Waves, Light and Sound
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Wave Basics: Amplitude, Wavelength, Frequency 20 questions
The graph shows a snapshot of a transverse wave travelling along a rope.
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State what is meant by the frequency of a wave, and give the unit it is measured in.
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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.
A wave source vibrates with a frequency of $f = 50\,\text{Hz}$. Calculate the period of the wave, using $T = \dfrac{1}{f}$.
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Step 1 — State the formula:
$$ T = \frac{1}{f} $$
Step 2 — Substitute:
$$ T = \frac{1}{50} $$
Answer: $T = 0.02\,\text{s}$
A wave source produces waves with frequency $f = 5\,\text{Hz}$ and wavelength $\lambda = 2\,\text{m}$, travelling through a fixed medium.
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Answer: $v = 10\,\text{m/s}$
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.
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Step 1 — Rearrange $v=f\lambda$ for $f$:
$$ f = \frac{v}{\lambda} $$
Step 2 — Substitute:
$$ f = \frac{340}{0.68} $$
Answer: $f = 500\,\text{Hz}$
Distinguish between a transverse wave and a longitudinal wave, giving one real example of each.
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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.
A wave machine produces $15$ complete waves in $3\,\text{s}$. Each wave has a wavelength of $0.5\,\text{m}$.
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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.
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- Set the dipper to the first test frequency and let the ripples become steady.
- 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.
- 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.
- Repeat for each test frequency, keeping water depth constant.
A student wants to investigate how the tension in a stretched spring affects the speed of a wave pulse sent along it.
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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.
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The table shows the displacement of a rope at different distances along it, at one instant in time.
| Distance along wave (m) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Displacement (cm) | 0 | 3 | 0 | -3 | 0 | 3 | 0 | -3 | 0 |
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A student uses a ripple tank to measure the wavelength of water waves at several different dipper frequencies.
| Frequency (Hz) | 2 | 4 | 8 |
|---|---|---|---|
| Wavelength (m) | 1.70 | 0.85 | 0.425 |
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The graph shows a wave on the sea surface, produced by a source vibrating at $f = 0.25\,\text{Hz}$.
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Answer: $v = 2\,\text{m/s}$
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) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 |
|---|---|---|---|---|---|
| Frequency measured (Hz) | 24 | 12 | 5 | 6 | 4.8 |
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Two students separately counted the number of complete waves produced by the same wave machine over the same $10\,\text{s}$ interval, by eye.
| Student | Number of complete waves counted in 10 s |
|---|---|
| X | 30 |
| Y | 34 |
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A student times how long a pendulum-driven wave source takes to complete $20$ full oscillations, repeating the measurement $5$ times.
| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Time for 20 oscillations (s) | 4.1 | 4.3 | 3.9 | 4.2 | 4.0 |
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Scientists tracked a tsunami wave crossing the open ocean after an undersea earthquake.
| Time after earthquake (min) | 0 | 10 | 20 | 30 | 40 |
|---|---|---|---|---|---|
| Distance travelled (km) | 0 | 120 | 240 | 362 | 478 |
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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.
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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").
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.
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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.
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.
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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
The diagram shows a sound wave travelling through air, drawn as vertical lines representing air particles at one instant.
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Explain why sound cannot travel through a vacuum (empty space with no particles), but light can.
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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.
State which property of a sound wave (frequency or amplitude) determines each of the following:
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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.
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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: $204\,\text{m}$
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.
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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: $v = 1500\,\text{m/s}$
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.
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A ship's sonar system sends out an ultrasound pulse of frequency $f = 50\,000\,\text{Hz}$, which travels through seawater at $v = 1500\,\text{m/s}$.
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A student wants to investigate how the tension in a guitar string affects the pitch of the note it produces when plucked.
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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.
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- Stand at the measured distance from the wall, facing it, in a quiet outdoor area.
- Clap sharply once, starting the stopwatch at the same instant.
- Stop the stopwatch the instant the echo is heard.
- Repeat the clap-and-time process at least $5$ times at the same distance, and record each time.
In the echo experiment above, the student starts and stops the stopwatch by hand, reacting to the clap and to hearing the echo.
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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.
| Note | Trace height (mm) |
|---|---|
| X | 12 |
| Y | 4 |
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A student stands $85\,\text{m}$ from a wall and claps, timing the echo $3$ times.
| Trial | 1 | 2 | 3 |
|---|---|---|---|
| Time for echo (s) | 0.50 | 0.52 | 0.48 |
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Answer: $v = 340\,\text{m/s}$
The traces show two different sounds, A and B, recorded over the same time interval.
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A class tested the highest frequency each of five volunteers of different ages could hear, using a frequency generator.
| Age (years) | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Highest audible frequency (Hz) | 19 500 | 18 000 | 52 000 | 15 000 | 13 000 |
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A student measures the sound intensity level from a loudspeaker at increasing distances.
| Distance from speaker (m) | 1 | 2 | 4 | 8 |
|---|---|---|---|---|
| Sound intensity level (dB) | 100 | 94 | 88 | 82 |
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A student stands $100\,\text{m}$ from a wall and times a clap's echo $5$ times.
| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Echo time for 100 m (s) | 0.61 | 0.58 | 0.60 | 0.57 | 0.59 |
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The table shows the approximate range of frequencies that different animals can hear.
| Animal | Lower limit (Hz) | Upper limit (Hz) |
|---|---|---|
| Human | 20 | 20 000 |
| Dog | 67 | 45 000 |
| Bat | 2 000 | 110 000 |
| Elephant | 1 | 20 000 |
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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.
