dashed normal is labelled normal. The ray rises from the lower left at a large angle to the normal and is reflected back down into the water at the same angle.waterairnormal
A ray in water at 60° to the normal is past the critical angle. All of it is reflected.

Worked example

Water has a refractive index of 1.33. What is the critical angle for light going from water into air?

  1. At the critical angle, sin c = 1 ÷ n.
  2. sin c = 1 ÷ 1.33 = 0.7519.
  3. The angle whose sine is 0.7519 is c = 48.8°.

Answer: 48.8°

Diamond has a very high index. Its critical angle is small, so light that enters a cut diamond is reflected inside again and again before it comes out. This is part of why diamonds sparkle.

An optical fibre is a thin, flexible thread of glass. Light that enters one end meets the wall at an angle greater than the critical angle. It is totally reflected, again and again, and follows the fibre round bends to the far end.

A glass fibre that runs straight and then bends upwards, shaded with dots and labelled glass fibre. A ray of light labelled light enters at the left end. It zigzags along the fibre, reflecting from the walls, and follows the bend to the top end.glass fibrelight
A ray follows a bent fibre by total internal reflection. Every meeting with the wall is past the critical angle.

A real fibre has a core of glass inside a coating of glass of slightly lower index. The total internal reflection happens where the core meets the coating. A fibre cannot be bent too sharply. The light would then meet the wall at less than the critical angle and leak out.

Fibres carry telephone calls and the internet as pulses of light, which you meet in the last chapter. Doctors also use bundles of fibres in a thin tube, an endoscope, to look inside the body without a large cut.

Try it

Test the law of reflection

You need: a small flat mirror, a torch, two cards held with a narrow gap between them, a sheet of white paper, a pencil and a ruler, something to hold the mirror upright, such as a lump of clay or a book.

Be safe: Never shine the torch into anyone's eyes, and never use the Sun or a mirror in sunlight for this. Tape the edges of a glass mirror, which can cut. Keep the mirror and the clay away from young children.

  1. Draw a straight line across the paper. Stand the mirror upright on the line, with its shiny face towards you. Draw a dashed line at right angles to the mirror, as the normal. Mark where it meets the mirror.
  2. Lay the torch behind the two cards. A thin beam should come through the gap. It should cross the paper at a slant and meet the mirror at the foot of the normal. Keep the beam low and flat on the paper.
  3. Mark two small dots on the beam that goes to the mirror, and two on the beam that leaves it. Take away the mirror and the torch, and draw lines through the dots.
  4. Mark a point on each line exactly 10 cm from the mirror point. Measure how far each of the two points is from the normal.
  5. If the two distances are equal, the two angles are equal.

What you will see: The two distances come out almost the same, within a millimetre or so. A point 10 cm along a ray is 10 cm times the sine of the angle away from the normal. So equal distances mean equal angles.

If you cannot try it, this is what happens.
A table with the columns Point and Distance from normal. Row 1: on the incident ray, 6.1 cm. Row 2: on the reflected ray, 6.0 cm.PointFrom normalincident ray6.1 cmreflected ray6.0 cm

Try it

The silver surface of water

You need: a clear glass or jar, filled with water, a table.

Be safe: Use a glass or jar that cannot tip over, away from the edge of the table. If you have a clear plastic cup, use that. Wipe up any spills.

  1. Stand the glass of water on a table, away from the edge.
  2. Crouch so that your eye is a little lower than the top of the water. Look up at the underside of the water surface, through the side of the glass.
  3. Look steeply upwards. Then look at a slant, nearly along the underside of the surface.

What you will see: Looking steeply up, you see through the surface to the ceiling. At a shallow slant, the surface looks like a silver mirror. You see the bottom of the glass or the objects in the water. This is total internal reflection.

If you cannot try it, this is what happens.
A table with the columns Look and You see. Row 1: steeply up, through it. Row 2: at a slant, a mirror.LookYou seesteeply upthrough itat a slanta mirror

A wrong idea many people have

A ray that makes 30° with a mirror has an angle of incidence of 30°.

Why it feels right: The angle you can see is the one between the ray and the mirror. It is the first angle we think to measure on a flat surface.

The test that breaks it: Draw the normal first. The ray makes 30° with the mirror, and the mirror makes 90° with the normal. So the ray makes 90° − 30° = 60° with the normal. The law of reflection uses the angle from the normal.

What is true: Angles of incidence, reflection and refraction are always measured from the normal, never from the surface.

A wrong idea many people have

Light always bends towards the normal when it changes material.

Why it feels right: Our first picture of refraction is a ray going into water or glass, and that ray bends towards the normal.

The test that breaks it: Look at the ray going from glass into air. It bends away from the normal, and the angle in the air is bigger than the angle in the glass.

What is true: A ray bends towards the normal when it goes into a material where light is slower. It bends away from the normal when it goes into a material where light is faster.

A wrong idea many people have

Total internal reflection can happen when light goes from air into glass, if the angle is large enough.

Why it feels right: A large angle seems to be all it takes, and light reflects from a glass window at a slant.

The test that breaks it: From air into glass, sin r = sin i ÷ 1.5. This is always less than 1, so a refracted ray always exists. A window at a slant also reflects a little, but some light always goes in.

What is true: Total internal reflection needs two things. The light goes from a higher index to a lower index. The angle of incidence is greater than the critical angle.

Key words

angle of incidence
the angle between the incident ray and the normal.
angle of reflection
the angle between the reflected ray and the normal.
virtual image
an image made where rays seem to come from, so it cannot be caught on a screen.
refractive index
the speed of light in a vacuum divided by its speed in the material. It is written n.
critical angle
the angle of incidence in a denser material at which the refracted ray runs along the edge. sin c = 1 ÷ n.
total internal reflection
all the light reflected inside a material, when it meets the edge beyond the critical angle.
optical fibre
a thin glass thread that carries light along it by total internal reflection.

Back to your guess

The answer is B: 70°, because scientists measure from the line at right angles to the mirror.

Scientists measure the angle from the normal, the line at 90° to the mirror. A ray that makes 20° with the surface makes 90° − 20° = 70° with the normal. That is the angle of incidence.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A ray of light makes an angle of 35° with the surface of a flat mirror. What is the angle of incidence?

    Tap your answer.
    1. Not this one. That is the angle with the mirror. The angle of incidence is measured from the normal, which is at 90° to the mirror. Try another.

    2. Not this one. You added 35° to 90°. Take it away instead: 90° − 35°. Try another.

    3. Yes, that is right. The normal is at 90° to the mirror. So the angle from the normal is 90° − 35° = 55°.

    See the answer

    55°

    The normal is at 90° to the mirror. So the angle from the normal is 90° − 35° = 55°.

    • If you chose A: That is the angle with the mirror. The angle of incidence is measured from the normal, which is at 90° to the mirror.
    • If you chose B: You added 35° to 90°. Take it away instead: 90° − 35°.
  2. An object stands 5 cm in front of a flat mirror, as in the picture. How far is the image from the object?

    A flat mirror seen from the side, labelled mirror. An upright arrow in front of it is labelled object. A dashed arrow behind the mirror is labelled image. Two rays go from the object to the mirror and reflect to an eye on the left, labelled eye. A mark under the picture from the object to the mirror is labelled 5 cm.objectimagemirroreye5 cm
    Tap your answer.
    1. Not this one. That is the distance from the object to the mirror. The image is also 5 cm behind the mirror, so the object and the image are further apart. Try another.

    2. Not this one. The image is not halfway. It is as far behind the mirror as the object is in front. Try another.

    3. Not this one. The image is as far behind the mirror as the object is in front, which is 5 cm, not 10 cm. Try another.

    4. Yes, that is right. The image is 5 cm behind the mirror, and the object is 5 cm in front of it. The distance between them is 5 + 5 = 10 cm.

    See the answer

    10 cm

    The image is 5 cm behind the mirror, and the object is 5 cm in front of it. The distance between them is 5 + 5 = 10 cm.

    • If you chose A: That is the distance from the object to the mirror. The image is also 5 cm behind the mirror, so the object and the image are further apart.
    • If you chose B: The image is not halfway. It is as far behind the mirror as the object is in front.
    • If you chose C: The image is as far behind the mirror as the object is in front, which is 5 cm, not 10 cm.
  3. A ray of light goes from air into water at a slant. What happens to it?

    Tap your answer.
    1. Yes, that is right. Light is slower in water than in air. A ray that goes into a slower material bends towards the normal.

    2. Not this one. The light does slow down, but a slower speed turns the ray towards the normal, not away from it. Try another.

    3. Not this one. Light is slower in water than in air. It slows down, and that turns the ray towards the normal. Try another.

    4. Not this one. Clear materials still change the speed of light. A ray at a slant bends at the edge. Go to S3.9.2. Try another.

    See the answer

    It bends towards the normal, because the light slows down.

    Light is slower in water than in air. A ray that goes into a slower material bends towards the normal.

    • If you chose B: The light does slow down, but a slower speed turns the ray towards the normal, not away from it.
    • If you chose C: Light is slower in water than in air. It slows down, and that turns the ray towards the normal.
    • If you chose D: Clear materials still change the speed of light. A ray at a slant bends at the edge. Go to S3.9.2.
  4. A ray of light travels along the normal as it goes from air into glass. What happens?

    Tap your answer.
    1. Not this one. It is already on the normal, so there is no angle to bend. The angle of incidence is 0°. Try another.

    2. Not this one. Total internal reflection needs light inside the glass going out, at a large angle. From air into glass, light always goes in. Try another.

    3. Yes, that is right. The angle of incidence is 0°, and sin 0° = 0, so the angle of refraction is also 0°. The ray does not bend, but it slows down in the glass.

    See the answer

    It goes straight on, at a lower speed.

    The angle of incidence is 0°, and sin 0° = 0, so the angle of refraction is also 0°. The ray does not bend, but it slows down in the glass.

    • If you chose A: It is already on the normal, so there is no angle to bend. The angle of incidence is 0°.
    • If you chose B: Total internal reflection needs light inside the glass going out, at a large angle. From air into glass, light always goes in.
  5. Which pair of conditions is needed for total internal reflection?

