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MYP 3 · Science

Waves, Light and Sound

80 questions across 4 sub-topics

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Wave Basics: Amplitude, Wavelength, Frequency Sound Waves and Hearing Reflection and Refraction of Light The Electromagnetic Spectrum (intro)

Wave Basics: Amplitude, Wavelength, Frequency 20 questions

QUESTION 1 2 marks Criterion A
Easy
0 2 4 6 8 10 12 Distance along the rope (cm) Displacement (cm) A λ

The graph shows a snapshot of a transverse wave travelling along a rope.

a. State the amplitude of the wave.
[1]
b. State the wavelength of the wave.
[1]
Show complete worked solution
(a)
The amplitude is the maximum displacement from the midline (rest position) to a crest. From the graph, this is $3\,\text{cm}$.
(b)
The wavelength is the distance between two adjacent crests (or any two adjacent, identical points on the wave). From the graph, this is $4\,\text{cm}$.
QUESTION 2 2 marks Criterion A
Easy

State what is meant by the frequency of a wave, and give the unit it is measured in.

Show complete worked solution

The frequency of a wave is the number of complete waves (oscillations) passing a fixed point per second. It is measured in hertz ($\text{Hz}$), where $1\,\text{Hz}$ means one complete wave per second.

QUESTION 3 2 marks Criterion A
Easy

A wave source vibrates with a frequency of $f = 50\,\text{Hz}$. Calculate the period of the wave, using $T = \dfrac{1}{f}$.

Show complete worked solution

Step 1 — State the formula:

$$ T = \frac{1}{f} $$

Step 2 — Substitute:

$$ T = \frac{1}{50} $$

Answer: $T = 0.02\,\text{s}$

QUESTION 4 4 marks Criterion A
Medium

A wave source produces waves with frequency $f = 5\,\text{Hz}$ and wavelength $\lambda = 2\,\text{m}$, travelling through a fixed medium.

a. Calculate the speed of the wave, using $v = f\lambda$.
[2]
b. The source is now made to vibrate twice as fast, so $f = 10\,\text{Hz}$, but the wave still travels through the same medium. State the new wave speed, and calculate the new wavelength.
[2]
Show complete worked solution
(a)
$$ v = f\lambda = 5 \times 2 $$

Answer: $v = 10\,\text{m/s}$

(b)
Wave speed depends on the medium the wave travels through, not on the source's frequency — since the medium is unchanged, the speed is still $v = 10\,\text{m/s}$. Using $v = f\lambda$ rearranged: $$ \lambda = \frac{v}{f} = \frac{10}{10} = 1\,\text{m} $$ Doubling the frequency has halved the wavelength, keeping the speed the same.
QUESTION 5 3 marks Criterion A
Medium

Sound travels through air at $v = 340\,\text{m/s}$. A particular sound wave has a wavelength of $\lambda = 0.68\,\text{m}$. Calculate its frequency.

Show complete worked solution

Step 1 — Rearrange $v=f\lambda$ for $f$:

$$ f = \frac{v}{\lambda} $$

Step 2 — Substitute:

$$ f = \frac{340}{0.68} $$

Answer: $f = 500\,\text{Hz}$

QUESTION 6 4 marks Criterion A
Medium

Distinguish between a transverse wave and a longitudinal wave, giving one real example of each.

Show complete worked solution

In a transverse wave, the particles (or field) vibrate at right angles (perpendicular) to the direction the wave travels — for example, a light wave or a wave on a shaken rope.

In a longitudinal wave, the particles vibrate parallel to (along the same line as) the direction the wave travels, creating compressions and rarefactions — for example, a sound wave.

QUESTION 7 6 marks Criterion A
Hard

A wave machine produces $15$ complete waves in $3\,\text{s}$. Each wave has a wavelength of $0.5\,\text{m}$.

a. Calculate the frequency of the waves.
[1]
b. Calculate the period of the waves.
[1]
c. Calculate the speed of the waves, in $\text{m/s}$.
[2]
d. Give the wavelength in centimetres instead, and hence state the wave speed in $\text{cm/s}$. Show that this is consistent with your answer to (c).
[2]
Show complete worked solution
(a)
$$ f = \frac{\text{number of waves}}{\text{time}} = \frac{15}{3} = 5\,\text{Hz} $$
(b)
$$ T = \frac{1}{f} = \frac{1}{5} = 0.2\,\text{s} $$
(c)
$$ v = f\lambda = 5 \times 0.5 = 2.5\,\text{m/s} $$
(d)
$$ 0.5\,\text{m} = 50\,\text{cm} $$ $$ v = f\lambda = 5 \times 50 = 250\,\text{cm/s} $$ Checking: $250\,\text{cm/s} \div 100 = 2.5\,\text{m/s}$, which matches part (c), confirming the answer is consistent regardless of the unit used.
QUESTION 8 7 marks Criterion B
Medium

A student wants to use a ripple tank to investigate how the frequency of a vibrating dipper affects the wavelength of the water waves it produces.

a. State the independent and dependent variables.
[2]
b. State one variable that should be controlled, and explain why.
[2]
c. Describe a method, including equipment, to measure the wavelength at each frequency.
[3]
Show complete worked solution
(a)
Independent variable: frequency of the dipper (set using the motor's controller). Dependent variable: wavelength of the ripples produced.
(b)
The depth of the water should be kept the same throughout. Water depth affects the speed of the ripples, so if it changed between trials, any change in wavelength could be caused by the changing depth rather than by the changing frequency — making the test unfair.
(c)
  1. Set the dipper to the first test frequency and let the ripples become steady.
  2. Use a strobe light flashing at the same frequency as the dipper to "freeze" the ripple pattern, or shine a lamp above the tank to project shadows of the wavefronts onto a screen below.
  3. Measure the distance across several wavefronts (e.g. across 5 wavelengths) using a ruler, then divide by 5 to find one wavelength — this reduces the effect of measurement error.
  4. Repeat for each test frequency, keeping water depth constant.
QUESTION 9 6 marks Criterion B
Medium

A student wants to investigate how the tension in a stretched spring affects the speed of a wave pulse sent along it.

a. Identify the independent, dependent, and one controlled variable for this investigation.
[3]
b. Describe how the student could measure the speed of the wave pulse for each tension tested.
[3]
Show complete worked solution
(a)
Independent: tension in the spring (e.g. changed by adding different hanging masses to stretch it). Dependent: speed of the wave pulse. Controlled: length of the spring used each time — a different length would change the distance (and therefore the time) the pulse must travel, making comparisons unfair.
(b)
Measure the length of the spring with a metre ruler. Give one end a sharp flick to send a single pulse along it, and use a stopwatch to time how long the pulse takes to travel to the other end (or use a video recording, played back frame-by-frame, for a more accurate time). Repeat each timing $3$ times and take a mean to reduce reaction-time error. Calculate speed using $v = \dfrac{\text{length of spring}}{\text{mean time}}$ for each tension tested.
QUESTION 10 8 marks Criterion B
Hard

A student measures the frequency of a vibrating guitar string by counting how many times it crosses its rest position in exactly $1\,\text{s}$, using a stopwatch and their own eyes.

a. Identify a weakness of counting oscillations over only $1\,\text{s}$.
[2]
b. Suggest an improvement to the method that would reduce this error, and explain why it works.
[3]
c. Suggest a further improvement that would remove human reaction time from the measurement almost entirely, and explain how it works.
[3]
Show complete worked solution
(a)
Starting and stopping the stopwatch involves human reaction time (typically $\pm0.2$–$0.3\,\text{s}$). Over a total time of only $1\,\text{s}$, this reaction-time error is a very large fraction of the measurement, so the counted frequency could be significantly wrong.
(b)
Instead, count the time taken for a larger number of oscillations, e.g. $20$ oscillations, then divide the time by $20$ to find the period of just one. The same $\pm0.2$–$0.3\,\text{s}$ reaction-time error is now spread across a much longer total time, so it makes up a much smaller percentage of the measurement, giving a far more accurate frequency.
(c)
Record the vibrating string with a slow-motion video camera (or use a motion sensor/microphone connected to a data logger and oscilloscope software). Counting the oscillations frame-by-frame afterwards (with a known frame rate), or reading the period directly off the recorded waveform, removes the need for a human to start/stop a stopwatch in real time, eliminating reaction-time error almost completely.
QUESTION 11 3 marks Criterion C
Easy

The table shows the displacement of a rope at different distances along it, at one instant in time.

Distance along wave (m)012345678
Displacement (cm)030-3030-30
a. State the amplitude shown by this data.
[1]
b. Use the table to find the wavelength of the wave, explaining how you found it.
[2]
Show complete worked solution
(a)
The largest displacement recorded is $3\,\text{cm}$, so the amplitude is $3\,\text{cm}$.
(b)
The pattern of displacements repeats every $4\,\text{m}$ (e.g. the displacement is $0$ at $0\,\text{m}$, $4\,\text{m}$ and $8\,\text{m}$, with an identical crest-trough pattern in between each time). So the wavelength is $4\,\text{m}$.
QUESTION 12 6 marks Criterion C
Medium

A student uses a ripple tank to measure the wavelength of water waves at several different dipper frequencies.

