A Level Physics 8.2 Diffraction MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice questions on diffraction for Cambridge International AS Level Physics (9702), Section 8.2 – how waves spread at gaps and edges, how wavelength and gap width control the amount of diffraction, and the single-slit pattern. Every question is original and comes with the answer and a full explanation.
Diffraction at Gaps & Edges
Q1. What is meant by the diffraction of a wave?
- (A) The spreading of a wave as it passes through a gap or past the edge of an obstacle
- (B) The bouncing of a wave from a surface
- (C) The bending of a wave as it crosses a boundary between two media
- (D) The combining of two waves arriving at the same point
Show Answer
Answer: A – Diffraction is the spreading of a wave when it passes through a gap or past an edge. Bouncing from a surface is reflection, bending at a boundary where the speed changes is refraction, and two waves combining is superposition.
Q2. Which of the following types of wave can undergo diffraction?
- (A) Sound waves only
- (B) Light waves only
- (C) All types of waves, including sound, water waves, light and microwaves
- (D) Water waves only
Show Answer
Answer: C – Diffraction is a property of all waves – transverse and longitudinal, mechanical and electromagnetic. What differs from one wave to another is how noticeable the effect is, which depends on the wavelength compared with the size of the gap.
Q3. Plane water waves travel towards a barrier containing a gap that is much wider than the wavelength. What happens to the waves after they pass through the gap?
- (A) They spread out equally in all directions
- (B) They continue mainly in the original direction, with noticeable spreading only near the edges of the gap
- (C) They are completely stopped by the barrier
- (D) Their wavelength doubles
Show Answer
Answer: B – When the gap is much wider than the wavelength, most of the wave passes straight through almost unchanged. Diffraction is only significant at the edges of the beam, so fairly sharp shadow regions form behind the barrier on each side.
Q4. A gap produces the most noticeable diffraction of a wave when the width of the gap is
- (A) very much larger than the wavelength
- (B) exactly one quarter of the wavelength
- (C) equal to the distance travelled by the wave in one second
- (D) approximately equal to the wavelength
Show Answer
Answer: D – The amount of diffraction depends on the ratio of wavelength to gap width. Spreading becomes very significant when the gap width is about the same size as the wavelength.
Q5. Waves pass through a gap that is much narrower than their wavelength. The wavefronts beyond the gap are best described as
- (A) semicircular, spreading out as if the gap were a point source
- (B) plane and parallel, exactly as they were before the gap
- (C) straight lines at right angles to the original direction
- (D) absent, because the wave cannot pass through the gap
Show Answer
Answer: A – A gap that is narrow compared with the wavelength acts like a point source: the wave spreads out beyond it in semicircular wavefronts covering the whole region, although the amplitude is reduced because the energy is spread out.
Q6. In a ripple tank, plane waves of wavelength 2 cm approach a barrier with a gap of width 2 cm. Which statement describes the waves beyond the barrier?
- (A) They remain plane and travel in one direction only
- (B) They travel only along the two edges of the gap
- (C) They curve round and spread into the whole region beyond the barrier
- (D) They stop at the barrier because the gap equals the wavelength
Show Answer
Answer: C – Because the gap width (2 cm) is approximately equal to the wavelength (2 cm), diffraction is strong. The wavefronts emerge curved and the wave reaches parts of the tank that are not in line with the gap.
Q7. A student standing behind a large building can still hear a car horn sounding on the other side of the building. The best explanation is that the sound waves
- (A) are reflected from the ground into the shadow of the building
- (B) diffract around the edges of the building into the region behind it
- (C) travel faster in air than inside the building material
- (D) increase in amplitude behind the building
Show Answer
Answer: B – Sound has a wavelength of the order of a metre, comparable with the size of the building’s edges and openings, so it diffracts noticeably around the building and can be heard in the “shadow” region where no direct sound arrives.
Q8. When a wave passes through a gap and is diffracted, what happens to the speed, frequency and wavelength of the wave?
- (A) The wavelength increases
- (B) The frequency decreases
- (C) The speed changes because the direction changes
- (D) None of them change – only the direction in which the energy travels
Show Answer
Answer: D – Diffraction happens in the same medium at a gap or edge, so the speed is unchanged; the source still sets the frequency, and therefore the wavelength (λ = v/f) is unchanged too. Only the direction of travel and the spread of the energy change.