    Tap your answer.
    1. Not this one. From air into glass, light always crosses. Total internal reflection is for light going from the denser material outwards. Try another.

    2. Not this one. Under the critical angle some light leaves the glass. The angle must be greater than the critical angle. Try another.

    3. Not this one. Glass has a higher index than water, so light going from water into glass bends towards the normal and always crosses. Try another.

    4. Yes, that is right. The light must go from a higher index to a lower index, such as glass to air. The angle of incidence must also be greater than the critical angle.

    See the answer

    Light goes from glass into air, and the angle of incidence is greater than the critical angle.

    The light must go from a higher index to a lower index, such as glass to air. The angle of incidence must also be greater than the critical angle.

    • If you chose A: From air into glass, light always crosses. Total internal reflection is for light going from the denser material outwards.
    • If you chose B: Under the critical angle some light leaves the glass. The angle must be greater than the critical angle.
    • If you chose C: Glass has a higher index than water, so light going from water into glass bends towards the normal and always crosses.
  6. Why does light stay inside a bent glass fibre?

    A glass fibre that runs straight and then bends upwards, shaded with dots and labelled glass fibre. A ray of light enters at the left end and zigzags along the fibre to the top end.glass fibre
    Tap your answer.
    1. Yes, that is right. The ray meets the wall at an angle of incidence greater than the critical angle. So all of it is reflected back inside. It repeats this all along the fibre.

    2. Not this one. Paint would not send the light back. The light is reflected, as in a mirror, by total internal reflection at the glass wall. Try another.

    3. Not this one. Light leaves glass freely at a small angle of incidence. It stays in only when it meets the wall beyond the critical angle. Try another.

    See the answer

    The ray meets the wall beyond the critical angle each time, so it is totally reflected.

    The ray meets the wall at an angle of incidence greater than the critical angle. So all of it is reflected back inside. It repeats this all along the fibre.

    • If you chose B: Paint would not send the light back. The light is reflected, as in a mirror, by total internal reflection at the glass wall.
    • If you chose C: Light leaves glass freely at a small angle of incidence. It stays in only when it meets the wall beyond the critical angle.
  7. A ray goes from air into glass of index 1.5. The angle of incidence is 45°. What is the angle of refraction?

    Tap your answer.
    1. Not this one. You divided the angle by 1.5. The rule uses sines: sin r = sin i ÷ n. Try another.

    2. Yes, that is right. sin r = sin 45° ÷ 1.5 = 0.7071 ÷ 1.5 = 0.4714. The angle whose sine is 0.4714 is 28.1°.

    3. Not this one. You multiplied the angle by 1.5. The ray bends towards the normal, so r must be smaller than 45°, and the rule uses sines. Try another.

    See the answer

    28.1°

    sin r = sin 45° ÷ 1.5 = 0.7071 ÷ 1.5 = 0.4714. The angle whose sine is 0.4714 is 28.1°.

    • If you chose A: You divided the angle by 1.5. The rule uses sines: sin r = sin i ÷ n.
    • If you chose C: You multiplied the angle by 1.5. The ray bends towards the normal, so r must be smaller than 45°, and the rule uses sines.
  8. From an earlier chapter

    A cliff returns the echo of a shout after 2 s. The speed of sound is 340 m/s. How far away is the cliff?

    Tap your answer.
    1. Not this one. That is the whole way there and back. The cliff is half that far away. Go to S4.10.4. Try another.

    2. Not this one. You halved twice. The echo time is the time for the whole way there and back, so halve the distance once. Go to S4.10.4. Try another.

    3. Yes, that is right. The sound goes 340 × 2 = 680 m there and back. The cliff is half of that: 340 m.

    See the answer

    340 m

    The sound goes 340 × 2 = 680 m there and back. The cliff is half of that: 340 m.

    • If you chose A: That is the whole way there and back. The cliff is half that far away. Go to S4.10.4.
    • If you chose B: You halved twice. The echo time is the time for the whole way there and back, so halve the distance once. Go to S4.10.4.
  9. A ray goes from air into glass of index 1.5. The angle of incidence is 50°. What is the angle of refraction, to one decimal place?

    See the answer

    30.7°

    sin r = sin 50° ÷ 1.5 = 0.7660 ÷ 1.5 = 0.5107. The angle whose sine is 0.5107 is 30.7°.

  10. Diamond has a refractive index of 2.42. What is the critical angle for light going from diamond into air, to one decimal place?

    See the answer

    24.4°

    sin c = 1 ÷ n = 1 ÷ 2.42 = 0.4132. The angle whose sine is 0.4132 is 24.4°.

Keep it

Copy this into your exercise book.

Angles are from the normal. Reflection: i = r. Slower material: bends towards the normal. Total internal reflection: denser to less dense, beyond the critical angle, sin c = 1 ÷ n.

Draw: A glass edge with three rays: one bending out, one along the edge, one reflected, and the normal as a dashed line.

S4.10.6 Chapter 6

Lenses

Held far from a window, a magnifying glass puts a small, upside down picture of the window on a sheet of paper. Why is the picture upside down?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

A lens is a clear piece of glass or plastic with curved faces. It bends light by refraction at each face. A converging lens is thicker in the middle and brings parallel rays together. A diverging lens is thinner in the middle and spreads them apart.

The principal focus, F, of a converging lens is the point where rays parallel to the axis meet after the lens. The axis is the line through the middle of the lens at right angles to its faces. The distance from the lens to F is the focal length, f.

A converging lens drawn as an upright shape, labelled lens. Five parallel rays arrive from the left, bend at the lens, and all pass through one point on the axis to the right of the lens, labelled focus. They spread out again beyond it. A dashed line is the axis.lensfocus
Parallel rays meet at the principal focus. A fatter lens bends them more, so its focus is nearer.

A lens works both ways, so it has a focus on each side, at the same distance f. A fatter, more curved lens has a shorter focal length. A real lens bends light at both faces. To keep the picture clear, we draw one bend, at one line through the middle.

To find the image of an object, draw an arrow for the object. Then trace three rays from its tip. These three rules are enough.

Rule one: a ray that travels parallel to the axis bends and goes through the focus on the far side. Rule two: a ray that goes through the centre of the lens carries straight on. Rule three: a ray that passes through the focus on the near side leaves the lens parallel to the axis.

Where the rays meet is the image of the tip. If the rays really cross there, the image is a real image. You can catch it on a screen. It is upside down. The picture of the window is a real image.

A converging lens with an object arrow on the left, labelled object, beyond twice the focal length. A mark F on each side of the lens is the focus, and a mark 2F on each side is twice as far. Three rays from the tip of the object bend at the lens and meet on the right, between F and 2F. A small upside down arrow there is labelled image. The lens is labelled.lensobjectimageFF2F2F
An object beyond 2F gives a real, upside down, smaller image between F and 2F.
A converging lens with an object arrow on the left, labelled object, exactly at 2F. Three rays from the tip bend at the lens and meet at 2F on the right. An upside down arrow of the same size there is labelled image. The marks F and 2F are shown on both sides. The lens is labelled.lensobjectimageFF2F2F
An object at 2F gives a real, upside down image of the same size, also at 2F.
A converging lens with an object arrow on the left, labelled object, between F and 2F. Three rays from the tip bend at the lens and meet far to the right, beyond 2F. A larger upside down arrow there is labelled image. The marks F are shown on both sides. The lens is labelled.lensobjectimageFF
An object between F and 2F gives a real, upside down, larger image beyond 2F.

Move the object closer, to a point inside F. The rays now leave the lens spreading out. They never cross on the far side. Look backwards along the rays, and they seem to come from one point on the object's side.

A converging lens with an object arrow on the left, labelled object, nearer to the lens than F. Three rays from the tip bend at the lens and spread out to the right. Dotted lines go back from the lens to the left and meet behind the object, where a larger upright dashed arrow is labelled image. The marks F are shown on both sides. The lens is labelled.lensobjectimageFF
An object inside F gives an upright, larger image on the same side as the object. This is a magnifying glass.

This image is a virtual image. Light does not really pass through it, so it cannot be caught on a screen. Your eye sees it because the rays seem to come from it. A virtual image from a converging lens is upright and larger. This is how a magnifying glass works.

A table with the columns Object, Image and Size. Row 1: beyond 2F, real, smaller. Row 2: at 2F, real, same. Row 3: 2F to F, real, larger. Row 4: inside F, virtual, larger.ObjectImageSizebeyond 2Frealsmallerat 2Frealsame2F to Freallargerinside Fvirtuallarger
The image from a converging lens. A real image is upside down. A virtual image is upright. An object at F gives no image: the rays leave parallel.

Diagrams give the shape of the answer. A rule gives the numbers. For a thin lens, 1/f = 1/u + 1/v. Here u is the distance from the object to the lens and v is the distance from the lens to the image. All are in the same unit.

This book counts distances to a real image as positive, and a virtual image has a negative v. Other books use other sign rules and write the formula another way. The physics is the same. To find v, rearrange: 1/v = 1/f − 1/u.

The magnification is the height of the image divided by the height of the object. It also equals v ÷ u when you ignore the sign. A magnification of 2 means the image is twice as tall as the object. It has no unit.

Worked example

An object is 4 cm from a converging lens of focal length 3 cm. Find the image distance and the magnification.

  1. For a real image, 1/v = 1/f − 1/u. Here f = 3 cm and u = 4 cm.
  2. 1/v = 1/3 − 1/4 = 4/12 − 3/12 = 1/12.
  3. So v = 12 cm. The image is real and 12 cm from the lens.
  4. The magnification is v ÷ u = 12 ÷ 4 = 3.

Answer: 12 cm and a magnification of 3

A converging lens with an object arrow on the left, labelled object, between F and 2F. Three rays from the tip bend at the lens and meet to the right, beyond 2F. A larger upside down arrow there is labelled image. The marks F are shown on both sides. The lens is labelled.lensobjectimageFF
Any object between F and 2F gives a larger real image, as in the worked example.

Now the diverging lens. It spreads parallel rays so that they seem to come from a focus on the object's side. The three rays bend the other way. A ray parallel to the axis leaves as if from the near focus. A ray through the centre goes straight on. A ray aimed at the focus on the far side leaves parallel to the axis.