Frequency (Hz)248
Wavelength (m)1.700.850.425
a. Calculate the wave speed using the $f = 2\,\text{Hz}$ data.
[2]
b. Calculate the wave speed using the $f = 8\,\text{Hz}$ data.
[2]
c. What do your answers to (a) and (b) show about the speed of the water waves in this ripple tank?
[2]
Show complete worked solution
(a)
$$ v = f\lambda = 2 \times 1.70 = 3.4\,\text{m/s} $$
(b)
$$ v = f\lambda = 8 \times 0.425 = 3.4\,\text{m/s} $$
(c)
Both calculations give the same wave speed, $3.4\,\text{m/s}$, even though the frequency and wavelength are both different. This confirms that the wave speed is constant for this water depth — it depends on the medium (the water), not on the frequency the dipper is set to; wavelength simply adjusts to compensate.
QUESTION 13 4 marks Criterion C
Medium
0 4 8 12 16 Distance (m) Displacement (m) A λ

The graph shows a wave on the sea surface, produced by a source vibrating at $f = 0.25\,\text{Hz}$.

a. Read the wavelength of the wave from the graph.
[1]
b. Calculate the speed of this wave.
[3]
Show complete worked solution
(a)
From the graph, the wavelength is $\lambda = 8\,\text{m}$.
(b)
$$ v = f\lambda = 0.25 \times 8 $$

Answer: $v = 2\,\text{m/s}$

QUESTION 14 5 marks Criterion C
Medium

A student measures the frequency of waves of different wavelengths travelling through the same stretched spring (wave speed $= 12\,\text{m/s}$ throughout).

Wavelength (m)0.51.01.52.02.5
Frequency measured (Hz)2412564.8
a. Identify the anomalous reading in the table.
[1]
b. Show, using a calculation, why this reading does not fit the pattern of the rest of the data.
[2]
c. State what should be done with this reading before drawing a conclusion from the data.
[2]
Show complete worked solution
(a)
The reading at $\lambda = 1.5\,\text{m}$, where $f = 5\,\text{Hz}$, is anomalous.
(b)
Using $v = f\lambda$, the expected frequency at $\lambda = 1.5\,\text{m}$ is: $$ f = \frac{v}{\lambda} = \frac{12}{1.5} = 8\,\text{Hz} $$ This is close to what the other rows give (e.g. $\lambda=1.0\,\text{m} \to f=12\,\text{Hz}$, $v=12\,\text{m/s}$; $\lambda=2.0\,\text{m} \to f=6\,\text{Hz}$, $v=12\,\text{m/s}$), but the recorded value of $5\,\text{Hz}$ gives $v = 5\times1.5=7.5\,\text{m/s}$, clearly inconsistent with the other rows.
(c)
The anomalous reading should be excluded from any average or graph, and that measurement should be repeated to check whether $5\,\text{Hz}$ was a genuine one-off error (e.g. a miscount) before finalising the results.
QUESTION 15 5 marks Criterion C
Medium

Two students separately counted the number of complete waves produced by the same wave machine over the same $10\,\text{s}$ interval, by eye.

StudentNumber of complete waves counted in 10 s
X30
Y34
a. Calculate the frequency each student would calculate from their count.
[2]
b. Suggest one reason the two students' counts might differ, even though they watched the same waves.
[1]
c. Suggest how the class could get a more reliable value for the frequency.
[2]
Show complete worked solution
(a)
Student X: $$ f_X = \frac{30}{10} = 3.0\,\text{Hz} $$ Student Y: $$ f_Y = \frac{34}{10} = 3.4\,\text{Hz} $$
(b)
Counting waves by eye is easy to get wrong, especially near the start/end of the $10\,\text{s}$ — a student may have miscounted a partial wave at the boundary, or simply lost count and guessed.
(c)
Have several students count independently (or the same student repeat the count several times), then calculate the mean of all the counts. Individual counting errors are unlikely to all be the same, so averaging several independent counts reduces the effect of any one mistake and gives a more reliable frequency.
QUESTION 16 7 marks Criterion C
Hard

A student times how long a pendulum-driven wave source takes to complete $20$ full oscillations, repeating the measurement $5$ times.

Trial12345
Time for 20 oscillations (s)4.14.33.94.24.0
a. Calculate the mean time for $20$ oscillations.
[2]
b. Calculate the period of one oscillation.
[2]
c. Calculate the frequency of the source, and explain why timing $20$ oscillations (rather than just $1$) gives a more precise value for the frequency.
[3]
Show complete worked solution
(a)
$$ \text{mean} = \frac{4.1+4.3+3.9+4.2+4.0}{5} = \frac{20.5}{5} = 4.1\,\text{s} $$
(b)
$$ T = \frac{4.1}{20} = 0.205\,\text{s} $$
(c)
$$ f = \frac{1}{T} = \frac{1}{0.205} = 4.88\,\text{Hz}\ (\text{3 s.f.}) $$ Timing $20$ oscillations spreads the fixed reaction-time error of starting/stopping the stopwatch across a much longer total time, so that error makes up a much smaller fraction of the measured time than if only $1$ oscillation were timed — giving a more precise period, and therefore frequency.
QUESTION 17 7 marks Criterion C
Hard

Scientists tracked a tsunami wave crossing the open ocean after an undersea earthquake.

Time after earthquake (min)010203040
Distance travelled (km)0120240362478
a. Calculate the wave's speed between $t=0$ and $t=10\,\text{min}$, in $\text{km/min}$.
[2]
b. Calculate the wave's speed between $t=30\,\text{min}$ and $t=40\,\text{min}$.
[2]
c. Evaluate whether the tsunami travelled at a genuinely constant speed across the whole ocean, using your calculated values.
[3]
Show complete worked solution
(a)
$$ v = \frac{120-0}{10-0} = 12.0\,\text{km/min} $$
(b)
$$ v = \frac{478-362}{40-30} = \frac{116}{10} = 11.6\,\text{km/min} $$
(c)
The two calculated speeds, $12.0\,\text{km/min}$ and $11.6\,\text{km/min}$, are close but not identical — a difference of about $3\%$. This is a small enough difference that it is unlikely to be a measurement error alone; tsunami wave speed genuinely depends on ocean depth, so the slight slowing shown here is most likely because the wave reached a region of shallower water later in its journey, rather than the speed being perfectly constant throughout.
QUESTION 18 5 marks Criterion D
Medium

After an undersea earthquake, scientists use the known speed of tsunami waves through open ocean (see the earlier question) to calculate how many minutes until the wave reaches a coastline, and issue a tsunami warning.

Discuss one benefit and one drawback of relying on wave-speed calculations for tsunami warning systems.

Show complete worked solution

Benefit: Because tsunami wave speed can be calculated fairly reliably from ocean depth data, scientists can predict, often with tens of minutes to hours of warning, when a wave will reach a populated coastline. This gives coastal communities time to evacuate to higher ground, potentially saving thousands of lives compared to having no warning system.

Drawback: The calculation depends on accurate, detailed data about ocean depth along the wave's whole path, and on correctly detecting the earthquake itself — if this data is incomplete or the earthquake is very close to shore, there may not be enough warning time, or the predicted arrival time could be wrong. False alarms can also occur, which may lead communities to distrust or ignore future warnings ("warning fatigue").

QUESTION 19 6 marks Criterion D
Medium

Earthquakes produce two types of seismic wave that travel at different speeds: faster P-waves, which cause little damage, arrive first, followed by slower but more damaging S-waves. Earthquake early-warning apps detect the P-wave and use it to send an alert seconds before the damaging S-wave arrives.

Discuss one benefit and one drawback of this technology.

Show complete worked solution

Benefit: Even a warning of just a few seconds is enough for people to take protective action — dropping, covering, and holding on, stepping away from windows, or for automatic systems to stop trains, open elevator doors, or pause surgery, significantly reducing injuries and damage from the more destructive S-waves and the shaking that follows.

Drawback: The warning time is often very short (seconds, not minutes) and depends on how far away the earthquake's origin is, so it may be too brief to be useful for earthquakes that start very close to a city. Access to the technology is also unequal — it relies on a dense network of sensors and on people owning smartphones with the app installed, meaning wealthier regions typically benefit far more than poorer ones, even though the risk from earthquakes is not limited to wealthy areas.

QUESTION 20 6 marks Criterion D
Hard

Radio waves are used for mobile phone calls, WiFi, GPS navigation, television broadcasts, and emergency services communication, all sharing the same limited range of frequencies known as the radio spectrum.

Evaluate the impact of relying so heavily on radio-frequency waves for communication, discussing both a benefit and a concern.

Show complete worked solution

Benefit: Radio waves allow instant wireless communication over large distances without needing physical cables, enabling mobile phones, GPS navigation, emergency service coordination, and internet access almost anywhere. This has transformed how people work, travel, and respond to emergencies, and has made information and connectivity available to people in remote areas that cables would be impractical to reach.