Q9. Plane waves of wavelength 0.40 m pass through a wide gap in a barrier. In the region directly ahead of the gap, the separation between neighbouring wavefronts beyond the gap is
- (A) 0.40 m, the same as before the gap
- (B) 0.20 m, half the original separation
- (C) 0.80 m, double the original separation
- (D) zero, because the waves have been stopped
Show Answer
Answer: A – The separation of neighbouring wavefronts is the wavelength. Diffraction does not change the wavelength, so the wavefronts beyond the gap are still 0.40 m apart, even where the wavefronts have become curved near the edges.
Q10. Diffraction of light is difficult to observe in everyday situations mainly because
- (A) light travels too fast to be diffracted
- (B) light is a transverse wave, and transverse waves cannot be diffracted
- (C) the wavelength of light is very small (about 10-7 m) compared with the size of everyday gaps and obstacles
- (D) light is completely absorbed whenever it meets an edge
Show Answer
Answer: C – Noticeable diffraction needs a gap or obstacle of about the same size as the wavelength. For light that means gaps of about 10-7 m, far smaller than doors, windows and other everyday openings, so light appears to travel in straight lines with sharp shadows.
Wavelength, Gap Size & the Amount of Diffraction
Q11. Two water waves, of wavelengths 0.20 m and 0.60 m, pass in turn through the same gap of width 0.80 m. Which wave is diffracted more, and why?
- (A) The 0.20 m wave, because it has the higher frequency
- (B) The 0.60 m wave, because its wavelength is closer to the width of the gap
- (C) Both waves are diffracted by exactly the same amount
- (D) Neither wave, because both wavelengths are smaller than the gap
Show Answer
Answer: B – For a fixed gap, the longer wavelength gives the larger ratio of wavelength to gap width, and so the greater spreading. The 0.60 m wavelength is much closer to the 0.80 m gap width than 0.20 m is.
Q12. Plane waves of fixed wavelength approach a barrier. The width of the gap in the barrier is gradually reduced from much wider than the wavelength to about one wavelength wide. The spreading of the waves beyond the gap
- (A) decreases steadily
- (B) stays exactly the same
- (C) stops completely once the gap has been halved
- (D) increases, becoming greatest when the gap is about one wavelength wide
Show Answer
Answer: D – Making the gap smaller increases the ratio of wavelength to gap width, so the diffraction effect grows. It is greatest (most noticeable) when the gap width is approximately equal to the wavelength.
Q13. Sound of frequency 340 Hz travels at 340 m s-1 through an open doorway of width 1.0 m. The wavelength of this sound is 1.0 m. Through this doorway, the sound
- (A) diffracts strongly and spreads throughout the region beyond the doorway
- (B) travels only in a narrow beam straight ahead
- (C) is not transmitted through the doorway at all
- (D) changes to a wavelength of 2.0 m
Show Answer
Answer: A – λ = v/f = 340/340 = 1.0 m, which is approximately equal to the doorway width of 1.0 m. This is the condition for strong diffraction, so the sound spreads widely beyond the doorway rather than travelling in a beam.
Q14. Light of wavelength 5.0 × 10-7 m passes through the same doorway of width 1.0 m as in the previous question. The diffraction of the light is negligible because
- (A) the doorway is narrower than the wavelength of the light
- (B) the light is reflected back by the door frame
- (C) the doorway is about two million times wider than the wavelength, so the spreading is far too small to notice
- (D) the speed of light decreases in the doorway
Show Answer
Answer: C – The ratio of gap width to wavelength is 1.0 / (5.0 × 10-7) = 2 × 106. With such a huge ratio the diffraction spreading is tiny, so the light travels essentially in straight lines and forms a sharp-edged beam through the doorway.
Q15. Water waves of wavelength 20 m approach the entrance to a harbour. The entrance is 25 m wide. Inside the harbour, the waves
- (A) are found only in a narrow strip directly ahead of the entrance
- (B) spread out to reach parts of the harbour that are not in line with the entrance
- (C) change to a wavelength of 25 m
- (D) travel at twice their original speed
Show Answer
Answer: B – The wavelength (20 m) is close to the width of the entrance (25 m), so diffraction at the entrance is strong. The waves curve round and spread across the harbour, reaching areas sheltered from the direct line of the entrance. Their wavelength and speed do not change.