A diverging lens drawn as an upright line with arrow heads pointing inwards, labelled lens. An object arrow on the left is labelled object. Three rays from its tip spread out after the lens. Dotted lines go back to the left and meet at the tip of a small upright dashed arrow between the lens and the object, labelled image. The marks F are on both sides.lensobjectimageFF
A diverging lens always gives an upright, smaller, virtual image between the lens and the object.

Now turn to the eye. The cornea and the lens of the eye act together as a converging lens. They make a real, upside down image of what you look at on the retina at the back of the eye. The brain turns the picture the right way up.

A muscle changes the shape of the eye lens. To look at something close, the lens is made fatter, so its focal length is shorter. To look at something far, it is made thinner. This focusing is why a clear eye sees both near and far.

In short sight, the eye bends light too much, or the eyeball is too long. Rays from a far object cross in front of the retina, and the image on the retina is blurred. Near things are clear. A diverging lens in front of the eye spreads the rays a little and moves the image back onto the retina.

A short sighted eye seen from the side, with the retina, labelled retina, at the back and the eye lens, labelled lens, near the front. Three parallel rays arrive from the left, from a far object, labelled far object. They bend at the lens and cross at a point in front of the retina. They then spread out again, so they reach the retina at three different places.lensretinafar object
Short sight without glasses. The rays from a far object meet in front of the retina, so the picture on the retina is blurred.
The same short sighted eye with a diverging lens in front of it, labelled glasses. Three parallel rays from a far object, labelled far object, spread a little at the glasses, bend at the eye lens, labelled lens, and meet exactly at one point on the retina, labelled retina.lensretinaglassesfar object
Short sight corrected with a diverging lens. The rays now meet on the retina.

In long sight, the eye bends light too little, or the eyeball is too short. Rays from a near object would meet behind the retina, so the image on the retina is blurred. Far things are clear. A converging lens in front of the eye helps to bend the rays and moves the image forward onto the retina.

An eye seen from the side, with an object arrow on the left, labelled object. A converging glasses lens stands in front of the eye. The rays from the tip and the foot of the object come together a little at the glasses lens, bend at the eye lens, labelled lens, and meet exactly on the retina, labelled retina. The small upside down image there is labelled image.lensretinaimageobject
Long sight corrected with a converging lens. The image is back on the retina. The object is drawn near, as a book would be.

Opticians give the strength of a lens as its power: P = 1 ÷ f, with f in metres. The unit is the dioptre, D. A converging lens has a positive power and a diverging lens a negative one. A short focal length is a high power.

A short sighted person sees clearly only as far as a distance called the far point. A diverging lens in front of the eye should make a faraway object seem to be at the far point. So its focal length equals the far point distance, with the sign of a diverging lens.

Worked example

A short sighted person sees clearly only up to 0.5 m. What power of lens do they need?

  1. The lens must make faraway things look as if they are 0.5 m away. So it is a diverging lens with focal length 0.5 m.
  2. The size of the power is 1 ÷ f = 1 ÷ 0.5 = 2 D.
  3. A diverging lens has a negative power, so the lens is −2 D.

Answer: The lens is −2 D

Try it

Make a picture with a lens

You need: a lens from an old pair of reading glasses, or a magnifying glass, a sheet of white paper, a ruler, a room with a window or a lamp across the room.

Be safe: Never look at the Sun, and never point a lens at the Sun. A lens focuses sunlight into a hot spot that can burn skin, paper and eyes. Choose a window or a lamp, and keep the Sun out of the view. Do not look into a bright lamp through the lens.

  1. Hold the paper up as a screen, with its back to the window. Hold the lens between the window and the paper.
  2. Move the lens towards and away from the paper until a sharp small picture of the window appears on the paper.
  3. Measure the distance from the lens to the paper. This is about the focal length f, because the window is so far away.
  4. Work out the power: P = 1 ÷ f, with f in metres.

What you will see: The picture on the paper is upside down and smaller than the window. It is a real image. The window is far away, so the image forms very near the focus.

If you cannot try it, this is what happens.
A table with the columns Measure and Value. Row 1: lens to paper, 12 cm. Row 2: focal length f, 0.12 m. Row 3: power 1 ÷ f, 8.3 D.MeasureValuelens to paper12 cmfocal length f0.12 mpower 1 ÷ f8.3 D

A wrong idea many people have

A converging lens always makes an upside down image.

Why it feels right: The first picture we see, a window on a screen, is upside down. Most pictures of lenses show an upside down image.

The test that breaks it: Hold a magnifying glass close to some print. The print looks larger and the right way up. That image is virtual, and the object is inside the focal length.

What is true: A real image from a converging lens is upside down. When the object is inside F, the image is virtual, upright and larger.

A wrong idea many people have

A short sighted person needs a lens that magnifies, so a converging lens.

Why it feels right: We call a magnifying glass a lens that helps us see, and a short sighted person sees things badly.

The test that breaks it: In short sight, the rays from a far object cross in front of the retina. A converging lens would bring them together even sooner, and make it worse.

What is true: Short sight needs a diverging lens, to spread the rays. Long sight needs a converging lens, to bring them together sooner.

A wrong idea many people have

A lens has one focus.

Why it feels right: Pictures of parallel rays arriving from one side show only one point where the rays meet.

The test that breaks it: Turn the lens round and send the rays from the other side. They meet at a point the same distance from the lens on the other side.

What is true: A lens has a focus on each side, at the same distance f. Rule three uses the focus on the object's side.

Key words

principal focus
the point where rays that arrive parallel to the axis meet after a converging lens.
focal length
the distance from the lens to its principal focus. Its symbol is f.
real image
an image where rays really cross. It can be caught on a screen, and it is upside down.
virtual image
an image where the rays only seem to cross. It cannot be caught on a screen.
magnification
the height of the image divided by the height of the object.
short sight
a fault in which far things are blurred. A diverging lens corrects it.
power
the strength of a lens, P = 1 ÷ f with f in metres.
dioptre
the unit of the power of a lens, written D.

Back to your guess

The answer is A: Rays from the top of the window cross after the lens and reach the bottom of the picture.

Rays from the top of the window go through the lens, bend, and cross. After crossing they are below the axis, so they land at the bottom of the picture. Rays from the bottom land at the top. So the picture is upside down.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A ray travels parallel to the axis towards a converging lens. Where does it go after the lens?

    Tap your answer.
    1. Not this one. Only a ray through the centre of the lens goes straight on. A ray that is off the centre is bent towards the focus. Try another.

    2. Not this one. A ray leaves a lens parallel to the axis only if it passed through the near focus. A ray that arrives parallel is bent. Try another.

    3. Not this one. The bend carries the ray across the axis, to the focus on the far side of the lens. Try another.

    4. Yes, that is right. Rule one: a ray that arrives parallel to the axis goes through the focus on the far side of a converging lens.

    See the answer

    Through the focus on the far side

    Rule one: a ray that arrives parallel to the axis goes through the focus on the far side of a converging lens.

    • If you chose A: Only a ray through the centre of the lens goes straight on. A ray that is off the centre is bent towards the focus.
    • If you chose B: A ray leaves a lens parallel to the axis only if it passed through the near focus. A ray that arrives parallel is bent.
    • If you chose C: The bend carries the ray across the axis, to the focus on the far side of the lens.
  2. The picture shows an object beyond 2F from a converging lens. What kind of image does the lens make?

    A converging lens with an object arrow on the left, labelled object, beyond twice the focal length. The marks F on both sides are the focus, and the marks 2F on both sides are twice as far. Three rays from the tip of the object are drawn as far as the lens only. The lens is labelled.lensobjectFF2F2F
    Tap your answer.
    1. Not this one. That is for an object inside F. Here the object is beyond 2F, and the rays do cross on the far side. Try another.

    2. Yes, that is right. An object beyond 2F gives a real image between F and 2F on the other side. It is upside down and smaller than the object.

    3. Not this one. A real image from a converging lens is always upside down. This one is smaller, because the object is beyond 2F. Try another.

    4. Not this one. A virtual image from a converging lens is upright. This object is beyond 2F, so the image is real. Try another.

    See the answer

    Real, upside down and smaller

    An object beyond 2F gives a real image between F and 2F on the other side. It is upside down and smaller than the object.

    • If you chose A: That is for an object inside F. Here the object is beyond 2F, and the rays do cross on the far side.
    • If you chose C: A real image from a converging lens is always upside down. This one is smaller, because the object is beyond 2F.
    • If you chose D: A virtual image from a converging lens is upright. This object is beyond 2F, so the image is real.
  3. An object is 20 cm from a converging lens of focal length 15 cm. How far from the lens is the image?

    Tap your answer.
    1. Not this one. You added 15 and 20. The rule works with reciprocals: 1/v = 1/f − 1/u. Try another.

    2. Not this one. You added the reciprocals. The rule has a minus: 1/v = 1/f − 1/u. Try another.

    3. Yes, that is right. 1/v = 1/15 − 1/20 = 4/60 − 3/60 = 1/60. So v = 60 cm.

    See the answer

    60 cm

    1/v = 1/15 − 1/20 = 4/60 − 3/60 = 1/60. So v = 60 cm.

    • If you chose A: You added 15 and 20. The rule works with reciprocals: 1/v = 1/f − 1/u.
    • If you chose B: You added the reciprocals. The rule has a minus: 1/v = 1/f − 1/u.
  4. An object is 3 cm from a converging lens of focal length 5 cm. Which describes the image?

    Tap your answer.
    1. Yes, that is right. The object is nearer than F. The rays spread out after the lens, and look as if they come from a larger, upright, virtual image. This is how a magnifying glass works.

    2. Not this one. The object is inside F, so the rays leave the lens spreading out and they do not cross on the far side. Try another.

    3. Not this one. A small real image needs the object beyond 2F. This object is inside F. Try another.

    4. Not this one. No image forms only when the object is exactly at F. Inside F there is a virtual image. Try another.

    See the answer

    Virtual, upright and larger

    The object is nearer than F. The rays spread out after the lens, and look as if they come from a larger, upright, virtual image. This is how a magnifying glass works.