Concern: Because so many different services (phones, WiFi routers, radio, television, satellites) all need their own slice of the same limited radio spectrum, overcrowding and interference between signals is an ongoing challenge, requiring careful regulation of which frequencies each service is allowed to use. There has also been public concern about possible long-term health effects of continuous exposure to radio-frequency waves from phones and masts — although major scientific reviews have found no confirmed harm at the low power levels used, the concern has still shaped where phone masts can be built and how devices are designed and tested.

Sound Waves and Hearing 20 questions

QUESTION 1 3 marks Criterion A
Easy
compression rarefaction wavelength (λ) direction of vibration is left-right (parallel to wave travel)

The diagram shows a sound wave travelling through air, drawn as vertical lines representing air particles at one instant.

a. What name is given to a region where the particles are bunched close together?
[1]
b. What name is given to a region where the particles are spread further apart?
[1]
c. How is the wavelength of this sound wave defined, in terms of these regions?
[1]
Show complete worked solution
(a)
This is called a compression.
(b)
This is called a rarefaction.
(c)
The wavelength is the distance from one compression to the next adjacent compression (or equivalently, from one rarefaction to the next).
QUESTION 2 2 marks Criterion A
Easy

Explain why sound cannot travel through a vacuum (empty space with no particles), but light can.

Show complete worked solution

Sound is a longitudinal mechanical wave — it travels by making particles of a medium (like air, water, or a solid) vibrate and pass the vibration on to neighbouring particles. In a vacuum there are no particles to vibrate, so sound cannot be transmitted. Light, on the other hand, does not need a medium to travel — it is an electromagnetic wave and can travel through empty space, which is why we can see distant stars and the Sun's light reaches us through the vacuum of space.

QUESTION 3 2 marks Criterion A
Easy

State which property of a sound wave (frequency or amplitude) determines each of the following:

a. The pitch of the sound (how high or low it sounds).
[1]
b. The loudness of the sound.
[1]
Show complete worked solution
(a)
Frequency determines pitch — a higher frequency gives a higher-pitched sound.
(b)
Amplitude determines loudness — a larger amplitude gives a louder sound.
QUESTION 4 3 marks Criterion A
Medium

A hiker shouts towards a distant cliff face and hears the echo after $1.2\,\text{s}$. The speed of sound in air is $340\,\text{m/s}$.

Calculate the distance from the hiker to the cliff face.

Show complete worked solution

Step 1 — Note that the sound travels to the cliff AND back, so halve the total distance:

$$ \text{total distance} = v \times t = 340 \times 1.2 = 408\,\text{m} $$

Step 2 — Divide by 2 for the one-way distance:

$$ \text{distance to cliff} = \frac{408}{2} $$

Answer: $204\,\text{m}$

QUESTION 5 3 marks Criterion A
Medium

A submarine sends a sonar pulse straight down. It reflects off the seabed, which is $1500\,\text{m}$ below the submarine, and returns after a total time of $2.0\,\text{s}$.

Calculate the speed of sound in the seawater.

Show complete worked solution

Step 1 — Find the total distance travelled by the pulse (down AND back up):

$$ \text{total distance} = 2 \times 1500 = 3000\,\text{m} $$

Step 2 — Apply $v = \dfrac{\text{distance}}{\text{time}}$:

$$ v = \frac{3000}{2.0} $$

Answer: $v = 1500\,\text{m/s}$

QUESTION 6 3 marks Criterion A
Medium

The normal range of human hearing is approximately $20\,\text{Hz}$ to $20\,000\,\text{Hz}$. For each frequency below, state whether it is infrasound, within the normal human hearing range, or ultrasound.

a. $10\,\text{Hz}$
[1]
b. $2000\,\text{Hz}$
[1]
c. $50\,000\,\text{Hz}$
[1]
Show complete worked solution
(a)
This is below $20\,\text{Hz}$, so it is infrasound.
(b)
This is between $20\,\text{Hz}$ and $20\,000\,\text{Hz}$, so it is within the normal human hearing range.
(c)
This is above $20\,000\,\text{Hz}$, so it is ultrasound.
QUESTION 7 6 marks Criterion A
Hard

A ship's sonar system sends out an ultrasound pulse of frequency $f = 50\,000\,\text{Hz}$, which travels through seawater at $v = 1500\,\text{m/s}$.

a. Calculate the period of the ultrasound pulse.
[2]
b. The pulse takes $0.08\,\text{s}$ to travel to the seabed and back. Calculate the depth of the seabed below the ship.
[2]
c. Explain why sonar uses ultrasound rather than sound within the normal human hearing range for this purpose.
[2]
Show complete worked solution
(a)
$$ T = \frac{1}{f} = \frac{1}{50\,000} = 0.00002\,\text{s} = 2\times10^{-5}\,\text{s} $$
(b)
$$ \text{total distance} = v \times t = 1500 \times 0.08 = 120\,\text{m} $$ $$ \text{depth} = \frac{120}{2} = 60\,\text{m} $$
(c)
Ultrasound (above $20\,000\,\text{Hz}$) has a very short wavelength, which allows it to reflect well off relatively small or precise targets (such as the seabed or a fish shoal) and gives more detailed, accurate readings. It also does not add audible noise for the crew or interfere with other on-board equipment or communications that use audible sound.
QUESTION 8 6 marks Criterion B
Medium

A student wants to investigate how the tension in a guitar string affects the pitch of the note it produces when plucked.

a. Identify the independent, dependent, and one controlled variable for this investigation.
[3]
b. Describe how the student could measure the frequency produced for each tension tested.
[3]
Show complete worked solution
(a)
Independent: tension in the string (adjusted using the tuning peg). Dependent: pitch of the note produced (measured as frequency, using a tuner app or frequency-measuring app). Controlled: length and thickness (gauge) of the string used each time — changing these would also change the pitch, independently of tension, making the test unfair.
(b)
Pluck the string in the same way each time (e.g. same finger position and force) and use a smartphone tuner app or a microphone connected to frequency-analysis software to record the frequency of the note produced. Repeat each measurement $3$ times at each tension and take a mean, to reduce the effect of any variation in how the string is plucked.
QUESTION 9 7 marks Criterion B
Medium

A student wants to measure the speed of sound in air using an echo method: standing a measured distance from a large flat wall, clapping once, and timing how long the echo takes to return.

a. State the equipment the student would need, and describe how the distance to the wall should be measured.
[2]
b. Describe the method the student should follow to collect the timing data.
[2]
c. State one variable that should be controlled during this investigation, and explain why.
[3]
Show complete worked solution
(a)
Equipment needed: a large flat wall (to reflect the sound), a trundle wheel or long tape measure to measure the distance from the student to the wall, and a stopwatch. The distance should be measured along a straight line directly towards the centre of the wall, and should be large (e.g. $50$–$100\,\text{m}$) so the echo time is long enough to time reasonably accurately.
(b)
  1. Stand at the measured distance from the wall, facing it, in a quiet outdoor area.
  2. Clap sharply once, starting the stopwatch at the same instant.
  3. Stop the stopwatch the instant the echo is heard.
  4. Repeat the clap-and-time process at least $5$ times at the same distance, and record each time.
(c)
Air temperature should be controlled (or at least recorded and kept similar between repeats). The speed of sound in air changes with temperature (it travels slightly faster in warmer air), so if the temperature changed noticeably between repeats, the timed results would not be directly comparable, and the calculated speed of sound would be less reliable.
QUESTION 10 7 marks Criterion B
Hard

In the echo experiment above, the student starts and stops the stopwatch by hand, reacting to the clap and to hearing the echo.

a. Identify the main source of error in this method, and explain why it has a large effect on the result.
[2]
b. Suggest how repeating the experiment and calculating a mean improves the reliability of the result.
[2]
c. Suggest a way of removing human reaction time from the measurement almost completely, and explain how it would work.
[3]
Show complete worked solution
(a)
The main source of error is human reaction time when starting and stopping the stopwatch (typically $\pm0.2$–$0.3\,\text{s}$ combined). Since the whole echo only takes a fraction of a second to travel to the wall and back, this reaction-time error is a large fraction of the total measured time, so it has a big effect on the calculated speed.
(b)
Reaction-time errors will sometimes make an individual timing too long and sometimes too short, essentially at random. Taking a mean of several repeated timings allows these random errors to partly cancel out, giving a value closer to the true echo time than any single measurement alone.
(c)
Use a microphone connected to a data logger (or a computer running sound-recording software) placed next to the student. The software can automatically detect the sharp spike in sound level from the clap, and then the second spike from the echo, and calculate the precise time between them electronically. This removes the need for a person to react to the sounds and start/stop a stopwatch, eliminating reaction-time error almost entirely.
QUESTION 11 3 marks Criterion C
Easy

A microphone connected to an oscilloscope displays the trace height (a measure of amplitude) for two notes played on the same instrument at the same frequency.

NoteTrace height (mm)
X12
Y4
a. Which note, X or Y, is louder? Explain your answer using the data.
[2]
b. If note X were instead played at a higher pitch than note Y, which property of the wave would this change reflect?
[1]
Show complete worked solution
(a)
Note X is louder. Trace height on an oscilloscope shows amplitude, and amplitude determines loudness — X has a much larger trace height ($12\,\text{mm}$) than Y ($4\,\text{mm}$), so X has a larger amplitude and sounds louder.
(b)
A change in pitch reflects a change in frequency, not amplitude.
QUESTION 12 5 marks Criterion C
Medium

A student stands $85\,\text{m}$ from a wall and claps, timing the echo $3$ times.

