Types of electromagnetic waves - AQA GCSE Physics Exam Questions

Practise 31 exam-style questions on Types of electromagnetic waves for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

31 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The diagram shows the electromagnetic spectrum. Two of the groups of waves are labelled A and B.

Name group A and group B.

Model answer

A is microwaves.
B is ultraviolet.

Explanation

From radio waves on the left, the order is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. A sits between radio waves and infrared, so it is microwaves; B sits between visible light and X-rays, so it is ultraviolet. A useful memory aid is that the wavelength gets shorter (and the frequency higher) from left to right.

Mark scheme

1 mark
A = microwaves
1 mark
B = ultraviolet
allow UV

Question 2

2 marks

Two electromagnetic waves, P and Q, are travelling through a vacuum.

The frequency of wave Q is twice the frequency of wave P.

Compare the wavelength of wave Q with the wavelength of wave P. Give a reason for your answer.

Model answer

The wavelength of Q is half the wavelength of P. All electromagnetic waves travel at the same speed in a vacuum, and wave speed = frequency × wavelength, so if the frequency doubles the wavelength must halve.

Explanation

Because v is the same for every electromagnetic wave in a vacuum, f × λ is constant, so frequency and wavelength are inversely proportional. Doubling one halves the other.

Mark scheme

1 mark
the wavelength of Q is half the wavelength of P
do not accept 'shorter' alone for this mark
1 mark
both waves travel at the same speed and wave speed = frequency × wavelength
allow wavelength is inversely proportional to frequency (as speed is constant); allow 'shorter' plus this reason for 1 mark

Question 3

3 marks

A long-wave radio station broadcasts radio waves with a wavelength of 1500 m.
A speed camera uses microwaves with a wavelength of 3.0 cm.

Both waves travel at the same speed.

Calculate how many times greater the frequency of the microwaves is than the frequency of the radio waves.

Model answer

3.0 cm = 0.030 m
The speed is the same, so frequency is inversely proportional to wavelength.
Ratio of frequencies = 1500 ÷ 0.030 = 50 000
The microwave frequency is 50 000 times greater.

Explanation

Since v = f λ and v is the same for both waves, the ratio of the frequencies equals the inverse ratio of the wavelengths. The wavelengths must be in the same unit before dividing; using 3.0 instead of 0.030 is the most common error.

Mark scheme

1 mark
3.0 cm = 0.030 m
an answer of 50 000 scores 3 marks; allow 1500 m = 150 000 cm
1 mark
ratio = 1500 ÷ 0.030
allow calculating both frequencies with v = 3.0 × 10⁸ m/s: 2.0 × 10⁵ Hz and 1.0 × 10¹⁰ Hz
1 mark
50 000
allow 5 × 10⁴; allow ecf; 500 (no conversion of cm) scores 2 marks

Question 4

4 marks

During a thunderstorm, a flash of lightning happens 1.65 km away from an observer.

The speed of light is 3.0 × 10⁸ m/s.
The speed of sound in air is 330 m/s.

Use calculations to explain why the observer sees the lightning before hearing the thunder.

Model answer

1.65 km = 1650 m
Light: t = s ÷ v = 1650 ÷ 3.0 × 10⁸ = 5.5 × 10⁻⁶ s
Sound: t = 1650 ÷ 330 = 5.0 s
The light arrives almost instantly, but the sound takes 5 s, so the lightning is seen about 5 seconds before the thunder is heard.

Explanation

Light is an electromagnetic wave and travels about a million times faster than sound in air. The light delay over a few kilometres is only a few millionths of a second, so the gap between flash and thunder is almost entirely the time taken by the sound.

Mark scheme

1 mark
1.65 km = 1650 m and t = 1650 ÷ 3.0 × 10⁸
1 mark
time for light = 5.5 × 10⁻⁶ s
allow 0.0000055 s
1 mark
time for sound = 1650 ÷ 330 = 5.0 s
allow ecf from an incorrect distance conversion
1 mark
light arrives (almost) instantly, about 5 s before the sound / light travels much faster than sound
must be a comparison

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