    • If you chose B: The object is inside F, so the rays leave the lens spreading out and they do not cross on the far side.
    • If you chose C: A small real image needs the object beyond 2F. This object is inside F.
    • If you chose D: No image forms only when the object is exactly at F. Inside F there is a virtual image.
  5. Which lens corrects short sight?

    Tap your answer.
    1. Not this one. In short sight the rays already meet too soon. A converging lens would make them meet even sooner. Try another.

    2. Yes, that is right. In short sight, rays from a far object meet in front of the retina. A diverging lens spreads them, so that they meet on the retina.

    3. Not this one. Flat glass does not bend rays that arrive straight on. The eye needs rays that are spread a little. Try another.

    See the answer

    A diverging lens

    In short sight, rays from a far object meet in front of the retina. A diverging lens spreads them, so that they meet on the retina.

    • If you chose A: In short sight the rays already meet too soon. A converging lens would make them meet even sooner.
    • If you chose C: Flat glass does not bend rays that arrive straight on. The eye needs rays that are spread a little.
  6. A long sighted person cannot see near things clearly. What goes wrong in the eye?

    Tap your answer.
    1. Not this one. That is what happens in short sight. In long sight the eye bends the rays too little. Try another.

    2. Not this one. The picture is blurred, not too small. The rays do not meet on the retina. Try another.

    3. Yes, that is right. The eye bends light too little, or the eyeball is too short. So the rays from a near object would meet behind the retina. A converging lens in glasses helps.

    See the answer

    The rays from a near object would meet behind the retina.

    The eye bends light too little, or the eyeball is too short. So the rays from a near object would meet behind the retina. A converging lens in glasses helps.

    • If you chose A: That is what happens in short sight. In long sight the eye bends the rays too little.
    • If you chose B: The picture is blurred, not too small. The rays do not meet on the retina.
  7. A lens has a focal length of 25 cm. What is its power?

    Tap your answer.
    1. Not this one. That is the focal length in cm, not the power. The power is 1 divided by the focal length in metres. Try another.

    2. Not this one. 0.25 is the focal length in metres. The power is 1 divided by it. Try another.

    3. Not this one. You worked out 1 ÷ 25, with f in cm. Change f to metres first: 25 cm is 0.25 m. Try another.

    4. Yes, that is right. First change to metres: 25 cm = 0.25 m. Then P = 1 ÷ f = 1 ÷ 0.25 = 4 D.

    See the answer

    4 D

    First change to metres: 25 cm = 0.25 m. Then P = 1 ÷ f = 1 ÷ 0.25 = 4 D.

    • If you chose A: That is the focal length in cm, not the power. The power is 1 divided by the focal length in metres.
    • If you chose B: 0.25 is the focal length in metres. The power is 1 divided by it.
    • If you chose C: You worked out 1 ÷ 25, with f in cm. Change f to metres first: 25 cm is 0.25 m.
  8. From an earlier chapter

    A dense crystal has a refractive index of 2.0. What is the critical angle for light going from it into air?

    Tap your answer.
    1. Yes, that is right. sin c = 1 ÷ n = 1 ÷ 2.0 = 0.5. The angle whose sine is 0.5 is 30°.

    2. Not this one. That is 90° − 30°. The critical angle is the angle whose sine is 1 ÷ n, which is 0.5. Go to S4.10.5. Try another.

    3. Not this one. That is the angle whose tangent is 0.5. The rule uses the sine: sin c = 1 ÷ n. Go to S4.10.5. Try another.

    See the answer

    30°

    sin c = 1 ÷ n = 1 ÷ 2.0 = 0.5. The angle whose sine is 0.5 is 30°.

    • If you chose B: That is 90° − 30°. The critical angle is the angle whose sine is 1 ÷ n, which is 0.5. Go to S4.10.5.
    • If you chose C: That is the angle whose tangent is 0.5. The rule uses the sine: sin c = 1 ÷ n. Go to S4.10.5.
  9. An object is 20 cm from a converging lens of focal length 12 cm. How far from the lens is the image?

    See the answer

    30 cm

    1/v = 1/12 − 1/20 = 5/60 − 3/60 = 2/60 = 1/30. So v = 30 cm.

  10. A lens has a focal length of 40 cm. What is its power in dioptres?

    See the answer

    2.5 D

    Change to metres: 40 cm = 0.4 m. Then P = 1 ÷ f = 1 ÷ 0.4 = 2.5 D.

Keep it

Copy this into your exercise book.

Three rays: parallel through the far focus, through the centre straight, through the near focus out parallel. 1/f = 1/u + 1/v. Short sight: diverging. Long sight: converging. P = 1 ÷ f.

Draw: A converging lens with its axis and F marks, an object arrow, and the three rays meeting at an upside down image.

S4.10.7 Chapter 7

The electromagnetic spectrum

In empty space, a radio wave and a gamma ray leave a star at the same moment. Which reaches the Earth first?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

  • You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Light is not the only kind of wave that crosses empty space. Radio waves, the heat from a fire, and the X-rays at a hospital are all waves of the same family. They are electromagnetic waves.

An electromagnetic wave is a pattern of electric and magnetic fields that changes in step and moves through space. It is a transverse wave. It needs no medium, so it crosses a vacuum, and it carries energy.

All electromagnetic waves travel at the same speed in a vacuum: 3 × 10⁸ m/s, which is 300 000 km/s. Radio waves, light and gamma rays all travel at this speed. In air the speed is almost the same. In glass and water it is lower.

A table with the columns Wave and Speed in a vacuum. Row 1: radio waves, 3 × 10⁸ m/s. Row 2: visible light, 3 × 10⁸ m/s. Row 3: gamma rays, 3 × 10⁸ m/s.WaveSpeedradio waves3 × 10⁸ m/svisible light3 × 10⁸ m/sgamma rays3 × 10⁸ m/s
Every electromagnetic wave goes at the same speed in a vacuum.

What makes the waves differ is their frequency and wavelength. They are linked by the wave equation. For electromagnetic waves in a vacuum we write c for the speed: c = fλ. So the longer the wavelength, the lower the frequency.

The whole family is the electromagnetic spectrum. It has seven main kinds. In order of falling wavelength and rising frequency they are radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.

Seven rows in order from the top. Row 1: radio, over 1 m. Row 2: microwave, 1 mm to 1 m. Row 3: infrared, 700 nm to 1 mm. Row 4: visible, 400 to 700 nm. Row 5: ultraviolet, 10 to 400 nm. Row 6: X-ray, 0.01 to 10 nm. Row 7: gamma, under 0.01 nm. A long arrow down the left side is labelled frequency rises. A long arrow down the right side is labelled wavelength falls.frequency riseswavelength fallsradioover 1 mmicrowave1 mm to 1 minfrared700 nm to 1 mmvisible400 to 700 nmultraviolet10 to 400 nmX-ray0.01 to 10 nmgammaunder 0.01 nm
The electromagnetic spectrum. The edges are not sharp, and different books put them a little differently. Some books count microwaves as the short end of radio. X-rays and gamma rays overlap.

The wavelengths run from kilometres to far smaller than an atom. A nanometre, nm, is 10⁻⁹ m, a billionth of a metre. The frequencies run in the opposite order, from a few kilohertz to more than 10¹⁹ Hz. Visible light is only a thin slice.

One megahertz, MHz, is 1 000 000 Hz, which is 10⁶ Hz. FM radio stations are often near 100 MHz. Mobile phones and Wi-Fi use frequencies from several hundred to several thousand megahertz.

Worked example

An FM radio station broadcasts at 100 MHz. Find the wavelength of its waves.

  1. Change the frequency to hertz: 100 MHz = 100 × 10⁶ = 1 × 10⁸ Hz.
  2. Rearrange c = fλ to λ = c ÷ f.
  3. λ = 3 × 10⁸ ÷ (1 × 10⁸) = 3 m.

Answer: 3 m

Worked example

Blue green light has a wavelength of 500 nm. Find its frequency. Use 1 nm = 10⁻⁹ m.

  1. Change the wavelength to metres: 500 nm = 500 × 10⁻⁹ = 5 × 10⁻⁷ m.
  2. Rearrange c = fλ to f = c ÷ λ.
  3. f = 3 × 10⁸ ÷ (5 × 10⁻⁷) = 0.6 × 10¹⁵ Hz.
  4. In standard form, f = 6 × 10¹⁴ Hz.

Answer: 6 × 10¹⁴ Hz

Now go through the seven kinds, starting at the long wave end. Each has uses that come from how it passes through things and how things absorb it.

Radio waves have the longest wavelengths. They carry radio and television, signals to aircraft and ships, and the signals between spacecraft and Earth. Some stars and galaxies also send them out, and radio telescopes collect them.

Microwaves are shorter. A microwave oven uses them because water in food absorbs them and heats up. Mobile phones, satellites, Wi-Fi and radar use them too. The oven is shielded so that the microwaves stay inside.

Infrared waves are given out by every warm object. Heaters, toasters and your own body give them out. A television remote control sends a code in infrared. Cameras that sense infrared make pictures of heat, and they work in the dark.

Visible light is the only part we see. It lets us see, and it is used in lamps, cameras, lasers and optical fibres. Its wavelengths run from about 700 nm for red light to about 400 nm for violet light.

A table with the columns Colour and λ in nm. Row 1: red, 700. Row 2: orange, 620. Row 3: yellow, 580. Row 4: green, 530. Row 5: blue, 470. Row 6: violet, 410.Colourλ in nmred700orange620yellow580green530blue470violet410
Typical wavelengths of the colours, rounded. The colours blend into each other, and indigo lies between blue and violet.

Ultraviolet is a little shorter than violet light, and our eyes cannot see it. It is used to sterilise water and surfaces, and to show security marks on notes and cards. Lamps that make some materials glow use it. Sunlight contains it.

X-rays pass through soft parts of the body and are blocked more by bone. Doctors and dentists use them to take pictures of bones and teeth. Airports use them to look inside bags.

Gamma rays have the shortest wavelengths and are very penetrating. They are used to kill cancer cells, to sterilise medical equipment, and to follow a tracer through the body. They come from the nuclei of atoms.