Trial123
Time for echo (s)0.500.520.48
a. Calculate the mean echo time.
[2]
b. Use the mean time to calculate the speed of sound in air.
[3]
Show complete worked solution
(a)
$$ \text{mean} = \frac{0.50+0.52+0.48}{3} = \frac{1.50}{3} = 0.50\,\text{s} $$
(b)
The sound travels to the wall and back, a total of $2 \times 85 = 170\,\text{m}$: $$ v = \frac{170}{0.50} $$

Answer: $v = 340\,\text{m/s}$

QUESTION 13 4 marks Criterion C
Medium
Sound A Sound B Time

The traces show two different sounds, A and B, recorded over the same time interval.

a. Which sound is louder? Explain your reasoning.
[2]
b. Which sound has the higher pitch? Explain your reasoning.
[2]
Show complete worked solution
(a)
Sound A is louder. A has a taller trace (a greater height above and below its centre line), meaning it has a greater amplitude, and amplitude determines loudness.
(b)
Sound B has the higher pitch. B completes more full wave cycles in the same amount of time, meaning it has a higher frequency, and frequency determines pitch.
QUESTION 14 5 marks Criterion C
Medium

A class tested the highest frequency each of five volunteers of different ages could hear, using a frequency generator.

Age (years)1020304050
Highest audible frequency (Hz)19 50018 00052 00015 00013 000
a. Identify the anomalous result, explaining why it does not fit the pattern.
[2]
b. Suggest a likely explanation for this anomalous result.
[1]
c. State what should be done with this result before drawing a conclusion about hearing and age.
[2]
Show complete worked solution
(a)
The $30$-year-old's result of $52\,000\,\text{Hz}$ is anomalous. Every other reading follows a clear downward trend with age (hearing typically declines with age), and no reading is anywhere near $52\,000\,\text{Hz}$ — this is far above the upper limit of human hearing (about $20\,000\,\text{Hz}$), which is not physically possible for a person to hear.
(b)
It was most likely a recording or equipment error — for example, the volunteer may have mistakenly indicated they could hear a frequency they couldn't, or the frequency generator's display/setting was misread when the result was recorded.
(c)
The anomalous reading should be excluded from the analysis, and that volunteer's test should be repeated to obtain a reliable value before any conclusion about the relationship between age and hearing range is drawn.
QUESTION 15 5 marks Criterion C
Medium

A student measures the sound intensity level from a loudspeaker at increasing distances.

Distance from speaker (m)1248
Sound intensity level (dB)100948882
a. Describe the pattern shown by the data.
[2]
b. Using this pattern, predict the sound intensity level at $16\,\text{m}$.
[1]
c. Explain, in terms of sound energy, why intensity decreases as distance from the source increases.
[2]
Show complete worked solution
(a)
As the distance from the speaker doubles each time ($1\to2\to4\to8\,\text{m}$), the sound intensity level decreases by a constant $6\,\text{dB}$ each time ($100\to94\to88\to82$).
(b)
Continuing the pattern of $-6\,\text{dB}$ for each doubling of distance: $$ 82 - 6 = 76\,\text{dB} $$
(c)
The sound energy leaving the speaker spreads outward over an increasingly larger area as it travels further from the source. Since the same total energy is spread over more area, the amount of energy reaching any given point (and therefore the intensity detected there) gets smaller with increasing distance.
QUESTION 16 7 marks Criterion C
Hard

A student stands $100\,\text{m}$ from a wall and times a clap's echo $5$ times.

Trial12345
Echo time for 100 m (s)0.610.580.600.570.59
a. Calculate the mean echo time.
[2]
b. Calculate the speed of sound using the mean time.
[2]
c. The accepted speed of sound in air is $340\,\text{m/s}$. Calculate the percentage difference between the student's result and this accepted value, and comment on the reliability of the experiment.
[3]
Show complete worked solution
(a)
$$ \text{mean} = \frac{0.61+0.58+0.60+0.57+0.59}{5} = \frac{2.95}{5} = 0.59\,\text{s} $$
(b)
$$ v = \frac{2 \times 100}{0.59} = \frac{200}{0.59} = 339\,\text{m/s}\ (\text{3 s.f.}) $$
(c)
$$ \% \text{ difference} = \frac{340-339}{340}\times100 \approx 0.3\% $$ This is a very small percentage difference, showing that despite the echo times varying slightly between trials, taking a mean of $5$ repeats produced a result extremely close to the accepted value — suggesting the method was reliable and any reaction-time errors mostly cancelled out.
QUESTION 17 7 marks Criterion C
Hard

The table shows the approximate range of frequencies that different animals can hear.

AnimalLower limit (Hz)Upper limit (Hz)
Human2020 000
Dog6745 000
Bat2 000110 000
Elephant120 000
a. Using the table, identify which animal can hear the highest-frequency sounds, and state its upper limit.
[2]
b. Using the table, identify which animal can hear frequencies humans cannot, at the very low end of the range, and name this type of sound.
[2]
c. A dog whistle produces a sound at $25\,000\,\text{Hz}$. Using the data, explain why this whistle is silent to humans but can be heard by a dog.
[3]
Show complete worked solution
(a)
The bat can hear the highest frequencies, up to $110\,000\,\text{Hz}$ ($110\,\text{kHz}$), which is far into the ultrasound range.
(b)
The elephant can hear down to $1\,\text{Hz}$, well below the human lower limit of $20\,\text{Hz}$. Sound below $20\,\text{Hz}$ is called infrasound.
(c)
$25\,000\,\text{Hz}$ is above the human upper hearing limit of $20\,000\,\text{Hz}$, so it lies in the ultrasound range for humans and is completely inaudible to them. However, $25\,000\,\text{Hz}$ is below the dog's upper limit of $45\,000\,\text{Hz}$, so it falls within the dog's hearing range and the dog can hear it clearly.
QUESTION 18 5 marks Criterion D
Medium

Ultrasound scanning is widely used in hospitals, including to monitor a baby's development during pregnancy, by sending ultrasound pulses into the body and detecting the echoes that reflect off internal structures.

Discuss one benefit and one drawback of using ultrasound imaging in medicine.

Show complete worked solution

Benefit: Unlike X-rays, ultrasound does not use ionising radiation, so it is considered very safe to use repeatedly, including on unborn babies and pregnant patients, with no known harmful side effects at the intensities used. It also produces images in real time, allowing doctors to see movement, such as a beating heart, immediately.

Drawback: Ultrasound images generally have lower resolution and detail than other scanning methods such as MRI, and the quality of the image depends heavily on the skill of the person operating the probe. Ultrasound also struggles to pass through bone or air-filled spaces (like the lungs), so it is not suitable for imaging every part of the body.

QUESTION 19 6 marks Criterion D
Medium

Typical sound levels include: normal conversation at $60\,\text{dB}$, busy city traffic at $85\,\text{dB}$, and music through headphones at high volume at $100$–$110\,\text{dB}$. Prolonged exposure above about $85\,\text{dB}$ is known to risk permanent hearing damage.

Discuss one benefit and one drawback of the widespread use of personal headphones for listening to music.

Show complete worked solution

Benefit: Headphones allow people to enjoy music, podcasts, and calls privately and conveniently, anywhere, without disturbing others — supporting entertainment, relaxation, focus while working or studying, and accessible communication, all without needing external speakers.

Drawback: Many people listen at volumes close to or above $100\,\text{dB}$, well above the $85\,\text{dB}$ threshold linked to hearing damage, often for long periods. Repeated exposure at these levels can cause permanent, irreversible hearing loss or tinnitus (persistent ringing in the ears), especially among younger people who may not realise the damage is accumulating until symptoms appear later in life.

QUESTION 20 6 marks Criterion D
Hard

Sonar (sound navigation and ranging) is widely used by ships to navigate safely, locate shoals of fish for the fishing industry, and by navies to detect submarines, by sending out pulses of sound and analysing the echoes.

Evaluate the impact of sonar technology, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: Sonar allows ships to accurately map the seabed and detect obstacles, greatly improving navigational safety, and allows fishing fleets to locate fish stocks efficiently, and navies and rescue teams to detect submarines or sunken vessels — all without needing to see through the water, which is usually impossible at depth.

Concern: Many marine mammals, such as whales and dolphins, rely on their own natural echolocation (a biological form of sonar) to navigate, communicate, and find food. Loud, high-powered sonar used by ships and navies can interfere with this, disorient the animals, or even cause physical harm to their hearing, and has been linked by some researchers to mass whale strandings. This has led to calls for stricter regulation of sonar use, particularly in areas known to be important marine mammal habitats.

