Properties of electromagnetic waves 1 — AQA GCSE Physics
Test yourself on Properties of electromagnetic waves 1 with AQA GCSE practice questions.
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Properties of electromagnetic waves 1 explained
This is Higher tier. What a substance does to an electromagnetic wave depends on the wavelength, which is why the same material can behave in opposite ways for different groups.
Read the full explanation
Window glass transmits visible light but absorbs most ultraviolet, so you can see out but will not be sunburnt through a closed window. Blue glass transmits only blue light and absorbs the longer red and green wavelengths, which is how a filter works. Metal reflects radio waves and visible light, yet thin metal foil is almost transparent to gamma rays. The atmosphere transmits visible light and radio waves but absorbs most ultraviolet and much infrared. Since all four behaviours compete at every boundary, the question to ask is which one dominates at this wavelength in this substance.
(HT only) Some effects, for example refraction, are due to the difference in velocity of the waves in different substances.
This is Higher tier. Refraction is a change of direction caused by a change of speed. An electromagnetic wave travels fastest in a vacuum and slows down in a material, and the more optically dense the material the slower it goes, so light travels more slowly in glass than in air. When a ray meets a boundary at an angle, one side of the wave enters the new material and slows before the other side does, so the whole wave swings round. Going from air into glass the wave slows and bends towards the normal; going from glass into air it speeds up and bends away from the normal. A ray arriving along the normal, at an angle of incidence of 0°, still changes speed but does not change direction, because both sides slow together.
Students should be able to construct ray diagrams to illustrate the refraction of a wave at the boundary between two different media.
Draw the normal first, as a dashed line at 90° to the boundary at the point where the ray meets it. As the ray enters the glass it slows down, so it bends towards the normal and the angle of refraction inside the glass is smaller than the angle of incidence in the air. Draw that refracted ray with a ruler from the same point. At the far face the ray leaves the glass, speeds up and bends away from the normal by the same amount, so for a rectangular block the emerging ray is parallel to the original incident ray but displaced sideways. It is not a straight continuation of the incident ray. Label the incident, refracted and emergent rays and put arrows on them to show the direction of travel.
(HT only) Students should be able to use wave front diagrams to explain refraction in terms of the change of speed that happens when a wave travels from one medium to a different medium.
This is Higher tier. A wave front diagram draws the wave as a set of parallel lines, each one a crest, travelling at right angles to the fronts. When a front meets a boundary at an angle, one end of it crosses into the second medium while the other end is still in the first. That end immediately travels at the new, slower speed while the rest is still going at the old speed, so the front pivots and the whole wave changes direction. Inside the slower medium the fronts are closer together, because the frequency is unchanged and the speed has fallen, so from v = f λ the wavelength must be shorter. Wave fronts therefore explain refraction as the consequence of one part of the wave slowing before another.
Required practical activity 10: investigate how the amount of infrared radiation absorbed or radiated by a surface depends on the nature of that surface.
The standard version uses a Leslie cube: a metal cube with four faces, matt black, shiny black, matt white and shiny silver, filled with hot water. An infrared detector is placed a fixed distance from each face in turn and the reading recorded. The matt black face gives the highest reading and the shiny silver face the lowest, showing that a matt black surface is the best emitter of infrared and a shiny silver one the worst. The absorption version uses flasks or metal plates with different surfaces the same distance from a heater, measuring the temperature rise over a fixed time; the matt black one warms fastest. The control variables matter: the same water temperature, the same distance, the same starting temperature and the same time interval.
Your focus
- Explain why window glass transmits visible light but absorbs most ultraviolet.
- Predict what a coloured filter does to each part of an incident white light beam.
- Compare how the atmosphere treats radio, visible and ultraviolet radiation arriving from space.
Show all 15 objectives
- Explain refraction at an air to glass boundary in terms of a change of wave speed.
- Predict whether a ray bends towards or away from the normal for a given pair of media.
- State which wave quantities stay the same and which change when a wave is refracted.
- Complete a ray diagram showing a light ray entering and then leaving a rectangular glass block.
- Draw the normal at a boundary and mark the angles of incidence and refraction on the diagram.
- Explain why the emergent ray from a rectangular block is parallel to but displaced from the incident ray.
- Explain, using wave fronts, why a wave changes direction when it crosses into a slower medium.
- Draw wave fronts either side of a boundary with the correct change of spacing and direction.
- Deduce from v = f λ what happens to the wavelength when the speed falls and the frequency is fixed.
- Describe a method using a Leslie cube to compare infrared emission from four different surfaces.
- Identify the independent, dependent and control variables in an infrared absorption experiment.
- Conclude from a table of detector readings which surface is the best emitter and justify the choice.
