Electromagnetic spectrum: properties, refraction, hazards and uses (RP10) – GCSE Physics exam questions with mark scheme

1Grade 1–34 marks

The groups of waves in the electromagnetic spectrum are shown below in order of increasing frequency.

(a) Name the parts labelled A and B. [2]

(b) Give one use of the waves in part A. [1]

(a) A: microwaves (allow microwave)(1)
(a) B: ultraviolet (allow UV)(1)
(b) any one from: satellite communications / cooking (heating) food (allow mobile phone signals / Wi-Fi / radar)(1)
(c) any one from: premature ageing of the skin / (increased risk of) skin cancer (allow cancer / sunburn / damage to eyes)(1)
Answer: (a) A microwaves, B ultraviolet (b) e.g. cooking food (c) e.g. increased risk of skin cancer
Examiner tip: Learn the order radio, microwave, infrared, visible, UV, X-ray, gamma, and check whether the list runs by increasing frequency or increasing wavelength before filling the gaps.

2Grade 6–7Higher5 marks

A ray of light travels from air into a glass block, as shown.

(a) Explain, in terms of wave fronts, why the light changes direction as it enters the glass. [4]

(b) Describe what happens to a ray that enters the glass along the normal. [1]

(a) light travels more slowly in glass than in air(1)
(a) one side/end of the wave front enters the glass first and slows down first(1)
(a) while the rest of the wave front is still travelling faster in air(1)
(a) so the wave front changes direction / bends towards the normal(1)
(b) passes straight through without changing direction (allow speed decreases but direction unchanged)(1)
Answer: (a) Glass slows light; part of the wave front slows first while the rest is still faster, so the wave front turns towards the normal. (b) It does not change direction.
Examiner tip: You must link the change of direction to different parts of the same wave front travelling at different speeds.

3Grade 4–55 marks

A student investigates how the type of surface affects the infrared radiation emitted. She fills a metal cube with hot water. Each vertical face of the cube has a different surface: matt black, shiny black, matt white and shiny silver. She measures the infrared radiated from each face using an infrared detector.

(a) Identify the independent variable. [1]

(b) Give two control variables. [2]

(a) type of surface (colour / texture of the face)(1)
(b) any two from: distance between the detector and the face / starting temperature of the water (allow readings taken at the same time after filling, or quickly one after another) / same infrared detector / area of the face (or size of detector window facing it) / no draughts or other heat sources nearby (ignore 'same amount of water' / 'same equipment' alone)(1)
(b) second control variable from the list above(1)
(c) matt black(1)
(c) (dark) matt black surfaces are the best emitters of infrared radiation (allow black surfaces are good emitters)(1)
Answer: (a) type of surface (b) e.g. distance from detector, water temperature (c) matt black — best emitter
Examiner tip: Name the control variable precisely: 'the distance from the detector to the surface', not just 'distance'.

4Grade 6–7Higher5 marks

A radio transmitter produces radio waves using oscillating currents in an electrical circuit.

(a) Explain how radio waves can produce an electrical signal in a receiver aerial. [2]

(b) A radio station broadcasts waves with a wavelength of 250 m. The speed of radio waves is $3.0 \times 10^{8}$ m/s.

Calculate the frequency of the waves in kHz. [3]

(a) the radio waves are absorbed by the aerial and cause electrons in it to oscillate(1)
(a) producing an alternating current (allow induced a.c. / alternating potential difference) with the same frequency as the radio wave(1)
(b) $f = \dfrac{3.0 \times 10^{8}}{250}$ (allow $3.0 \times 10^{8} = f \times 250$)(1)
(b) $1.2 \times 10^{6}$ Hz(1)
(b) 1200 kHz (allow 1.2 × 10³ kHz; a correct answer with no working scores 3 marks)(1)
Answer: (a) Electrons in the aerial oscillate, giving an a.c. at the same frequency. (b) 1200 kHz
Examiner tip: 1 kHz = 1000 Hz — divide by 1000 to convert Hz to kHz.

5Grade 1–3Foundation6 marks

Electromagnetic waves form a continuous spectrum.

(a) Complete the sentence. Choose the answer from the box.

longitudinal | sound | transverse

All electromagnetic waves are __________ waves. [1]

(b) Which group of electromagnetic waves can be detected by the human eye? [1]

Type of wave		Use
infrared		energy efficient lamps
visible light		thermal imaging camera
ultraviolet		fibre optic communications
		satellite communications

(d) Which statement is true for all electromagnetic waves travelling through a vacuum? Tick (✓) one box.