Reflection and Refraction of Light 20 questions

QUESTION 1 2 marks Criterion A
Easy
normal incident ray reflected ray i r plane mirror

The diagram shows a ray of light striking a plane mirror.

a. State the law of reflection.
[1]
b. The angle of incidence shown is $42\degree$. State the angle of reflection.
[1]
Show complete worked solution
(a)
The angle of incidence equals the angle of reflection ($i = r$), and both are measured from the normal.
(b)
By the law of reflection, the angle of reflection also equals $42\degree$.
QUESTION 2 3 marks Criterion A
Easy

Define each of the following terms used to describe reflection at a mirror:

a. Incident ray
[1]
b. Reflected ray
[1]
c. The normal
[1]
Show complete worked solution
(a)
The incident ray is the ray of light travelling towards the mirror, before it is reflected.
(b)
The reflected ray is the ray of light that bounces off the mirror and travels away from it.
(c)
The normal is an imaginary line drawn at $90\degree$ (perpendicular) to the mirror's surface, at the point where the ray strikes it. Angles of incidence and reflection are always measured from the normal, not from the mirror surface itself.
QUESTION 3 2 marks Criterion A
Easy

Light travels from air into glass, a denser medium, hitting the surface at an angle.

a. What happens to the speed of the light as it enters the glass?
[1]
b. What happens to the direction of the light ray as it enters the glass?
[1]
Show complete worked solution
(a)
The light slows down — light travels more slowly in glass than in air.
(b)
The ray bends towards the normal.
QUESTION 4 4 marks Criterion A
Medium
air glass normal incident ray refracted ray i r

The diagram shows a ray of light entering a glass block from air.

a. Name the angles labelled $i$ and $r$ on the diagram.
[2]
b. State and explain which of the two angles is larger.
[2]
Show complete worked solution
(a)
$i$ is the angle of incidence (in the air, between the incident ray and the normal). $r$ is the angle of refraction (in the glass, between the refracted ray and the normal).
(b)
The angle of incidence, $i$, is larger than the angle of refraction, $r$. This is because glass is a denser (optically) medium than air, so light travelling from air into glass slows down and bends towards the normal, making the angle it makes with the normal smaller.
QUESTION 5 4 marks Criterion A
Medium

Total internal reflection is used in optical fibres to transmit light signals over long distances.

a. Describe what total internal reflection is.
[2]
b. Name one everyday application of total internal reflection.
[2]
Show complete worked solution
(a)
Total internal reflection occurs when a ray of light travelling inside a denser medium (like glass) strikes the boundary with a less dense medium (like air) at an angle greater than the material's critical angle — instead of refracting out, all of the light is reflected back into the denser medium, with no light escaping.
(b)
Optical fibres, used to carry internet and telephone signals as pulses of light over long distances (also accepted: medical endoscopes, or reflective road signs/cat's eyes).
QUESTION 6 4 marks Criterion A
Medium

Describe four properties of the image formed by a plane (flat) mirror.

Show complete worked solution

The image formed by a plane mirror is:

  1. Virtual — it cannot be projected onto a screen, since the light rays only appear to come from behind the mirror.
  2. The same size as the object.
  3. Laterally inverted — left and right are swapped (e.g. text appears back-to-front).
  4. The same distance behind the mirror as the object is in front of it.
QUESTION 7 6 marks Criterion A
Hard
air glass normal incident ray refracted ray i r

A ray of light hits a glass block (refractive index $n = 1.5$) at an angle of incidence $i = 60\degree$, as shown. The angle of refraction can be found using $$ n = \frac{\sin i}{\sin r} $$

a. Calculate the angle of refraction, $r$.
[3]
b. The glass block has two parallel surfaces. Describe what happens to the ray's direction as it exits the glass block back into the air, and how the exit ray compares to the original incident ray.
[3]
Show complete worked solution
(a)
Rearranging: $$ \sin r = \frac{\sin i}{n} = \frac{\sin 60\degree}{1.5} = \frac{0.866}{1.5} = 0.577 $$ $$ r = \sin^{-1}(0.577) $$

Answer: $r \approx 35\degree$

(b)
As the ray leaves the glass and re-enters the air (a less dense medium), it bends away from the normal at the second surface, refracting back to an angle of $60\degree$ from the normal there — the same angle it entered at. Because the two surfaces of the block are parallel, the ray that exits is parallel to the original incident ray, just shifted sideways slightly (laterally displaced) from where it would have gone if the block were not there.
QUESTION 8 6 marks Criterion B
Medium

A student wants to test whether the law of reflection ($i = r$) holds true for all angles of incidence, using a ray box, a plane mirror, a protractor, and a sheet of paper.

a. State the independent and dependent variables.
[2]
b. Describe a method the student could use to collect and check this data for at least five different angles of incidence.
[4]
Show complete worked solution
(a)
Independent variable: angle of incidence (set by aiming the ray box). Dependent variable: angle of reflection (measured with the protractor).
(b)
  1. Place the mirror upright on a sheet of paper and draw a line along its back edge, marking the point where the ray will strike it.
  2. Draw a normal line at $90\degree$ to the mirror at this point, using a protractor.
  3. Direct a single ray from the ray box at a chosen angle of incidence (e.g. $20\degree$) from the normal, and mark two points along the incident ray and two along the reflected ray on the paper.
  4. Remove the ray box and mirror, and join the marked points with a ruler to draw the full ray paths; use the protractor to measure the angle of reflection.
  5. Repeat for at least five different angles of incidence (e.g. $20\degree, 30\degree, 40\degree, 50\degree, 60\degree$), and compare each measured angle of reflection to the angle of incidence used.
QUESTION 9 6 marks Criterion B
Medium

A student wants to investigate how the angle of incidence affects the angle of refraction when light passes from air into a rectangular glass block.

a. Identify the independent, dependent, and one controlled variable for this investigation.
[3]
b. Describe how the student could measure the angle of refraction for each angle of incidence tested.
[3]
Show complete worked solution
(a)
Independent: angle of incidence in air. Dependent: angle of refraction in the glass. Controlled: the glass block used (same type of glass each time) — a different type of glass would bend light by a different amount, regardless of the angle of incidence, making comparisons unfair.
(b)
Place the glass block on paper and trace around it. Direct a ray from a ray box into one face at a chosen angle of incidence, marking where the ray enters and where it exits the far side of the block with small dots. After removing the block, draw a straight line connecting the entry point to the exit point to show the path of the ray inside the glass, draw the normal at the entry point, and measure the angle between this line and the normal using a protractor. Repeat for several different angles of incidence.
QUESTION 10 8 marks Criterion B
Hard

In the ray-box experiments above, students mark the ray's path using small pencil dots on paper, then join them with a ruler afterwards.

a. Identify a source of error in this method, and explain how it could affect the measured angles.
[3]
b. Suggest an improvement to the method that would reduce this error.
[3]
c. Suggest a way of removing this source of error almost entirely.
[2]
Show complete worked solution
(a)
A real ray of light from a ray box has some width, and marking its exact centre with a dot involves judgement (parallax) error — if the two dots marking a ray are not placed precisely along its true centre line, the ruled line (and therefore the measured angle) will be slightly wrong. This error is worse if the two dots marking one ray are placed close together, since a small placement error has a larger effect on the angle of a short line than a long one.
(b)
Mark the two dots for each ray as far apart as possible along its visible path (e.g. one dot close to the mirror/block and one near the edge of the paper) — this reduces the effect of any small error in dot placement on the final measured angle. Using a thin ray box slit to produce a narrower beam would also make the ray's true centre easier to judge accurately.
(c)
Photograph the ray paths on the paper from directly above (avoiding any angle that could distort the image) and use image-analysis or angle-measuring software on the photo, rather than marking and ruling by hand — or use a laser and a protractor board with a built-in scale, so the angle can be read directly without needing to draw and measure the ray's path at all.
QUESTION 11 3 marks Criterion C
Easy

A student measured the angle of reflection for five different angles of incidence at a plane mirror.

Angle of incidence (°)1025405570
Angle of reflection (°)1025405570
a. Describe the pattern shown by the data.
[2]
b. Which law of physics does this data confirm?
[1]
Show complete worked solution
(a)
For every angle tested, the angle of reflection is exactly equal to the angle of incidence ($i = r$ in every row).
(b)
This confirms the law of reflection.
QUESTION 12 5 marks Criterion C
Medium

A student measured the angle of refraction in a glass block for three different angles of incidence.

Angle of incidence in air (°)204060
Angle of refraction in glass (°)132535
a. Describe the general relationship shown between the angle of incidence and the angle of refraction.
[1]
b. Calculate the difference between the angle of incidence and the angle of refraction for each row.
[2]
c. Using your answers to (b), describe how the amount of bending changes as the angle of incidence increases.
[2]
Show complete worked solution
(a)
As the angle of incidence increases, the angle of refraction also increases, but the angle of refraction is always smaller than the angle of incidence.
(b)
$20\degree - 13\degree = 7\degree$; $\quad 40\degree - 25\degree = 15\degree$; $\quad 60\degree - 35\degree = 25\degree$
(c)
The difference between the two angles increases as the angle of incidence increases ($7\degree \to 15\degree \to 25\degree$), showing that light is bent by a greater amount (in absolute terms) the more steeply (at a larger angle from the normal) it strikes the glass surface.
QUESTION 13 4 marks Criterion C
Medium
normal incident ray reflected ray i = 55° r = 30° plane mirror

A student drew this ray diagram to represent a ray of light reflecting off a plane mirror, measuring the angles shown.

a. Explain why this diagram cannot be physically correct.
[2]
b. State what the angle of reflection should be, and suggest one likely cause of the student's error.
[2]
Show complete worked solution
(a)
The diagram shows an angle of incidence of $55\degree$ but an angle of reflection of only $30\degree$. This breaks the law of reflection, which requires the angle of incidence to always equal the angle of reflection ($i = r$) for a plane mirror — the two angles shown are not equal.
(b)
The angle of reflection should be $55\degree$, the same as the angle of incidence. A likely cause is that the student measured the reflected ray's angle from the mirror's surface rather than from the normal, or made a measurement/protractor-reading error when drawing or measuring the reflected ray.
QUESTION 14 5 marks Criterion C
Medium

A ray of light was directed into three different materials, each time at the same angle of incidence, $i = 50\degree$, and the angle of refraction was measured.