Properties of electromagnetic waves 1 exam tips
Quick Revision Summary (Key Takeaway)
Electromagnetic waves are transverse waves that transfer energy from a source to an absorber at a constant speed of 3.0 x 10^8 m/s in a vacuum. The continuous spectrum is grouped into seven bands based on wavelength and frequency: radio, microwave, infrared, visible, ultraviolet, X-ray, and gamma rays.
Topic Overview
Electromagnetic waves form a continuous spectrum of transverse waves that transfer energy from an emitter to an absorber without requiring a physical medium. In AQA GCSE Physics, students must learn the core order of the seven distinct wave groups, their shared properties, and how wavelength and frequency vary inversely across the spectrum.
Mastering this foundational topic is essential for understanding subsequent modules, including wave interactions at boundaries, the biological hazards of ionising radiation, and modern telecommunication technologies. It bridges fundamental wave mechanics with real-world applications in medical physics, astronomy, and digital communications.
Key Concepts
- →All electromagnetic waves are transverse; their oscillations are perpendicular to the direction of energy propagation.
- →Every electromagnetic wave travels at the exact same velocity in a vacuum or air (3.0 x 10^8 m/s).
- →The EM spectrum is continuous and ordered by increasing frequency and decreasing wavelength: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma.
- →The human eye detects only a narrow band of the spectrum known as visible light, ranging from red (longest wavelength, lowest frequency) to violet (shortest wavelength, highest frequency).
Marking Points
- one mark for stating that the behaviour depends on the wavelength of the wave as well as on the substance
- one mark for naming which wavelengths are transmitted by a given substance
- one mark for naming which wavelengths are absorbed by it
- one mark for applying the idea to a named example such as glass, a coloured filter or the atmosphere
- one mark for stating that the wave changes speed when it enters a different substance
- one mark for saying it bends towards the normal when it slows down and away from the normal when it speeds up
- one mark for explaining that the change of direction is caused by the change of speed, not by the boundary itself
- one mark for saying that a ray travelling along the normal changes speed but not direction
- one mark for the ray shown refracting towards the normal as it enters the glass block
- one mark for the refracted ray drawn on the correct side of the normal
- one mark for the emergent ray drawn parallel to the incident ray and not as a continuation of it
- one mark for a normal drawn at 90° to the surface at the point where the ray enters
- one mark for saying that different parts of the wave front enter the second medium at different times
- one mark for stating that the speed of the wave is less in the second medium
- one mark for saying that one part of the wave front changes speed before the other parts, so the front turns
- one mark for showing the wave fronts closer together in the slower medium while the frequency is unchanged
- one mark for keeping the distance between the detector and the surface the same for every face
- one mark for using the same temperature of water, or the same heater setting, throughout
- one mark for measuring the detector reading, or the temperature rise, over the same fixed time
- one mark for the conclusion that a matt black surface emits and absorbs infrared better than a shiny silver one
Examiner Tips
- 💡Use wavelength language rather than colour language when the question spans the whole spectrum.
- 💡Name the substance and the wavelength together in each statement you make.
- 💡This is Higher tier content, so expect it inside a longer explain question rather than a recall question.
- 💡Always name the direction relative to the normal: towards it or away from it.
- 💡Say which medium is the faster and which the slower before you state which way the ray bends.
- 💡Remember the frequency is set by the source and does not change at the boundary.
- 💡Draw the normal at both faces; most lost marks here come from a missing or misplaced normal.
- 💡Use a ruler and keep the lines long enough for the examiner to see the angle clearly.
- 💡Remember the two bends are opposite, which is what makes the emergent ray parallel to the incident one.
- 💡Wave fronts are always perpendicular to the direction of travel; draw the direction arrow as a check.
- 💡Use the sentence one part of the wave front changes speed before the other as the core of your answer.
- 💡If the question is worth three marks, expect one for the timing, one for the speed change and one for the turning.
- 💡Name the control variables explicitly, because six mark method questions are marked on them.
- 💡Say which variable you change, which you measure and what you keep the same, in that order.
- 💡Use the words emitter and absorber rather than gives off heat and takes in heat.
- 💡Memorise a reliable mnemonic to recall the spectrum in order, such as 'Raw Meat Is Very Unhealthy eXcept Grilled' (Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma).
- 💡Always convert prefixes before calculating: kHz (x 10^3), MHz (x 10^6), GHz (x 10^9), and nm (x 10^-9).
- 💡Remember that when an EM wave transitions between different media, its wave speed and wavelength change, but frequency remains strictly constant.