☐ They all have the same frequency.
☐ They all have the same wavelength.
☐ They all travel at the same speed.
☐ They all transfer the same amount of energy.

[1]

(a) transverse(1)
(b) visible light (allow light; ignore named colours alone)(1)
(c) infrared → thermal imaging camera(1)
(c) visible light → fibre optic communications(1)
(c) ultraviolet → energy efficient lamps
(more than one line from a wave type: that wave scores 0)(1)
(d) They all travel at the same speed (more than one box ticked: 0 marks)(1)
Answer: (a) transverse (b) visible light (c) infrared – thermal imaging camera; visible – fibre optics; ultraviolet – energy efficient lamps (d) same speed
Examiner tip: Satellite communications is a distractor here: it is a use of microwaves, so do not join it to any of the three waves given.

6Grade 4–55 marks

A communications satellite sends television signals to a dish on the roof of a house. The signals are carried by microwaves.

The frequency of the microwaves is 12 GHz.

The speed of electromagnetic waves is $3.0 \times 10^{8}$ m/s.

(a) Calculate the wavelength of the microwaves. Give your answer in centimetres. [3]

(b) The satellite is 36 000 km above the dish.

Calculate the time taken for the signal to travel from the satellite to the dish. [2]

(a) $3.0 \times 10^{8} = 1.2 \times 10^{10} \times \lambda$ (frequency converted to Hz and correct substitution; allow $\lambda = \dfrac{3.0 \times 10^{8}}{12 \times 10^{9}}$)(1)
(a) $\lambda = 0.025$ m(1)
(a) $\lambda = 2.5$ cm (allow 2.5 cm with no working shown for 3 marks; allow correct conversion to cm of an incorrect value in m)(1)
(b) $36\,000\,000 = 3.0 \times 10^{8} \times t$ (distance converted to m and correct substitution)(1)
(b) $t = 0.12$ s (allow 0.12 with no working for 2 marks)(1)
Answer: (a) 2.5 cm (b) 0.12 s
Examiner tip: Convert GHz to Hz (× 10⁹) and km to m (× 1000) before substituting – most lost marks here are power-of-ten slips.

7Grade 4–54 marks

The diagram shows a ray of light travelling through a glass block towards the boundary between the glass and the air.

A, B and C show three possible paths for the ray after it leaves the glass.

(a) What is the name of the dashed line drawn at 90° to the boundary? [1]

(b) Which path, A, B or C, shows the ray after it enters the air? Give a reason for your answer. [2]

Determine the angle of incidence. [1]

(a) normal(1)
(b) C(1)
(b) the ray bends away from the normal (when it passes from glass into air) (allow the light speeds up / travels faster in air; reason only scores if C is chosen)(1)
(c) 30° (the angle of incidence is measured from the normal)(1)
Answer: (a) normal (b) C – light bends away from the normal as it speeds up entering air (c) 30°
Examiner tip: Angles of incidence and refraction are always measured from the normal, not from the surface – 60° is the trap answer in (c).

8Grade 6–7Higher5 marks

Different types of electromagnetic wave are absorbed and transmitted by the body by different amounts.

The table shows the percentage of three types of wave transmitted through 1 cm of soft tissue and through 1 cm of bone.

Type of wave	Percentage transmitted through 1 cm of soft tissue	Percentage transmitted through 1 cm of bone
visible light	less than 1	less than 1
X-rays	68	8
gamma rays	87	75

(a) Use data from the table to explain why X-rays are used to produce an image of a broken bone. [3]

(b) Suggest why gamma rays would produce a poor image of a broken bone. Use data from the table. [1]

(a) most X-rays pass through / are transmitted by soft tissue (68%) (allow soft tissue absorbs few X-rays)(1)
(a) most X-rays are absorbed by bone (only 8% transmitted) (allow 92% absorbed)(1)
(a) so there is a large difference between bone and tissue, so the bone shows up (as a shadow / white area) on the detector / film and a crack shows where X-rays pass through(1)
(b) bone and soft tissue transmit similar percentages (75% and 87%), so there is little contrast between them (allow most gamma rays pass through both)(1)
(c) visible light is absorbed (allow scattered) by the skin / soft tissue so does not reach the bone / pass through the body (allow less than 1% transmitted)(1)
Answer: (a) Tissue transmits most X-rays (68%) but bone absorbs most (only 8% transmitted), giving a clear contrast/shadow of the bone. (b) Similar transmission through bone and tissue, so little contrast. (c) Visible light is absorbed by the skin/tissue.
Examiner tip: Comparing bone with tissue is the key idea – quoting one number on its own does not explain why an image forms.