A table with the columns Wave and A use. Row 1: radio, broadcasting. Row 2: microwave, ovens, phones. Row 3: infrared, remote controls. Row 4: visible, seeing, fibres. Row 5: ultraviolet, sterilising. Row 6: X-ray, bone images. Row 7: gamma, treating cancer.WaveA useradiobroadcastingmicrowaveovens, phonesinfraredremote controlsvisibleseeing, fibresultravioletsterilisingX-raybone imagesgammatreating cancer
A use for each kind of electromagnetic wave. The next chapter looks at calls and at health.

X-rays and gamma rays overlap in wavelength. They differ in where they come from. X-rays come from fast moving electrons, and gamma rays from the nuclei of atoms. For the same wavelength they behave in the same way.

A graph with the axes displacement and distance, drawn without numbers. Two waves of the same height. The solid wave is short, with many waves, and its key says gamma ray. The dashed wave is long, with one wave, and its key says radio wave.displacementdistancegamma rayradio wave
Not to scale. A gamma ray wave is far shorter than a radio wave. Both go at the same speed.

Try it

Feel infrared from a warm mug

You need: a mug of hot water, hot from a kettle but not boiling, poured by an adult, a card or a book, your hand.

Be safe: Use water that is hot but not boiling, and let an adult pour it. Stand the mug where it cannot be knocked over, and keep your hand to the side, never over the steam.

  1. Stand the mug on a table. Hold the back of your hand beside the mug, a few centimetres from the side, at the height of the mug. Do not touch it.
  2. Notice the warmth on your hand.
  3. Slide the card in between your hand and the mug. Notice what happens to the warmth.
  4. Take the card away, and notice the warmth come back.

What you will see: The warmth you feel at the side of the mug comes mostly from infrared. Warm air rises, so it does not reach your hand at the side. The card is opaque to infrared, so it stops it.

If you cannot try it, this is what happens.
A table with the columns Between and You feel. Row 1: nothing, warmth. Row 2: a card, no warmth.BetweenYou feelnothingwarmtha cardno warmth

Try it

See a remote control with a phone camera

You need: a television remote control or similar remote, a phone with a camera, in a dim room.

Be safe: The light from a remote control is weak and safe. Never look at the Sun, a laser or a very bright lamp through a camera or in any other way.

  1. Point the end of the remote at your eye from a metre away and press a button. Look for a light.
  2. Now look at the end of the remote through the phone camera screen, and press a button again.
  3. Try the front camera and the back camera.

What you will see: To your eye there is nothing to see, because the remote sends infrared. Many phone cameras can see infrared, and show a flickering light. Some cameras filter it out, so yours may show nothing.

If you cannot try it, this is what happens.
A table with the columns Looking with and You see. Row 1: your eye, nothing. Row 2: a phone camera, a flicker.Looking withYou seeyour eyenothinga phone cameraa flicker

A wrong idea many people have

Gamma rays travel faster than radio waves, because they have more energy.

Why it feels right: In daily life, things with more energy often go faster, and gamma rays sound like the most powerful waves.

The test that breaks it: A burst of gamma rays from a far galaxy can also give a flash of light. Both travel for billions of years, yet the flash can arrive within seconds of the gamma rays. If gamma rays were faster, they would arrive long before the light.

What is true: Every electromagnetic wave travels at 3 × 10⁸ m/s in a vacuum. The energy of a gamma ray goes with its frequency, not its speed.

A wrong idea many people have

Radio waves are sound waves.

Why it feels right: We hear a radio, so it seems that the radio waves must be sound that travels through the air.

The test that breaks it: Sound cannot cross a vacuum, but radio waves cross space to reach spacecraft far from the Earth. Radio waves are electromagnetic. The radio set turns them back into sound.

What is true: Radio waves are electromagnetic waves, like light. They are transverse and need no medium. The loudspeaker makes the sound.

A wrong idea many people have

A wave with a longer wavelength has a higher frequency.

Why it feels right: Both are measures of how big the wave is, and big things can seem to go together.

The test that breaks it: Use c = fλ with c fixed. A radio wave of 3 m has f = 3 × 10⁸ ÷ 3 = 1 × 10⁸ Hz. A wave of 1 m has f = 3 × 10⁸ Hz. The shorter wave has the higher frequency.

What is true: For electromagnetic waves in a vacuum, f and λ go opposite ways. A longer wavelength means a lower frequency.

Key words

electromagnetic waves
transverse waves of electric and magnetic fields. They need no medium and go at 3 × 10⁸ m/s in a vacuum.
electromagnetic spectrum
the family of electromagnetic waves, in order of wavelength.
radio waves
the longest wavelength electromagnetic waves.
microwaves
electromagnetic waves a little shorter than radio waves, used in ovens, phones and Wi-Fi.
infrared
electromagnetic waves given out by warm things, just longer than red light.
ultraviolet
electromagnetic waves just shorter than violet light.
X-rays
short wavelength electromagnetic waves that pass through soft parts of the body.
gamma rays
the shortest wavelength electromagnetic waves. They come from the nuclei of atoms.

Back to your guess

The answer is D: They arrive together.

All electromagnetic waves travel at the same speed in a vacuum, about 3 × 10⁸ m/s. Radio waves and gamma rays differ in frequency and wavelength, not in speed. They arrive together.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The rows list the kinds of electromagnetic wave, with the longest wavelength at the top and the shortest at the bottom. Which kind has the highest frequency?

    Seven rows in order from the top, with the names radio, microwave, infrared, visible, ultraviolet, X-ray and gamma. There are no arrows and no numbers.radiomicrowaveinfraredvisibleultravioletX-raygamma
    Tap your answer.
    1. Not this one. Radio waves have the longest wavelength and so the lowest frequency, because c = fλ. Try another.

    2. Yes, that is right. The list runs from the longest wavelength and lowest frequency to the shortest wavelength and highest frequency. Gamma rays are last, so they have the highest frequency.

    3. Not this one. Visible light is in the middle of the spectrum. Waves further down have higher frequencies. Try another.

    4. Not this one. Infrared has a lower frequency than visible light. The frequency rises down the list. Try another.

    See the answer

    Gamma

    The list runs from the longest wavelength and lowest frequency to the shortest wavelength and highest frequency. Gamma rays are last, so they have the highest frequency.

    • If you chose A: Radio waves have the longest wavelength and so the lowest frequency, because c = fλ.
    • If you chose C: Visible light is in the middle of the spectrum. Waves further down have higher frequencies.
    • If you chose D: Infrared has a lower frequency than visible light. The frequency rises down the list.
  2. What is the speed of all electromagnetic waves in a vacuum?

    Tap your answer.
    1. Not this one. That is the speed of sound in air, a wave of a different kind. Go to S4.10.4. Try another.

    2. Not this one. That is one hundred times too slow. Light goes at 3 × 10⁸ m/s, which is 300 000 km/s. Try another.

    3. Not this one. That is one hundred times too fast. Nothing goes faster than 3 × 10⁸ m/s. Try another.

    4. Yes, that is right. Light, radio waves and all the other electromagnetic waves travel at 3 × 10⁸ m/s in a vacuum.

    See the answer

    3 × 10⁸ m/s

    Light, radio waves and all the other electromagnetic waves travel at 3 × 10⁸ m/s in a vacuum.

    • If you chose A: That is the speed of sound in air, a wave of a different kind. Go to S4.10.4.
    • If you chose B: That is one hundred times too slow. Light goes at 3 × 10⁸ m/s, which is 300 000 km/s.
    • If you chose C: That is one hundred times too fast. Nothing goes faster than 3 × 10⁸ m/s.
  3. A radio wave and an X-ray travel through space. Which statement is true?

    Tap your answer.
    1. Not this one. Frequency does not set the speed in a vacuum. Every electromagnetic wave goes at the same speed there. Try another.

    2. Not this one. Wavelength does not set the speed in a vacuum. A longer wavelength comes with a lower frequency, and the speed stays the same. Try another.

    3. Yes, that is right. Both are electromagnetic waves. In a vacuum they travel at the same speed, 3 × 10⁸ m/s. Only the frequency and wavelength differ.

    See the answer

    They go at the same speed.

    Both are electromagnetic waves. In a vacuum they travel at the same speed, 3 × 10⁸ m/s. Only the frequency and wavelength differ.

    • If you chose A: Frequency does not set the speed in a vacuum. Every electromagnetic wave goes at the same speed there.
    • If you chose B: Wavelength does not set the speed in a vacuum. A longer wavelength comes with a lower frequency, and the speed stays the same.
  4. A radio station broadcasts at 60 MHz. What is the wavelength of its waves? Use c = 3 × 10⁸ m/s.

    Tap your answer.
    1. Not this one. You divided the frequency by c. The wavelength is c ÷ f, the other way round. Try another.

    2. Not this one. You wrote 60 MHz as 6 × 10⁶ Hz. 60 MHz is 60 × 10⁶ = 6 × 10⁷ Hz. Try another.

    3. Not this one. You multiplied c by f. The wavelength is c divided by f. Try another.

    4. Yes, that is right. 60 MHz = 60 × 10⁶ = 6 × 10⁷ Hz. Then λ = c ÷ f = 3 × 10⁸ ÷ (6 × 10⁷) = 5 m.

    See the answer

    5 m

    60 MHz = 60 × 10⁶ = 6 × 10⁷ Hz. Then λ = c ÷ f = 3 × 10⁸ ÷ (6 × 10⁷) = 5 m.

    • If you chose A: You divided the frequency by c. The wavelength is c ÷ f, the other way round.
    • If you chose B: You wrote 60 MHz as 6 × 10⁶ Hz. 60 MHz is 60 × 10⁶ = 6 × 10⁷ Hz.
    • If you chose C: You multiplied c by f. The wavelength is c divided by f.
  5. The microwaves in an oven have a wavelength of 12 cm. What is their frequency? Use c = 3 × 10⁸ m/s.

    Tap your answer.
    1. Not this one. You used 12 instead of 0.12 m. Change cm to m first: 12 cm = 0.12 m. Try another.