MaterialWaterGlassDiamond
Angle of refraction for i = 50°35°30°19°
a. Which material bends (refracts) light the most, and which the least? Use the data to support your answer.
[2]
b. Using this data, suggest which of the three materials light travels slowest through, explaining your reasoning.
[3]
Show complete worked solution
(a)
Diamond bends light the most — it produces the smallest angle of refraction ($19\degree$), meaning the ray bends furthest from its original direction. Water bends light the least — it produces the largest angle of refraction ($35\degree$), closest to the original angle of incidence.
(b)
Light bends more when entering a medium that slows it down more. Since diamond causes the greatest bending (largest change from $50\degree$ to $19\degree$), it must slow the light down the most of the three materials, meaning light travels slowest through diamond.
QUESTION 15 5 marks Criterion C
Medium

The critical angle is the angle of incidence (inside a medium) above which total internal reflection occurs instead of refraction. The table shows the critical angle for three materials.

MaterialWaterGlassDiamond
Critical angle49°42°24°
a. State what total internal reflection means.
[1]
b. Using the table, which material would experience total internal reflection over the widest range of angles?
[2]
c. Explain why this property makes diamond useful for gemstone cutting, where sparkle is desirable.
[2]
Show complete worked solution
(a)
All of the light striking the boundary from inside the denser medium is reflected back into it, with none escaping by refraction.
(b)
Diamond, since it has the smallest critical angle ($24\degree$) — total internal reflection happens for any angle of incidence greater than the critical angle, so a smaller critical angle means a much wider range of angles (from $24\degree$ up to $90\degree$) cause total internal reflection.
(c)
Because diamond's critical angle is so small, light entering a cut diamond is very likely to strike an internal face at an angle greater than the critical angle and undergo total internal reflection repeatedly, bouncing around inside the stone before eventually exiting towards the viewer — producing the bright, sparkling appearance diamonds are known for.
QUESTION 16 6 marks Criterion C
Hard

A student set the angle of incidence to exactly $45\degree$ and measured the angle of reflection $5$ times, using a protractor each time.

Trial12345
Measured angle of reflection (°) for i = 45°4446454743
a. Calculate the mean measured angle of reflection.
[2]
b. Compare the mean to the value predicted by the law of reflection, and comment on the reliability of the student's measurements.
[2]
c. Suggest one reason individual readings might vary slightly from exactly $45\degree$, even with a correctly working mirror.
[2]
Show complete worked solution
(a)
$$ \text{mean} = \frac{44+46+45+47+43}{5} = \frac{225}{5} = 45\degree $$
(b)
The law of reflection predicts an angle of reflection of exactly $45\degree$ (equal to the $45\degree$ angle of incidence). The mean of the $5$ trials is also exactly $45\degree$, showing excellent agreement — the small trial-to-trial variation ($43\degree$ to $47\degree$) most likely comes from small errors in reading the protractor, but averaging the repeats has cancelled this out well.
(c)
Small parallax or reading errors when lining up the protractor with the marked ray on the paper, or slight imprecision in marking the exact position of the thin light ray with a pencil dot, can cause the measured angle to vary by a degree or two between repeated trials.
QUESTION 17 7 marks Criterion C
Hard

A student measured angles of incidence and refraction for light entering a glass block, and looked up the sine of each angle.

i (°)sin ir (°)sin rsin i / sin r
200.342130.225?
400.643250.423?
600.866350.574?
a. Calculate $\dfrac{\sin i}{\sin r}$ for each row, giving your answers to 2 decimal places.
[3]
b. What do you notice about the value of $\dfrac{\sin i}{\sin r}$ across the three very different angles tested?
[2]
c. This constant ratio is called the refractive index of the glass. Evaluate whether this data supports the idea that the refractive index is a fixed property of a given material, rather than depending on the angle of incidence used.
[2]
Show complete worked solution
(a)
Row 1: $\dfrac{0.342}{0.225} = 1.52$
Row 2: $\dfrac{0.643}{0.423} = 1.52$
Row 3: $\dfrac{0.866}{0.574} = 1.51$
(b)
Even though the angle of incidence is very different in each row ($20\degree, 40\degree, 60\degree$), the ratio $\dfrac{\sin i}{\sin r}$ comes out to almost exactly the same value each time (about $1.5$).
(c)
Yes — since $\dfrac{\sin i}{\sin r}$ comes out very close to $1.5$ regardless of which angle of incidence was tested, this strongly supports the conclusion that the refractive index is a fixed property of the glass itself, not something that changes depending on the angle the light happens to enter at. The tiny variation between rows ($1.52, 1.52, 1.51$) is small enough to be explained by rounding and measurement error.
QUESTION 18 5 marks Criterion D
Medium

Optical fibres use total internal reflection to carry pulses of light, encoding phone calls, television, and internet data, over very long distances through thin glass or plastic strands.

Discuss one benefit and one drawback of relying on optical fibre technology for telecommunications.

Show complete worked solution

Benefit: Optical fibres can carry enormous amounts of data at extremely high speed over very long distances with very little signal loss, and are not affected by electrical or magnetic interference the way old copper cables are — this has made fast, reliable global internet and communication possible on a scale that older technologies could not support.

Drawback: Installing optical fibre networks is expensive and disruptive, requiring cables to be laid underground or undersea across huge distances, and the thin glass fibres are relatively fragile and can be damaged (e.g. by construction work), requiring specialist equipment and expertise to repair. This means fibre access has been rolled out unevenly, with rural and lower-income areas often gaining access much later than wealthier or denser urban areas.

QUESTION 19 6 marks Criterion D
Medium

Medical endoscopes use bundles of thin optical fibres, relying on total internal reflection, to let doctors see inside the human body through a small incision or natural opening, without needing open surgery.

Discuss one benefit and one drawback of using endoscopes in medicine.

Show complete worked solution

Benefit: Endoscopes allow doctors to examine or operate inside the body through a very small opening, which is far less invasive than traditional open surgery — this generally means less pain, a lower risk of infection, and a much faster recovery time for the patient.

Drawback: Endoscopic equipment is expensive and requires specially trained staff to operate safely and interpret the images correctly, meaning it is not equally available in all hospitals worldwide, particularly in poorer regions. There is also still some risk of discomfort, injury, or infection associated with inserting the endoscope, even though this risk is lower than with open surgery.

QUESTION 20 6 marks Criterion D
Hard

Concentrated solar power plants use large arrays of mirrors to reflect and focus sunlight onto a central tower, generating intense heat that is used to produce steam and generate electricity, as a renewable alternative to fossil fuels.

Evaluate the impact of this technology, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: Concentrated solar power generates electricity from sunlight, a renewable resource, without producing the greenhouse gas emissions associated with burning fossil fuels. Unlike some other forms of solar power, the heat it captures can be stored and used to generate electricity even after the Sun has set, helping to provide a more continuous, reliable supply of clean energy.

Concern: These plants require very large areas of land for the mirror arrays, which is often desert habitat that can still be home to wildlife — the intense concentrated light and heat around the central tower has also been reported to injure or kill birds and insects that fly through the focused beam. Building and maintaining such large reflective installations is also costly, and finding suitable large, sunny, flat sites is not possible everywhere.

The Electromagnetic Spectrum (intro) 20 questions

QUESTION 1 2 marks Criterion A
Easy
Gamma X-ray UV Visible Infrared Microwave Radio wavelength increasing, frequency & energy decreasing frequency & energy increasing The electromagnetic spectrum

The diagram shows the electromagnetic spectrum, arranged in order of wavelength.

a. Name the region of the spectrum with the longest wavelength.
[1]
b. Name the region of the spectrum with the highest frequency.
[1]
Show complete worked solution
(a)
Radio waves have the longest wavelength.
(b)
Gamma rays have the highest frequency.
QUESTION 2 2 marks Criterion A
Easy

State the speed at which all electromagnetic waves travel through a vacuum, and explain what makes this fact unusual.

Show complete worked solution

All electromagnetic waves travel through a vacuum at the same speed, $c = 3\times10^{8}\,\text{m/s}$ (the speed of light). This is unusual because it applies to every type of electromagnetic wave — radio waves, infrared, visible light, X-rays, and so on — despite them having enormously different wavelengths and frequencies; unlike sound, their speed in a vacuum does not depend on wavelength or frequency at all.