9Grade 6–76 marks

A student wants to investigate how the type of surface affects how much infrared radiation the surface absorbs.

She has:

two identical metal cans, one painted matt black and one painted shiny silver
an infrared heater
two thermometers
a measuring cylinder
a metre rule
a stopwatch
cold water.

Describe a method the student could use. Include how she should make sure the results are valid. [6]

measure the same volume of cold water into each can using the measuring cylinder(1)
put a thermometer in each can and record the starting temperature of the water(1)
place both cans the same distance from the heater (measured with the metre rule), e.g. one on each side of the heater(1)
each can facing the heater with its painted surface(1)
switch on the heater and start the stopwatch(1)
record the temperature of the water in each can after the same time, e.g. every minute for 10 minutes(1)
calculate the temperature increase for each can(1)
the can with the larger temperature increase absorbed more infrared(1)
control variables: same starting temperature of water, same can size/material, same volume of water, same distance, same time(1)
stir the water before reading the thermometer / read the thermometer at eye level(1)
repeat the experiment and calculate a mean temperature rise(1)
safety: do not touch the heater or the cans as they get hot(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically ordered method that could be followed by someone else to get valid results. Includes measuring the temperature rise of each can over the same time and controls the key variables (volume of water, distance from the heater, starting temperature).
Level 2	3–4	A method with most of the main steps, which would produce some results. Some control of variables is described, but steps may be out of order or lack detail.
Level 1	1–2	Some relevant steps are given, e.g. putting the cans near the heater and measuring temperature, but the method would not allow a valid comparison.
No relevant content	0	Nothing worthy of credit.

Answer: Same volume of cold water in each can, measure start temperatures, place both cans the same distance from the heater, switch on and record temperatures after the same time, compare temperature rises (larger rise = more IR absorbed); keep variables the same and repeat.
Examiner tip: Validity marks come from naming what is kept the same – especially the distance from the heater and the volume of water.

10Grade 6–7Higher7 marks

X-rays and gamma rays are both used in hospitals.

(a) Gamma rays and X-rays are produced by changes in atoms.

State which part of an atom gamma rays come from. [1]

(b) A patient has a CT scan, which uses X-rays. The patient receives a radiation dose of 8.0 mSv.

For every 1 Sv of radiation dose, the risk of developing a fatal cancer increases by about 5%.

1000 mSv = 1 Sv

Calculate the increase in the patient’s risk of developing a fatal cancer. [2]

(d) The radiographer who operates the CT scanner stands behind a lead-lined screen during every scan. The patient is not protected in this way.

Explain why it is reasonable for the patient, but not the radiographer, to be exposed to the X-rays. [2]

(a) (the) nucleus(1)
(b) 8.0 mSv = 0.0080 Sv(1)
(b) increase in risk = 0.0080 × 5 = 0.04% (allow 0.0004 or 1 in 2500; allow 0.04% with no working for 2 marks)(1)
(c) X-rays are ionising (radiation)(1)
(c) which can damage DNA / cause (gene) mutations in cells, so cells divide uncontrollably / cancer can develop (do not accept 'kills cells' as the route to cancer)(1)
(d) the patient has a single (small) dose and the benefit of the diagnosis outweighs the (very small) increase in risk(1)
(d) the radiographer would be exposed during many scans every day, so their total/cumulative dose (and risk) would be large, with no benefit to themselves(1)
Answer: (a) nucleus (b) 0.008 Sv × 5% = 0.04% (c) X-rays are ionising; they damage DNA/cause gene mutation, leading to cancer (d) Patient: one dose, benefit outweighs small risk; radiographer: repeated exposure would build up a large dose with no benefit.
Examiner tip: In (d) you must compare both people: one dose with a medical benefit versus a dose that would build up over many scans with no benefit.

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