    2. Not this one. You multiplied c by the wavelength. The frequency is c divided by the wavelength. Try another.

    3. Yes, that is right. 12 cm = 0.12 m. Then f = c ÷ λ = 3 × 10⁸ ÷ 0.12 = 2.5 × 10⁹ Hz.

    See the answer

    2.5 × 10⁹ Hz

    12 cm = 0.12 m. Then f = c ÷ λ = 3 × 10⁸ ÷ 0.12 = 2.5 × 10⁹ Hz.

    • If you chose A: You used 12 instead of 0.12 m. Change cm to m first: 12 cm = 0.12 m.
    • If you chose B: You multiplied c by the wavelength. The frequency is c divided by the wavelength.
  6. A hospital wants a picture of a broken bone. Which kind of electromagnetic wave does it use?

    Tap your answer.
    1. Yes, that is right. X-rays pass through soft parts of the body but are blocked more by bone. That makes a picture of the bone.

    2. Not this one. Radio waves are used for broadcasting. They are too long in wavelength to make a picture of a bone. Try another.

    3. Not this one. Microwaves are absorbed by water and used in ovens and Wi-Fi. They do not make pictures of bones. Try another.

    See the answer

    X-rays

    X-rays pass through soft parts of the body but are blocked more by bone. That makes a picture of the bone.

    • If you chose B: Radio waves are used for broadcasting. They are too long in wavelength to make a picture of a bone.
    • If you chose C: Microwaves are absorbed by water and used in ovens and Wi-Fi. They do not make pictures of bones.
  7. Which statement about radio waves is true?

    Tap your answer.
    1. Not this one. Sound needs a medium, and radio waves do not. They cross a vacuum, which sound cannot do. Go to S4.10.4. Try another.

    2. Not this one. Radio waves are transverse. They are electromagnetic waves, like light. Go to S4.10.1. Try another.

    3. Not this one. Radio waves have the longest wavelength. Gamma rays have the shortest. Try another.

    4. Yes, that is right. Radio waves are electromagnetic. They are transverse and need no medium, so they can cross space. The radio set turns them into sound.

    See the answer

    They are transverse electromagnetic waves that can cross a vacuum.

    Radio waves are electromagnetic. They are transverse and need no medium, so they can cross space. The radio set turns them into sound.

    • If you chose A: Sound needs a medium, and radio waves do not. They cross a vacuum, which sound cannot do. Go to S4.10.4.
    • If you chose B: Radio waves are transverse. They are electromagnetic waves, like light. Go to S4.10.1.
    • If you chose C: Radio waves have the longest wavelength. Gamma rays have the shortest.
  8. From an earlier chapter

    A wave has a period of 0.02 s. What is its frequency?

    Tap your answer.
    1. Not this one. 0.02 s is the period. The frequency is 1 divided by the period. Go to S4.10.2. Try another.

    2. Yes, that is right. f = 1 ÷ T = 1 ÷ 0.02 = 50 Hz.

    3. Not this one. You may have worked out 1 ÷ 0.2. Work out 1 ÷ 0.02. Go to S4.10.2. Try another.

    See the answer

    50 Hz

    f = 1 ÷ T = 1 ÷ 0.02 = 50 Hz.

    • If you chose A: 0.02 s is the period. The frequency is 1 divided by the period. Go to S4.10.2.
    • If you chose C: You may have worked out 1 ÷ 0.2. Work out 1 ÷ 0.02. Go to S4.10.2.
  9. A signal has a frequency of 150 MHz. What is the wavelength? Use c = 3 × 10⁸ m/s.

    See the answer

    2 m

    150 MHz = 150 × 10⁶ = 1.5 × 10⁸ Hz. Then λ = c ÷ f = 3 × 10⁸ ÷ (1.5 × 10⁸) = 2 m.

  10. Light has a wavelength of 550 nm. Find its frequency in terahertz, THz, to the nearest whole number. Use c = 3 × 10⁸ m/s, 1 nm = 10⁻⁹ m and 1 THz = 10¹² Hz.

    See the answer

    545 THz

    550 nm = 5.5 × 10⁻⁷ m. Then f = c ÷ λ = 3 × 10⁸ ÷ (5.5 × 10⁻⁷) = 5.45 × 10¹⁴ Hz. In terahertz that is 545 THz.

Keep it

Copy this into your exercise book.

Radio, microwave, infrared, visible, ultraviolet, X-ray, gamma: wavelength falls, frequency rises. All go at 3 × 10⁸ m/s in a vacuum. c = fλ.

Draw: Seven boxes in a row in order, with an arrow under them labelled frequency rises and one above labelled wavelength falls.

S4.10.8 Chapter 8

Waves in communication and health

Amina calls her cousin across the sea, and her voice arrives in a fraction of a second. What carries it?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess
  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

A wave can carry energy. By changing the wave in a pattern, it can also carry a message. The plain wave that is changed is the carrier wave. A radio station, a mobile phone and a fibre cable all work this way.

In an analogue signal, the changes copy the message smoothly. In amplitude modulation the height of the carrier follows the message. In frequency modulation the frequency of the carrier follows it.

Two waves one above the other. The top wave is a slow smooth wave labelled message. The wave below it is a much faster wave labelled wave sent, whose height rises and falls in step with the top wave.messagewave sent
Amplitude modulation. The height of the fast wave follows the message.
Two waves one above the other. The top wave is a slow smooth wave labelled message. The wave below it, labelled wave sent, keeps one height. Its waves bunch up where the message is high and spread out where it is low.messagewave sent
Frequency modulation. The frequency of the fast wave follows the message.

Phone calls and the internet use digital signals. In a digital signal the message is turned into numbers, written with only two digits, 1 and 0. Each digit is a bit. A bit is sent as a pulse, high for 1 and low for 0.

Two rows of pulses one above the other. The top row is labelled bits, with the digits 1 0 1 1 0 1 0 0 above it and a high step for each 1 and a low step for each 0. The row below is labelled pulses sent and has the same high and low steps.10110100bitspulses sent
A digital signal. Only two levels are used, so each bit can be told from the others.

Digital signals have an advantage on a long journey. A wave gets weaker and picks up noise on the way. A digital signal has only two levels. So even a weak, noisy pulse can be read as a 1 or a 0. It can be cleaned up and sent on as a fresh pulse. An analogue signal cannot be cleaned, because its noise looks like part of the message.

Many senders share the air. Each uses its own frequency. A radio set tunes to one frequency, which picks out one station and ignores the rest. Mobile phones also share the air by using different frequencies.

Now follow a call. The phone sends its code as microwaves to a mast. The mast passes it on, mostly as pulses of light in a glass optical fibre cable. The light is kept in by total internal reflection, which you met in the chapter on reflection and refraction. The fibre carries it over land and under the sea.

Four things one above the next, joined by arrows. The first is labelled phone, then mast, then far mast, then far phone. The arrow between the phone and the mast is labelled microwaves. The arrow between the mast and the far mast is labelled light in fibre. The arrow between the far mast and the far phone is labelled microwaves.phonemicrowavesmastlight in fibrefar mastmicrowavesfar phone
One way a call crosses a sea. The fibre between the masts runs over land and under the sea. The kind of wave changes along the way.

Light in a fibre is usually infrared, not visible light. A fibre can carry far more information than a copper wire, loses little energy, and is not disturbed by electrical interference. Most of the internet between cities and across oceans travels this way.

Microwaves pass through the atmosphere, so dishes on the ground can send them to a satellite and back. They travel in straight lines, so two dishes on hills need a clear view of each other. Wi-Fi also uses microwaves, to link a router to a phone or a computer.

A high layer of the air reflects some long radio waves. They can then travel round the curve of the Earth. This is why some radio stations can be heard from far away. Light and microwaves are not reflected in this way.

All of these signals travel at the speed of the wave. In a vacuum, microwaves and radio waves travel at 3 × 10⁸ m/s. Light in glass fibre is slower, about 2 × 10⁸ m/s. The time a signal takes is the distance divided by the speed.

Worked example

A signal crosses an undersea fibre cable 8 000 km long. Light in the fibre travels at 2 × 10⁸ m/s. How long does the signal take?

  1. Change the distance to metres: 8 000 km = 8 000 000 m.
  2. The time is t = d ÷ v = 8 000 000 ÷ 200 000 000 = 0.04 s.

Answer: 0.04 s

Two things one above the other, joined by an arrow. The first is labelled dish and the second is labelled dish. The arrow between them is labelled microwaves, with 60 km under it.dishmicrowaves60 kmdish
Two hilltop dishes linked by microwaves, 60 km apart.

Worked example

Two hilltop dishes are 60 km apart. Microwaves travel between them at 3 × 10⁸ m/s. How long does a signal take?

  1. Change the distance to metres: 60 km = 60 000 m.
  2. The time is t = d ÷ c = 60 000 ÷ 300 000 000 = 0.0002 s.
  3. That is 0.2 ms, much shorter than a blink.

Answer: 0.0002 s

Now think about health. When an electromagnetic wave is absorbed, it hands over energy. It does this in tiny packets. The energy of one packet is higher for a higher frequency. A wave with twice the frequency has packets with twice the energy.

Radio waves, microwaves, infrared and visible light have packets with too little energy to knock electrons off atoms. Their main effect is heating. Strong microwaves heat water and can burn, so a microwave oven is shielded and its door has a safety switch. Strong infrared also burns.

Ultraviolet, X-rays and gamma rays have packets with much more energy. The shortest ultraviolet, and all X-rays and gamma rays, can knock electrons off atoms. This is called ionising. They are ionising radiation. Even the longer ultraviolet can harm a cell by changing its chemistry.

An atom as a dot with a dashed ring and two electrons on it. A wavy ray labelled ray arrives from the left. Short double arrows beside the electrons show that they only shake. The atom is labelled atom and one electron is labelled electron. Nothing leaves the atom.rayatomelectron
A ray with low energy packets only makes the atom shake. It warms.
An atom as a dot with a dashed ring, a plus sign beside it and one electron on the ring. A wavy ray labelled ray arrives from the left. A second electron, labelled electron, has been knocked out and moves away along an arrow. The atom is labelled atom and the changed atom is labelled ion.+rayatomelectronion
A ray with high energy packets knocks an electron out. The atom becomes an ion.

Inside a living cell, the nucleus holds the DNA, which is the instruction for the cell. Ionising radiation can damage DNA. A cell can often repair small damage. A badly damaged cell dies. Sometimes a damaged cell lives on and copies its changed DNA. A change in DNA is a mutation.