QUESTION 3 2 marks Criterion A
Easy

State one everyday use for each of the following types of electromagnetic wave:

a. Microwaves
[1]
b. Infrared
[1]
Show complete worked solution
(a)
Cooking food in a microwave oven (also accepted: mobile phone/WiFi signals, satellite communication).
(b)
TV remote controls (also accepted: thermal imaging cameras, short-range wireless data transfer, heaters/grills).
QUESTION 4 3 marks Criterion A
Medium

An FM radio station broadcasts radio waves with a wavelength of $\lambda = 3\,\text{m}$. All electromagnetic waves travel at $c = 3\times10^{8}\,\text{m/s}$ in air. Calculate the frequency of this radio station, using $c = f\lambda$.

Show complete worked solution

Step 1 — Rearrange for $f$:

$$ f = \frac{c}{\lambda} $$

Step 2 — Substitute:

$$ f = \frac{3\times10^{8}}{3} $$

Answer: $f = 1\times10^{8}\,\text{Hz} = 100\,\text{MHz}$, a typical FM radio frequency.

QUESTION 5 3 marks Criterion A
Medium

A medical X-ray machine produces X-rays with a frequency of $f = 3\times10^{18}\,\text{Hz}$. Calculate the wavelength of these X-rays.

Show complete worked solution

Step 1 — Rearrange $c=f\lambda$ for $\lambda$:

$$ \lambda = \frac{c}{f} $$

Step 2 — Substitute:

$$ \lambda = \frac{3\times10^{8}}{3\times10^{18}} $$

Answer: $\lambda = 1\times10^{-10}\,\text{m}$, a typical X-ray wavelength.

QUESTION 6 4 marks Criterion A
Medium

Explain why ultraviolet (UV) radiation carries far more energy than radio waves, referring to wavelength and frequency in your answer.

Show complete worked solution

Across the electromagnetic spectrum, shorter wavelength corresponds to higher frequency, and higher frequency electromagnetic waves carry more energy. UV radiation has a very short wavelength and therefore a very high frequency, compared to radio waves, which have an extremely long wavelength and very low frequency. This is why UV radiation carries enough energy to damage skin cells and cause sunburn, while radio waves (used for broadcasting and communication) are considered low-energy and harmless at everyday exposure levels.

QUESTION 7 6 marks Criterion A
Hard

A beam of green light has a frequency of $f = 6\times10^{14}\,\text{Hz}$.

a. Calculate the wavelength of this light.
[2]
b. State which part of the electromagnetic spectrum this wavelength belongs to.
[1]
c. If the wavelength of this light were doubled to $1000\,\text{nm}$, state which part of the spectrum it would now belong to, and describe what would happen to its frequency and its energy.
[3]
Show complete worked solution
(a)
$$ \lambda = \frac{c}{f} = \frac{3\times10^{8}}{6\times10^{14}} $$

Answer: $\lambda = 5\times10^{-7}\,\text{m} = 500\,\text{nm}$

(b)
$500\,\text{nm}$ is within the visible light region (consistent with it appearing green).
(c)
$1000\,\text{nm}$ ($1\times10^{-6}\,\text{m}$) lies in the infrared region, just beyond the red end of visible light. Doubling the wavelength halves the frequency (since $c=f\lambda$ is fixed, $f = \frac{3\times10^{8}}{1\times10^{-6}} = 3\times10^{14}\,\text{Hz}$, exactly half of the original $6\times10^{14}\,\text{Hz}$). Since energy increases with frequency, this lower frequency also means lower energy.
QUESTION 8 6 marks Criterion B
Medium

A student wants to investigate how the material a wall is made from affects how much a WiFi (microwave) signal is weakened as it passes through.

a. Identify the independent, dependent, and one controlled variable for this investigation.
[3]
b. Describe a method for measuring the effect of each material on the signal.
[3]
Show complete worked solution
(a)
Independent: material the wall/barrier is made from (e.g. wood, brick, glass, metal sheet — tested one at a time). Dependent: WiFi signal strength measured on the far side (using a smartphone app that reads signal strength in dBm). Controlled: distance between the WiFi router and the measuring device, and the thickness of each material tested — a different distance or thickness would itself change the signal strength, regardless of which material was used, making the comparison unfair.
(b)
Set up the WiFi router at a fixed position and distance from a laptop or phone used to measure signal strength. First measure the signal strength with no barrier present. Then, for each material tested, place a sample of the same size and thickness directly between the router and the device, and record the new signal strength using the same app. Repeat each measurement $3$ times and take a mean, to reduce the effect of any random fluctuation in the signal.
QUESTION 9 6 marks Criterion B
Medium

A student wants to investigate whether the colour of a surface affects how much infrared radiation it absorbs, using an infrared lamp and identical metal cans painted different colours.

a. Identify the independent, dependent, and one controlled variable for this investigation.
[3]
b. Describe a method for collecting the temperature data for each can.
[3]
Show complete worked solution
(a)
Independent: colour of the can's surface (e.g. matt black, white, silver). Dependent: temperature rise of the water inside the can. Controlled: distance of each can from the infrared lamp, and the exposure time — a can placed closer to the lamp, or left under it longer, would absorb more energy regardless of its colour, making the comparison unfair.
(b)
Fill each identical can with the same volume of water at the same starting temperature, and record this starting temperature with a thermometer. Place one can at a fixed distance from the infrared lamp and switch the lamp on for a fixed time (e.g. $5$ minutes). Record the final water temperature, and calculate the temperature rise. Repeat for each differently coloured can, keeping the distance, exposure time, and starting temperature the same each time.
QUESTION 10 8 marks Criterion B
Hard

A student tests how well different sunscreens block UV radiation using UV-sensitive beads, which change colour more when exposed to more UV radiation. A drop of each sunscreen is spread over a bead, and the beads are placed under a UV lamp for a fixed time. The student judges "how much each bead changed colour" by eye, giving a score from $1$ to $10$.

a. Identify a weakness of judging colour change "by eye" in this method.
[3]
b. Suggest an improvement to the method that would make the results more objective, and explain how it works.
[3]
c. State one variable that must be controlled between beads for the comparison between sunscreens to be fair.
[2]
Show complete worked solution
(a)
Judging colour change by eye is subjective — different people (or even the same person on different occasions) may score the same shade of colour change differently, since there is no precise, objective scale being used. This makes the results difficult to reproduce reliably and hard to compare fairly between sunscreens or between different students' results.
(b)
Use a colorimeter or light sensor connected to a data logger to measure the exact colour/brightness of each bead numerically, both before and after UV exposure, instead of estimating a score by eye. This gives a precise, repeatable, numerical measurement of colour change that does not depend on any one person's judgement.
(c)
The distance of each bead from the UV lamp, and the exposure time, must be kept the same for every bead tested (also accepted: the amount/thickness of sunscreen applied) — otherwise differences in colour change could be caused by unequal UV exposure rather than by how effective each sunscreen actually is.
QUESTION 11 3 marks Criterion C
Easy

Three identical cans of water, painted different colours, were placed under an infrared lamp for $5$ minutes.

Surface colourBlackWhiteSilver
Temperature rise after 5 min (°C)1242
a. Which surface absorbed infrared radiation best?
[1]
b. Which surface absorbed infrared radiation the worst, and what happened to most of the infrared radiation that hit it instead of being absorbed?
[2]
Show complete worked solution
(a)
Black — it had the largest temperature rise ($12\degree\text{C}$).
(b)
Silver absorbed the least ($2\degree\text{C}$ rise). Most of the infrared radiation hitting the shiny silver surface was instead reflected away rather than absorbed.
QUESTION 12 6 marks Criterion C
Medium

A student calculated the frequency of three electromagnetic waves from their wavelength, using $c = 3\times10^{8}\,\text{m/s}$.

Wave typeWavelengthStated frequency
Radio10 m3 × 10? Hz
Microwave0.01 m3 × 10? Hz
Infrared0.00001 m3 × 10¹³ Hz
a. Check each stated frequency by calculating $f = \dfrac{c}{\lambda}$ for all three rows.
[3]
b. Identify which row contains an error.
[1]
c. State the correct frequency for this row.
[2]
Show complete worked solution
(a)
Radio: $f = \dfrac{3\times10^{8}}{10} = 3\times10^{7}\,\text{Hz}$ ? matches.
Microwave: $f = \dfrac{3\times10^{8}}{0.01} = 3\times10^{10}\,\text{Hz}$ — does not match the stated $3\times10^{8}\,\text{Hz}$.
Infrared: $f = \dfrac{3\times10^{8}}{0.00001} = 3\times10^{13}\,\text{Hz}$ ? matches.
(b)
The microwave row is incorrect.
(c)
The correct frequency is $f = 3\times10^{10}\,\text{Hz}$.
QUESTION 13 4 marks Criterion C
Medium
Gamma X-ray UV Visible Infrared Microwave Radio wavelength increasing, frequency & energy decreasing frequency & energy increasing The electromagnetic spectrum

Use the electromagnetic spectrum diagram to answer the questions below.

a. Between infrared and ultraviolet, which has the shorter wavelength, and what does this tell you about its frequency?
[2]
b. Despite occupying very different positions on the spectrum, state one property that gamma rays and radio waves have in common.
[2]
Show complete worked solution
(a)
Ultraviolet has the shorter wavelength (it sits further towards the high-frequency end of the diagram than infrared). Since wavelength and frequency are inversely related for electromagnetic waves, a shorter wavelength means UV has a higher frequency than infrared.
(b)
Both gamma rays and radio waves are electromagnetic waves that travel at the same speed, $3\times10^{8}\,\text{m/s}$, in a vacuum — despite having enormously different wavelengths and frequencies (and therefore very different energies and uses).
QUESTION 14 5 marks Criterion C
Medium

A student measured WiFi signal strength (a more negative value means a weaker signal) after the signal passed through different numbers of internal walls.