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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.
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.
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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.
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.
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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
The diagram shows a ray of light striking a plane mirror.
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Define each of the following terms used to describe reflection at a mirror:
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Light travels from air into glass, a denser medium, hitting the surface at an angle.
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The diagram shows a ray of light entering a glass block from air.
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Total internal reflection is used in optical fibres to transmit light signals over long distances.
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Describe four properties of the image formed by a plane (flat) mirror.
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The image formed by a plane mirror is:
- Virtual — it cannot be projected onto a screen, since the light rays only appear to come from behind the mirror.
- The same size as the object.
- Laterally inverted — left and right are swapped (e.g. text appears back-to-front).
- The same distance behind the mirror as the object is in front of it.
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 $$ n = \frac{\sin i}{\sin r} $$
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Answer: $r \approx 35\degree$
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.
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- 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.
- Draw a normal line at $90\degree$ to the mirror at this point, using a protractor.
- 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.
- 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.
- 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.
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.
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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.
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A student measured the angle of reflection for five different angles of incidence at a plane mirror.
| Angle of incidence (°) | 10 | 25 | 40 | 55 | 70 |
|---|---|---|---|---|---|
| Angle of reflection (°) | 10 | 25 | 40 | 55 | 70 |
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A student measured the angle of refraction in a glass block for three different angles of incidence.
| Angle of incidence in air (°) | 20 | 40 | 60 |
|---|---|---|---|
| Angle of refraction in glass (°) | 13 | 25 | 35 |
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A student drew this ray diagram to represent a ray of light reflecting off a plane mirror, measuring the angles shown.
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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.
| Material | Water | Glass | Diamond |
|---|---|---|---|
| Angle of refraction for i = 50° | 35° | 30° | 19° |
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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.
| Material | Water | Glass | Diamond |
|---|---|---|---|
| Critical angle | 49° | 42° | 24° |
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A student set the angle of incidence to exactly $45\degree$ and measured the angle of reflection $5$ times, using a protractor each time.
| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Measured angle of reflection (°) for i = 45° | 44 | 46 | 45 | 47 | 43 |
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A student measured angles of incidence and refraction for light entering a glass block, and looked up the sine of each angle.
| i (°) | sin i | r (°) | sin r | sin i / sin r |
|---|---|---|---|---|
| 20 | 0.342 | 13 | 0.225 | ? |
| 40 | 0.643 | 25 | 0.423 | ? |
| 60 | 0.866 | 35 | 0.574 | ? |
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Row 2: $\dfrac{0.643}{0.423} = 1.52$
Row 3: $\dfrac{0.866}{0.574} = 1.51$
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.
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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.
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.
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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.
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.
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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
The diagram shows the electromagnetic spectrum, arranged in order of wavelength.
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State the speed at which all electromagnetic waves travel through a vacuum, and explain what makes this fact unusual.
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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.
State one everyday use for each of the following types of electromagnetic wave:
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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 $c = f\lambda$.
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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.
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.
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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.
Explain why ultraviolet (UV) radiation carries far more energy than radio waves, referring to wavelength and frequency in your answer.
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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.
A beam of green light has a frequency of $f = 6\times10^{14}\,\text{Hz}$.
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Answer: $\lambda = 5\times10^{-7}\,\text{m} = 500\,\text{nm}$
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.
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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.
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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 $10$.
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Three identical cans of water, painted different colours, were placed under an infrared lamp for $5$ minutes.
| Surface colour | Black | White | Silver |
|---|---|---|---|
| Temperature rise after 5 min (°C) | 12 | 4 | 2 |
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A student calculated the frequency of three electromagnetic waves from their wavelength, using $c = 3\times10^{8}\,\text{m/s}$.
| Wave type | Wavelength | Stated frequency |
|---|---|---|
| Radio | 10 m | 3 × 10? Hz |
| Microwave | 0.01 m | 3 × 10? Hz |
| Infrared | 0.00001 m | 3 × 10¹³ Hz |
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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.
Use the electromagnetic spectrum diagram to answer the questions below.
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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 walls | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Signal strength (dBm) | -40 | -52 | -64 | -90 | -88 |
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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.
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
|---|---|---|---|---|---|---|---|
| UV index | 3 | 5 | 7 | 9 | 6 | 4 | 2 |
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A student repeated the infrared absorption experiment $3$ times for a black surface and $3$ times for a white surface, under identical conditions.
| Trial | 1 | 2 | 3 |
|---|---|---|---|
| Black surface temp. rise (°C) | 11 | 13 | 12 |
| White surface temp. rise (°C) | 4 | 3 | 5 |
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A student calculated the speed of three different electromagnetic waves using $v = f\lambda$, from measured wavelength and frequency data.
| Wave | Wavelength | Frequency | Calculated speed (student's answer) |
|---|---|---|---|
| 1 | 2 m | 1.5 × 10? Hz | 3 × 10? m/s |
| 2 | 5 × 10?? m | 6 × 10¹? Hz | 3 × 10? m/s |
| 3 | 0.03 m | 1 × 10¹? Hz | 3 × 10? m/s |
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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: $v = 1\times10^{10} \times 0.03 = 3\times10^{8}\,\text{m/s}$ ?
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.
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.
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.