If a mutation falls in the wrong place, the cell may grow and divide out of control. This is cancer. The more ionising radiation a body takes in, the dose, the greater the chance of harm. A very large dose can cause burns and radiation sickness, and can kill.

Seven rows in order from the top: radio, over 1 m. Microwave, 1 mm to 1 m. Infrared, 700 nm to 1 mm. Visible, 400 to 700 nm. Ultraviolet, 10 to 400 nm. X-ray, 0.01 to 10 nm. Gamma, under 0.01 nm. A thick box is drawn round the last three rows, ultraviolet, X-ray and gamma. A long arrow down the left is labelled frequency rises. A long arrow down the right is labelled wavelength falls.frequency riseswavelength fallsradioover 1 mmicrowave1 mm to 1 minfrared700 nm to 1 mmvisible400 to 700 nmultraviolet10 to 400 nmX-ray0.01 to 10 nmgammaunder 0.01 nm
The boxed waves can damage living cells. The frequency, and so the energy of each packet, rises down the list.

Ultraviolet from the Sun causes sunburn, ages the skin and harms the eyes. It is the main cause of skin cancer. A little sunlight on the skin helps the body make vitamin D, but a short time is enough. Darker skin has more of the pigment that absorbs ultraviolet, which gives some protection but not complete protection.

To protect your skin, stay in the shade near midday. Cover up with clothes and a hat. Wear sunglasses that block ultraviolet, and use sun cream. Thin cloud does not stop ultraviolet, so you can burn on a cloudy day.

X-rays are useful. A doctor orders one only when the benefit is greater than the small risk. The dose from one dental X-ray is small. The risk is small, but it is not zero. So staff stand behind a screen or leave the room. A lead apron may cover the parts of the body that are not being pictured. A woman who might be pregnant must tell the staff first.

Gamma rays kill cells too, which is why a narrow beam aimed from several directions treats cancer. The tumour gets a large dose and the tissue around it gets much less. Strong sources are kept in thick lead or concrete, and workers limit their time near them.

For people near ionising radiation, three rules help. Spend less time near it, stay further away, and put something thick between you and it, such as lead or concrete. Each one lowers the dose.

What about mobile phones and Wi-Fi? Their microwaves cannot ionise atoms. Their main effect on the body is a very small warming. Large studies have not found that phones cause harm. A cancer research agency of the World Health Organization lists radio frequency fields as "possibly" cancer causing. It uses that group when the evidence is limited. Studies go on.

One more danger is not about packets but about heat. A very bright beam of visible light can burn the retina at the back of the eye. Never look at the Sun, and never look into a laser beam, even for a moment. The damage can be permanent.

Try it

Send a message with a torch

You need: a torch, a friend at the other side of a dim room, or across a dark yard, paper and a pencil for each of you.

Be safe: Never shine the torch into anyone's eyes, and never use a laser pointer for this. Do it where nobody can trip in the dim light, and not near roads.

  1. Agree a code of two flashes for each letter. A is short short, B is short long, C is long short and D is long long. A short flash is a 0 and a long flash is a 1.
  2. Write a message of four letters, using only A, B, C and D. Do not show it to your friend.
  3. Send each letter by two flashes, with a pause between letters. Your friend writes down short or long for each flash.
  4. Your friend uses the code to read the letters. Compare with your message.
  5. Now hold a sheet of paper in front of the torch so that it is dimmer. Send the message again.

What you will see: The message is still read correctly when the torch is dim. Your friend only has to tell a short flash from a long one. A digital signal can be read even when it is weak.

If you cannot try it, this is what happens.
A table with the columns Letter and Flashes. Row 1: A, short short. Row 2: B, short long. Row 3: C, long short. Row 4: D, long long.LetterFlashesAshort shortBshort longClong shortDlong long

A wrong idea many people have

All electromagnetic waves are equally dangerous, or all are safe.

Why it feels right: The word "radiation" is used for all of them, and news stories treat it as one thing.

The test that breaks it: Infrared from a fire warms you, and visible light lets you see. Neither ionises atoms. A dental X-ray is a short exposure to a high energy wave. They are not the same kind of risk.

What is true: What a wave can do depends on its frequency and on the dose. Ultraviolet, X-rays and gamma rays can damage cells. Radio waves, microwaves, infrared and visible light cannot ionise, though strong ones can heat and burn.

A wrong idea many people have

A very small dose of X-rays does no harm at all.

Why it feels right: A dental or chest X-ray is safe enough for doctors to use, so it seems to have no risk.

The test that breaks it: Each packet of an X-ray can damage a molecule of DNA, and one change can start a cancer. The chance is very small for one X-ray, but it is not zero, and it grows with every exposure.

What is true: The risk from a small dose is small, not zero. Doctors use X-rays when the benefit is clearly greater, and they keep the dose as low as they can.

A wrong idea many people have

A strong enough radio signal would damage DNA like an X-ray.

Why it feels right: More power seems always to mean more danger, and "strong" sounds like "high energy".

The test that breaks it: Each radio packet has far too little energy to knock an electron off an atom. Making the beam stronger sends more packets, not bigger ones, so it cannot ionise.

What is true: A strong beam of radio waves or microwaves can heat and burn, but it cannot ionise. Only high frequency waves, with packets of high energy, ionise.

Key words

carrier wave
the plain wave that is changed to carry a message.
analogue signal
a signal in which the wave changes smoothly to copy the message.
digital signal
a signal sent as a code of 1s and 0s, as pulses.
bit
one binary digit, a 1 or a 0.
ionising radiation
waves or particles with enough energy to knock electrons off atoms. It can damage DNA.
mutation
a change in the DNA of a cell.
dose
how much ionising radiation a body takes in.

Back to your guess

The answer is A: Radio waves, microwaves and pulses of light in cables carry it.

The phone turns her voice into a code. Microwaves carry the code to a mast. Pulses of light in glass fibres carry it across land and under the sea. Sound does not travel all the way.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. How does a radio wave carry a message?

    Tap your answer.
    1. Not this one. Sound cannot travel inside an electromagnetic wave. The wave is changed in a pattern that copies the message, and the radio set turns the pattern back into sound. Try another.

    2. Not this one. The speed of a radio wave does not change. It is the height or the frequency that changes. Try another.

    3. Not this one. A radio wave stays a radio wave. The message is put on it by changing the wave. Try another.

    4. Yes, that is right. A carrier wave is changed in step with the message. In amplitude modulation its height follows the message, and in frequency modulation its frequency follows it.

    See the answer

    The wave is changed in step with the message, in its height or frequency.

    A carrier wave is changed in step with the message. In amplitude modulation its height follows the message, and in frequency modulation its frequency follows it.

    • If you chose A: Sound cannot travel inside an electromagnetic wave. The wave is changed in a pattern that copies the message, and the radio set turns the pattern back into sound.
    • If you chose B: The speed of a radio wave does not change. It is the height or the frequency that changes.
    • If you chose C: A radio wave stays a radio wave. The message is put on it by changing the wave.
  2. Why is a digital signal better than an analogue signal over a long distance?

    Tap your answer.
    1. Yes, that is right. A digital signal has only two levels. So a weak or noisy pulse can still be read as a 1 or a 0. It is then sent on as a fresh pulse. An analogue signal cannot be cleaned.

    2. Not this one. Both kinds travel at the same speed. The advantage is that a digital signal can be cleaned up on the way. Try another.

    3. Not this one. Every signal needs a wave or a current to carry it. A digital signal is a pattern of pulses on a wave. Try another.

    4. Not this one. It does get weaker. What helps is that with only two levels, it can be read and renewed before the noise swamps it. Try another.

    See the answer

    A weak, noisy digital signal can still be read as high or low pulses, and cleaned into fresh ones.

    A digital signal has only two levels. So a weak or noisy pulse can still be read as a 1 or a 0. It is then sent on as a fresh pulse. An analogue signal cannot be cleaned.

    • If you chose B: Both kinds travel at the same speed. The advantage is that a digital signal can be cleaned up on the way.
    • If you chose C: Every signal needs a wave or a current to carry it. A digital signal is a pattern of pulses on a wave.
    • If you chose D: It does get weaker. What helps is that with only two levels, it can be read and renewed before the noise swamps it.
  3. How does most of a phone call cross an ocean?

    Tap your answer.
    1. Yes, that is right. The code goes as pulses of light, usually infrared, in optical fibres under the sea. The light is kept inside by total internal reflection.

    2. Not this one. Sound in water travels far less well than a cable, and a phone does not make sound in the sea. Try another.

    3. Not this one. Sea water absorbs radio waves quickly. A cable under the sea does the job. Try another.

    4. Not this one. Old cables used copper wires. Modern ones use glass fibres and light, which carry much more. Try another.

    See the answer

    As pulses of light in glass fibres on the sea floor

    The code goes as pulses of light, usually infrared, in optical fibres under the sea. The light is kept inside by total internal reflection.

    • If you chose B: Sound in water travels far less well than a cable, and a phone does not make sound in the sea.
    • If you chose C: Sea water absorbs radio waves quickly. A cable under the sea does the job.
    • If you chose D: Old cables used copper wires. Modern ones use glass fibres and light, which carry much more.
  4. Which of these can ionise atoms?

    Tap your answer.
    1. Not this one. Infrared packets have too little energy to knock an electron off an atom. They warm things instead. Try another.

    2. Not this one. Radio packets have very little energy. Even a strong radio beam only heats, because it sends more packets, not bigger ones. Try another.

    3. Yes, that is right. Gamma rays have the highest frequency and so packets with the most energy. They can knock electrons off atoms. Infrared and radio waves cannot.

    See the answer

    Gamma rays

    Gamma rays have the highest frequency and so packets with the most energy. They can knock electrons off atoms. Infrared and radio waves cannot.

    • If you chose A: Infrared packets have too little energy to knock an electron off an atom. They warm things instead.
    • If you chose B: Radio packets have very little energy. Even a strong radio beam only heats, because it sends more packets, not bigger ones.
  5. Why can X-rays and gamma rays harm living cells, when radio waves cannot?