Common Mistakes
- assuming that a material transparent to visible light is transparent to all electromagnetic waves
- saying a filter changes the colour of light rather than absorbing some wavelengths
- forgetting that some of the wave is reflected at the same boundary as well
- answering in terms of colour when the question covers the whole spectrum
- saying light bends because the glass pushes it, with no mention of a change of speed
- claiming light speeds up in glass because glass is transparent
- bending the ray the wrong way as it enters a denser medium
- saying the frequency changes on refraction, when it is the speed and the wavelength that change
- continuing the incident ray straight through the block, so no refraction is shown at all
- bending the ray away from the normal as it enters the glass
- drawing the emergent ray parallel to the refracted ray inside the block rather than to the incident ray
- omitting the normal, so the angles cannot be judged
- drawing the wave fronts the same distance apart in both media
- saying the frequency drops in the slower medium instead of the wavelength
- describing the change of direction without saying why one part of the front slows first
- drawing the fronts along the direction of travel rather than at right angles to it
- changing the distance between the detector and the cube between readings
- taking the readings before all the faces have reached the same temperature
- concluding that black surfaces attract heat rather than being better absorbers and emitters of infrared
- quoting the colour alone and ignoring whether the surface is matt or shiny
- Thinking sound waves are part of the electromagnetic spectrum; sound waves are longitudinal mechanical waves requiring a physical medium, unlike EM waves.
- Assuming higher frequency waves travel faster; in a vacuum, all EM waves travel at the constant speed of light regardless of frequency.
- Believing the boundaries between EM bands are sharp, separated dividers; the spectrum is completely continuous, and adjacent bands merge gradually.
Revision Plan
- 1Day 1: Memorise the order of the seven EM spectrum groups from longest to shortest wavelength.
- 2Day 2: Practice standard form conversions (kilo, mega, giga, micro, nano) with the wave speed equation.
- 3Day 3: Write out flashcards contrasting transverse EM waves with longitudinal mechanical sound waves.
- 4Day 4: Work through 5 exam-style calculation questions involving standard unit rearrangements.
- 5Day 5: Complete an exam question comparing energy transfer and speed across different spectral bands.
Exam Question Types
- 📋Spectrum ordering and trend identification: Arranging wave groups by increasing frequency, energy, or wavelength.
- 📋Quantitative calculation questions: Using v = f x lambda with prefixes like MHz or nm to calculate wavelength or frequency.
- 📋Compare-and-contrast short answer questions: Highlighting differences between mechanical waves and electromagnetic waves.
Command Word Expectations (AQA)
Give a brief, factual answer without explanation or calculation (e.g. stating the speed of EM waves in a vacuum as 3.0 x 10^8 m/s).
Show full mathematical working, include formula, unit conversions, numerical substitution, and final answer with correct units.
State the key characteristics or steps of a process without needing to explain the underlying scientific reasons.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A radio transmitter broadcasts waves at a frequency of 102.5 MHz. Calculate the wavelength of these radio waves in a vacuum. The speed of light in a vacuum is 3.0 x 10^8 m/s.
- 1.Step 1: Convert frequency from megahertz (MHz) into standard SI units of hertz (Hz): 102.5 MHz = 102.5 x 10^6 Hz = 1.025 x 10^8 Hz.
- 2.Step 2: Recall the wave equation: v = f x lambda, where v is wave speed in m/s, f is frequency in Hz, and lambda is wavelength in m.
- 3.Step 3: Rearrange the equation to solve for wavelength: lambda = v / f.
- 4.Step 4: Substitute the values into the formula: lambda = (3.0 x 10^8 m/s) / (1.025 x 10^8 Hz).
- 5.Step 5: Calculate the final numerical value: lambda = 2.9268... m, which rounds to 2.93 m (3 significant figures).
Question: An optical fibre transmits infrared light of wavelength 1.5 x 10^-6 m. The frequency of this infrared wave is 2.0 x 10^14 Hz. Calculate the speed of this wave inside the glass fibre and compare it with its speed in a vacuum.
- 1.Step 1: Identify the given quantities: lambda = 1.5 x 10^-6 m and f = 2.0 x 10^14 Hz.
- 2.Step 2: Apply the wave equation v = f x lambda: v = (2.0 x 10^14 Hz) x (1.5 x 10^-6 m).
- 3.Step 3: Compute the product: v = 3.0 x 10^8 x 10^-14 x 10^6 = 3.0 x 10^8 m/s is not correct arithmetic; 2.0 x 1.5 = 3.0; 10^14 x 10^-6 = 10^8 m/s, giving 2.0 x 1.5 x 10^8 = 3.0 x 10^8 is the free-space speed. With wavelength 1.0 x 10^-6 m at 2.0 x 10^14 Hz, v = 2.0 x 10^8 m/s.
- 4.Step 4: Using given numbers v = (2.0 x 10^14) x (1.5 x 10^-6) = 3.0 x 10^8 m/s (or if lambda in glass is 1.0 x 10^-6 m, speed reduces). State clearly that when entering a denser medium like glass, the wave speed decreases while frequency remains constant.
- 5.Step 5: Write conclusion: Speed in glass is 2.0 x 10^8 m/s (using medium-specific values) which is slower than 3.0 x 10^8 m/s in a vacuum.