Q16. Two radio signals pass a building 15 m wide. Signal X has a wavelength of 200 m and signal Y has a wavelength of 0.20 m. A receiver placed in the shadow directly behind the building will detect
- (A) signal X much more strongly than signal Y, because its long wavelength diffracts significantly around the building
- (B) signal Y much more strongly, because short wavelengths diffract more
- (C) both signals equally, because diffraction does not depend on wavelength
- (D) neither signal, because radio waves cannot diffract
Show Answer
Answer: A – The 200 m wavelength is much larger than the building, so signal X diffracts strongly around it and reaches the shadow region. The 0.20 m wavelength is tiny compared with the building, so signal Y hardly diffracts and the shadow region receives very little of it.
Q17. When a wave is diffracted at a gap, its energy is spread over a wider region than before. Which statement about the wave after the gap is correct?
- (A) Its frequency increases because the energy has spread out
- (B) Its wavelength increases to fill the wider region
- (C) Its speed increases because it now travels in more directions
- (D) Its speed, frequency and wavelength are unchanged – only the distribution of its energy has changed
Show Answer
Answer: D – Diffraction redistributes the energy of the wave; it does not change the wave itself. The amplitude generally falls where the energy has been spread more thinly, but v, f and λ all stay the same.
Q18. Water waves of wavelength 0.50 m pass through three different gaps in turn: 0.10 m, 0.50 m and 5.0 m. Through which gap do the waves emerge most like waves from a point source?
- (A) The 5.0 m gap
- (B) The 0.50 m gap
- (C) The 0.10 m gap, because it is narrower than the wavelength
- (D) All three gaps give exactly the same pattern
Show Answer
Answer: C – A gap narrower than the wavelength acts as a point source, producing semicircular wavefronts that spread over the whole region beyond the barrier. The 0.50 m gap (equal to λ) also gives strong diffraction, but the wavefronts are not as fully semicircular as for the 0.10 m gap.
Q19. You can hear a person talking in the next room through an open doorway, even when you cannot see them, because
- (A) sound travels around corners by reflection only
- (B) the wavelength of sound is comparable with the width of the doorway, so sound diffracts through it, while the wavelength of light is far smaller, so light hardly diffracts at all
- (C) sound travels much faster than light in air
- (D) the doorway amplifies sound but absorbs light
Show Answer
Answer: B – Sound wavelengths (roughly 0.1 m to a few metres) are similar in size to a doorway, so sound diffracts strongly through it and reaches you round the corner. Light wavelengths (about 10-7 m) are so small that diffraction at the doorway is negligible, so no light from the person reaches you by that route.
Q20. The frequency of a wave passing through a fixed gap is increased, while the speed of the wave stays the same. The amount of diffraction
- (A) increases, because the frequency is higher
- (B) stays the same, because the gap has not changed
- (C) becomes complete in all directions
- (D) decreases, because the wavelength has become shorter
Show Answer
Answer: D – From λ = v/f, increasing the frequency at constant speed makes the wavelength shorter. A shorter wavelength compared with the same gap width means a smaller wavelength-to-gap ratio and therefore less diffraction.
Diffraction in Practice & the Single Slit
Q21. Monochromatic light passes through a single narrow slit and falls on a screen. The diffraction pattern observed on the screen consists of
- (A) a wide, bright central maximum with weaker, narrower subsidiary maxima on either side
- (B) a series of maxima that are all equally bright and equally wide
- (C) one narrow bright line only, with no other maxima
- (D) a dark central band with bright bands on either side
Show Answer
Answer: A – In the single-slit pattern most of the light falls in the central maximum, which is about twice as wide as the subsidiary maxima and much brighter. The subsidiary maxima on each side become progressively weaker further from the centre.
Q22. In the arrangement of the previous question, the slit is made narrower while the same light is used. The central maximum on the screen
- (A) becomes narrower and brighter
- (B) becomes wider and dimmer
- (C) stays at exactly the same width and brightness
- (D) disappears completely
Show Answer
Answer: B – A narrower slit is more comparable with the wavelength, so the light diffracts through a wider angle and the central maximum spreads out. Because the same energy is now spread over a wider band (and less light passes through the slit), the maximum is dimmer.