    Tap your answer.
    1. Yes, that is right. The energy of one packet rises with frequency. X-ray and gamma packets have enough energy to ionise atoms and damage DNA. Radio packets do not.

    2. Not this one. All electromagnetic waves go at the same speed in a vacuum. It is the energy of each packet that matters, and that rises with frequency. Try another.

    3. Not this one. They have a shorter wavelength. The longer the wavelength, the less the energy of each packet. Try another.

    4. Not this one. More packets each second means more energy in total. Each packet has the same energy, and that depends on the frequency. Try another.

    See the answer

    Each packet carries enough energy to damage DNA.

    The energy of one packet rises with frequency. X-ray and gamma packets have enough energy to ionise atoms and damage DNA. Radio packets do not.

    • If you chose B: All electromagnetic waves go at the same speed in a vacuum. It is the energy of each packet that matters, and that rises with frequency.
    • If you chose C: They have a shorter wavelength. The longer the wavelength, the less the energy of each packet.
    • If you chose D: More packets each second means more energy in total. Each packet has the same energy, and that depends on the frequency.
  6. Which is the best way to protect your skin from ultraviolet in strong sunlight?

    Tap your answer.
    1. Not this one. A tan is a sign of skin that has been damaged. It gives only a little protection. Try another.

    2. Not this one. Thin cloud lets much ultraviolet through, and you can burn. Cloud is not a safe shield. Try another.

    3. Not this one. Ultraviolet damages cells, and the damage builds up. Cover and shade protect the skin. Try another.

    4. Yes, that is right. Shade, clothing, a hat, sunglasses and sun cream all cut the ultraviolet that reaches your skin and eyes. Thin cloud does not stop it.

    See the answer

    Stay in the shade near midday, cover up, and use sun cream.

    Shade, clothing, a hat, sunglasses and sun cream all cut the ultraviolet that reaches your skin and eyes. Thin cloud does not stop it.

    • If you chose A: A tan is a sign of skin that has been damaged. It gives only a little protection.
    • If you chose B: Thin cloud lets much ultraviolet through, and you can burn. Cloud is not a safe shield.
    • If you chose C: Ultraviolet damages cells, and the damage builds up. Cover and shade protect the skin.
  7. A signal goes from a ground dish up to a satellite 36 000 km above, and back down to a second dish. How long does it take? Use c = 3 × 10⁸ m/s.

    Tap your answer.
    1. Yes, that is right. The whole way is 36 000 km up and 36 000 km down. That is 72 000 km. Then t = d ÷ c = 72 000 000 ÷ 300 000 000 = 0.24 s.

    2. Not this one. That is the time to go up only. The signal also has to come down again, so the whole way is 72 000 km. Try another.

    3. Not this one. That is double the time. The whole way is 36 000 km up and 36 000 km down, which is 72 000 km, no more. Try another.

    See the answer

    0.24 s

    The whole way is 36 000 km up and 36 000 km down. That is 72 000 km. Then t = d ÷ c = 72 000 000 ÷ 300 000 000 = 0.24 s.

    • If you chose B: That is the time to go up only. The signal also has to come down again, so the whole way is 72 000 km.
    • If you chose C: That is double the time. The whole way is 36 000 km up and 36 000 km down, which is 72 000 km, no more.
  8. From an earlier chapter

    Microwaves of frequency 3 × 10¹⁰ Hz are used in a radar. What is their wavelength? Use c = 3 × 10⁸ m/s.

    Tap your answer.
    1. Not this one. You divided the frequency by c. The wavelength is c divided by the frequency. Go to S4.10.7. Try another.

    2. Not this one. You multiplied c and f. The wavelength is c divided by f. Go to S4.10.7. Try another.

    3. Not this one. The power of ten is off by one. 3 × 10⁸ divided by 3 × 10¹⁰ is 10⁻², not 10⁻¹. Go to S4.10.7. Try another.

    4. Yes, that is right. λ = c ÷ f = 3 × 10⁸ ÷ (3 × 10¹⁰) = 0.01 m, which is 1 cm.

    See the answer

    0.01 m

    λ = c ÷ f = 3 × 10⁸ ÷ (3 × 10¹⁰) = 0.01 m, which is 1 cm.

    • If you chose A: You divided the frequency by c. The wavelength is c divided by the frequency. Go to S4.10.7.
    • If you chose B: You multiplied c and f. The wavelength is c divided by f. Go to S4.10.7.
    • If you chose C: The power of ten is off by one. 3 × 10⁸ divided by 3 × 10¹⁰ is 10⁻², not 10⁻¹. Go to S4.10.7.
  9. A signal crosses an undersea fibre cable 5 000 km long. Light in the fibre travels at 2 × 10⁸ m/s. How long does the signal take?

    See the answer

    0.025 s

    5 000 km = 5 000 000 m. Then t = d ÷ v = 5 000 000 ÷ 200 000 000 = 0.025 s.

  10. A Wi-Fi router uses microwaves of frequency 2.4 GHz. What is their wavelength in cm? Use c = 3 × 10⁸ m/s and 1 GHz = 10⁹ Hz.

    See the answer

    12.5 cm

    2.4 GHz = 2.4 × 10⁹ Hz. Then λ = c ÷ f = 3 × 10⁸ ÷ (2.4 × 10⁹) = 0.125 m, which is 12.5 cm.

Keep it

Copy this into your exercise book.

A message changes a carrier wave. Digital: pulses of 1 and 0. Calls: radio and microwaves, light in fibres. UV, X-rays and gamma rays ionise and can damage DNA. Dose matters.

Draw: A phone, a mast, a cable under the sea and a second phone, with the kind of wave written on each arrow.

Key words, A to Z

Every key word in this book, what it means, and the chapter that teaches it.

amplitude
the greatest displacement, measured from the rest line to a crest. S4.10.2
analogue signal
a signal in which the wave changes smoothly to copy the message. S4.10.8
angle of incidence
the angle between the incident ray and the normal. S4.10.5
angle of reflection
the angle between the reflected ray and the normal. S4.10.5
bit
one binary digit, a 1 or a 0. S4.10.8
carrier wave
the plain wave that is changed to carry a message. S4.10.8
compression
a place in a longitudinal wave where the material is pressed together. S4.10.1
crest
the highest point of a wave. S4.10.1
critical angle
the angle of incidence in a denser material at which the refracted ray runs along the edge. sin c = 1 ÷ n. S4.10.5
digital signal
a signal sent as a code of 1s and 0s, as pulses. S4.10.8
dioptre
the unit of the power of a lens, written D. S4.10.6
displacement
how far a piece of the medium is from its rest place, with a direction. It has a unit such as cm. S4.10.2
dose
how much ionising radiation a body takes in. S4.10.8
echo
sound that reflects from a surface and returns. It travels the distance there and back. S4.10.4
electromagnetic spectrum
the family of electromagnetic waves, in order of wavelength. S4.10.7
electromagnetic waves
transverse waves of electric and magnetic fields. They need no medium and go at 3 × 10⁸ m/s in a vacuum. S4.10.7
focal length
the distance from the lens to its principal focus. Its symbol is f. S4.10.6
frequency
the number of waves that pass a point each second. Its symbol is f. f = 1 ÷ T. S4.10.2
gamma rays
the shortest wavelength electromagnetic waves. They come from the nuclei of atoms. S4.10.7
hertz
the unit of frequency, written Hz. One hertz is one wave each second. S4.10.2
infrared
electromagnetic waves given out by warm things, just longer than red light. S4.10.7
infrasound
sound with a frequency below 20 Hz, too low for people to hear. S4.10.4
ionising radiation
waves or particles with enough energy to knock electrons off atoms. It can damage DNA. S4.10.8
longitudinal
a wave whose vibration is along the way it travels. S4.10.1
magnification
the height of the image divided by the height of the object. S4.10.6
microwaves
electromagnetic waves a little shorter than radio waves, used in ovens, phones and Wi-Fi. S4.10.7
mutation
a change in the DNA of a cell. S4.10.8
normal
a line at right angles (90°) to an edge, used to measure angles. S4.10.3
optical fibre
a thin glass thread that carries light along it by total internal reflection. S4.10.5
period
the time for one complete wave to pass a point, in s. Its symbol is T. S4.10.2
power
the strength of a lens, P = 1 ÷ f with f in metres. S4.10.6
principal focus
the point where rays that arrive parallel to the axis meet after a converging lens. S4.10.6
radio waves
the longest wavelength electromagnetic waves. S4.10.7
rarefaction
a place in a longitudinal wave where the material is spread apart. S4.10.1
real image
an image where rays really cross. It can be caught on a screen, and it is upside down. S4.10.6
refraction
the turning of a wave when its speed changes at a slanted edge. S4.10.3
refractive index
the speed of light in a vacuum divided by its speed in the material. It is written n. S4.10.5
rest line
the line on a wave graph where the medium lies when no wave is on it. S4.10.2
ripple tank
a shallow tray of water, seen from above, for studying water waves. S4.10.3
short sight
a fault in which far things are blurred. A diverging lens corrects it. S4.10.6
sonar
a way to find depths and objects in water by timing the echo of a pulse of sound. S4.10.4
total internal reflection
all the light reflected inside a material, when it meets the edge beyond the critical angle. S4.10.5
transverse
a wave whose vibration is at right angles to the way it travels. S4.10.1
trough
the lowest point of a wave. S4.10.1
ultrasound
sound with a frequency above 20 000 Hz, too high for people to hear. S4.10.4
ultraviolet
electromagnetic waves just shorter than violet light. S4.10.7
vacuum
a space with no particles in it. Sound cannot cross it, and light can. S4.10.4
virtual image
an image made where rays seem to come from, so it cannot be caught on a screen. S4.10.5, S4.10.6
wave
a disturbance that travels and carries energy. S4.10.1
wave equation
v = fλ. The speed of a wave is its frequency times its wavelength. S4.10.2
wavefront
a line joining the crests of a wave, seen from above. The wave travels at right angles to it. S4.10.3
wavelength
the distance between two matching points on neighbouring waves, in m. Its symbol is λ. S4.10.2
X-rays
short wavelength electromagnetic waves that pass through soft parts of the body. S4.10.7