Number of walls01234
Signal strength (dBm)-40-52-64-90-88
a. Identify the anomalous reading in the table.
[1]
b. Explain why this reading does not fit the pattern of the rest of the data.
[2]
c. Suggest a possible cause of this anomalous reading.
[2]
Show complete worked solution
(a)
The reading at $3$ walls, $-90\,\text{dBm}$, is anomalous.
(b)
Every other reading in the table weakens by almost exactly $12\,\text{dBm}$ for each extra wall ($-40\to-52\to-64$, a fall of $12$ each time), and the row for $4$ walls fits this trend if continued from $2$ walls ($-64 - 24 = -88$). Following this pattern, $3$ walls should give roughly $-76\,\text{dBm}$, not the much weaker $-90\,\text{dBm}$ that was recorded.
(c)
The third wall may have contained something that blocks or reflects microwave signals unusually strongly, such as metal pipes, wiring, or a metal-backed mirror inside it, or there may have been temporary interference from another wireless device at the moment that reading was taken.
QUESTION 15 5 marks Criterion C
Medium

The table shows the recorded UV index for each day of one week. A UV index of $8$ or above is considered "very high" risk.

DayMonTueWedThuFriSatSun
UV index3579642
a. On which day was the risk from UV radiation highest?
[1]
b. Calculate the mean UV index for the week.
[2]
c. Using the table, state which day(s) fall into the "very high" risk category, and suggest one precaution people should take on such a day.
[2]
Show complete worked solution
(a)
Thursday, with a UV index of $9$.
(b)
$$ \text{mean} = \frac{3+5+7+9+6+4+2}{7} = \frac{36}{7} = 5.14\ (\text{3 s.f.}) $$
(c)
Only Thursday (UV index $9$) falls into the "very high" category ($\geq8$). On such a day, people should apply high-SPF sunscreen, wear a hat and sunglasses, and avoid direct sun exposure during the middle of the day when UV levels are typically highest.
QUESTION 16 7 marks Criterion C
Hard

A student repeated the infrared absorption experiment $3$ times for a black surface and $3$ times for a white surface, under identical conditions.

Trial123
Black surface temp. rise (°C)111312
White surface temp. rise (°C)435
a. Calculate the mean temperature rise for each surface.
[2]
b. Evaluate the reliability of this data, using the size of the variation within each set of trials.
[3]
c. State the conclusion this data supports about the relationship between surface colour and infrared absorption.
[2]
Show complete worked solution
(a)
Black: $$ \frac{11+13+12}{3} = \frac{36}{3} = 12\degree\text{C} $$ White: $$ \frac{4+3+5}{3} = \frac{12}{3} = 4\degree\text{C} $$
(b)
For both surfaces, the three repeated trials are very close together (black: $11$–$13\degree\text{C}$, a range of only $2\degree\text{C}$; white: $3$–$5\degree\text{C}$, also a range of $2\degree\text{C}$). This small spread suggests the measurements were precise and reliable, and the difference between the two mean values ($12\degree\text{C}$ vs $4\degree\text{C}$) is far larger than the spread within either set — so the data strongly and reliably supports the conclusion that black absorbs infrared radiation better than white.
(c)
The data supports the conclusion that a black (dark, matt) surface absorbs infrared radiation much more effectively than a white surface, which reflects most of the infrared radiation instead of absorbing it.
QUESTION 17 7 marks Criterion C
Hard

A student calculated the speed of three different electromagnetic waves using $v = f\lambda$, from measured wavelength and frequency data.

WaveWavelengthFrequencyCalculated speed (student's answer)
12 m1.5 × 10? Hz3 × 10? m/s
25 × 10?? m6 × 10¹? Hz3 × 10? m/s
30.03 m1 × 10¹? Hz3 × 10? m/s
a. Recalculate the speed for all three waves to check the student's answers.
[3]
b. State the correct speed for Wave 2, and identify the likely type of mistake the student made.
[2]
c. Explain why checking that every calculated speed comes out close to $3\times10^{8}\,\text{m/s}$ is a useful and quick way to spot arithmetic errors like this one.
[2]
Show complete worked solution
(a)
Wave 1: $v = 1.5\times10^{8} \times 2 = 3\times10^{8}\,\text{m/s}$ ?
Wave 2: $v = 6\times10^{14} \times 5\times10^{-7} = 3\times10^{8}\,\text{m/s}$ — the student's stated answer of $3\times10^{7}\,\text{m/s}$ is wrong (out by a factor of $10$).
Wave 3: $v = 1\times10^{10} \times 0.03 = 3\times10^{8}\,\text{m/s}$ ?
(b)
The correct speed is $3\times10^{8}\,\text{m/s}$. The student's answer is smaller by exactly a factor of $10$, so the mistake was most likely a power-of-ten (decimal place) error when multiplying the standard form numbers, rather than a conceptual misunderstanding.
(c)
Because every electromagnetic wave travels at the same speed in a vacuum (or air), $c = 3\times10^{8}\,\text{m/s}$, any calculated speed that comes out noticeably different from this value must contain an arithmetic error, regardless of which specific wavelength and frequency were used — this makes it a fast, reliable way to check working without needing to redo the whole calculation from scratch.
QUESTION 18 5 marks Criterion D
Medium

X-rays are widely used in hospitals to produce images of bones and internal structures, helping doctors diagnose fractures and other conditions quickly. However, X-rays are a form of ionising radiation.

Discuss one benefit and one drawback of using X-rays in medicine.

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Benefit: X-ray imaging is quick, relatively low-cost, and gives doctors detailed images of bones and some internal structures almost immediately, without needing invasive surgery — this allows fractures, some tumours, and other internal problems to be diagnosed and treated rapidly, which can be critical in emergencies.

Drawback: Because X-rays are ionising radiation, they can damage living cells and increase the risk of cancer with repeated exposure over a lifetime. For this reason, hospitals strictly limit how many X-rays a patient receives, use lead aprons and shielding to protect other parts of the body and staff, and avoid X-rays for pregnant patients where possible, using safer imaging methods (such as ultrasound) instead when appropriate.

QUESTION 19 6 marks Criterion D
Medium

Mobile phones and WiFi routers use microwave and radio-frequency electromagnetic waves to give people constant access to calls, messaging, and the internet, almost everywhere.

Discuss one benefit and one concern raised by this widespread use of wireless technology.

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Benefit: Wireless communication gives people near-constant access to information, education, work, and emergency services, wherever they are, without needing physical cables — this has transformed education, business, and personal safety (for example, being able to call for help from almost anywhere).

Concern: With so many devices using the same limited ranges of frequencies, signal congestion and interference between networks is an ongoing challenge that requires careful regulation of which frequencies each type of device is allowed to use. There has also been public concern about possible long-term health effects of continuous exposure to radio-frequency radiation from phones and masts; although major scientific reviews have found no confirmed harm at the low power levels used, ongoing monitoring and research is still considered important as usage continues to grow.

QUESTION 20 7 marks Criterion D
Hard

Gamma radiation, the highest-energy and most ionising part of the electromagnetic spectrum, is used both to sterilise medical equipment (killing bacteria without heat) and, in carefully controlled doses, to treat cancer by destroying tumour cells (radiotherapy).

Evaluate the impact of using gamma radiation in medicine, discussing both a benefit and a drawback.

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Benefit: Gamma radiation can sterilise medical equipment (such as syringes and surgical instruments) thoroughly without using heat, which is essential for equipment that would be damaged by traditional heat sterilisation, helping prevent the spread of infection. In radiotherapy, carefully targeted and controlled doses of gamma radiation can destroy cancerous tumour cells, and has become one of the most important treatments for saving the lives of cancer patients worldwide.

Drawback: Gamma radiation is highly ionising and dangerous — it can damage or kill healthy living cells just as it can kill bacteria or cancer cells, so equipment producing or using it requires thick shielding (often lead or concrete) and strict safety procedures for the staff who work with it. In radiotherapy specifically, some healthy tissue surrounding a tumour is inevitably exposed too, causing side effects such as fatigue, skin irritation, or damage to nearby healthy organs, meaning doctors must carefully balance the dose needed to treat the cancer against the harm caused to the patient's healthy cells. Safe long-term disposal of radioactive sources used in this equipment is also a significant ongoing challenge.