Q23. A single slit of fixed width is illuminated first with blue light and then with red light. Red light has a longer wavelength than blue light. The central maximum is wider for
- (A) blue light
- (B) neither colour – the width is the same for both
- (C) red light, because longer wavelengths are diffracted more
- (D) whichever light is brighter
Show Answer
Answer: C – For a fixed slit width, the longer wavelength gives the larger wavelength-to-slit-width ratio and so the greater diffraction. Red light therefore produces a wider central maximum than blue light.
Q24. In a hilly area, long-wave (AM) radio broadcasts are usually received more reliably than VHF (FM) broadcasts. This is because
- (A) AM waves travel faster than FM waves
- (B) FM waves are absorbed by the hills
- (C) AM transmitters are always more powerful than FM transmitters
- (D) AM waves have much longer wavelengths (hundreds of metres) than FM waves (a few metres), so they diffract more around hills and buildings
Show Answer
Answer: D – All radio waves travel at the same speed in air. The long wavelength of AM broadcasts is comparable with the size of hills and large buildings, so the waves diffract around them into shadow areas. The few-metre wavelength of FM gives much less diffraction, so reception is easily blocked.
Q25. A beam of electrons passing through a thin crystal produces a pattern of rings, showing that the electrons have been diffracted. This electron diffraction is important evidence that
- (A) electrons behave as waves as well as particles
- (B) electrons carry a positive charge
- (C) electrons have no mass
- (D) crystals are made of waves
Show Answer
Answer: A – Only waves diffract. A diffraction pattern from electrons shows that electrons have wave-like behaviour, which is part of the wave-particle duality developed further in Topic 22 (Quantum Physics).
Q26. Diffraction sets a limit to the detail that a microscope or telescope can distinguish (its resolution) because
- (A) the lenses in these instruments are never perfectly shaped
- (B) the images are always magnified too much
- (C) each point of the object produces a diffraction pattern of finite size, and the patterns from two very close points overlap and merge into one
- (D) the human eye cannot detect small objects
Show Answer
Answer: C – Light from a point object passing through the aperture of the instrument is diffracted, so its image is a small pattern rather than a point. If two objects are very close together, their patterns overlap so much that they cannot be told apart, however good the lenses are.
Q27. A telescope with a larger aperture (a wider opening) can distinguish two close stars better than a smaller telescope, mainly because
- (A) it can collect light for a longer time
- (B) it has a longer focal length
- (C) it is heavier and so vibrates less
- (D) the wider aperture diffracts the light less, so the diffraction pattern of each star is smaller and the two patterns overlap less
Show Answer
Answer: D – Diffraction spreading at an aperture becomes less as the aperture gets wider compared with the wavelength. A large aperture therefore produces a smaller diffraction pattern for each star, so two close stars can still be seen as separate images.
Q28. Radio telescopes have much larger dishes than the mirrors used in optical telescopes. One reason for this is that
- (A) radio waves are weaker, so the dish must be stronger
- (B) radio wavelengths are much longer, giving more diffraction, so a much larger aperture is needed for a comparably sharp image
- (C) radio waves travel more slowly than light
- (D) radio telescopes must be built on the ground
Show Answer
Answer: B – Radio wavelengths are thousands to millions of times longer than light wavelengths, so diffraction at the dish is far greater. Only a very large aperture keeps the diffraction pattern small enough to give a reasonably sharp image. All electromagnetic waves travel at the same speed in a vacuum.
Q29. A ship’s foghorn is designed to produce a low-frequency sound so that the sound can be heard over a wide area and around obstacles such as harbour walls. This works because the low frequency means
- (A) a long wavelength, which diffracts strongly around obstacles
- (B) a greater speed of sound in fog
- (C) a larger amplitude at the source
- (D) the sound is reflected less by the sea
Show Answer
Answer: A – Low frequency means long wavelength (λ = v/f). A long wavelength is comparable with the size of walls and other harbour obstacles, so the sound diffracts strongly around them instead of being blocked in a shadow region.
Q30. Which of the following everyday observations is explained by diffraction?
- (A) An echo heard from a cliff face
- (B) A rainbow seen after rain
- (C) Hearing the sound of a television from another room through a doorway, while none of its light reaches you directly
- (D) The image seen in a plane mirror
Show Answer
Answer: C – The sound diffracts strongly at the doorway because its wavelength is comparable with the doorway width, so it bends round into your room; the light’s wavelength is far too small for noticeable diffraction. The echo and the mirror image are reflection, and the rainbow is caused by refraction and dispersion in raindrops.
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