A Level Physics 8.3 Interference MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 8.3 Interference. Practise superposition and coherence, Young’s double slit experiment and fringe calculations, with the full worked answer shown for every question.
Superposition & Coherence
Q1. Two waves overlap at a point in a medium. According to the principle of superposition, the resultant displacement at that point is
- (A) the product of the two individual displacements
- (B) the sum of the two individual displacements
- (C) always equal to the larger of the two displacements
- (D) zero, because the waves cancel each other out
Show Answer
Answer: B – When two or more waves meet, the resultant displacement at any point is the (vector) sum of the displacements that each wave would produce on its own. The waves then pass through each other unchanged.
Q2. Two sources are described as coherent when they emit waves that have
- (A) the same frequency and a constant phase difference
- (B) the same amplitude and the same speed
- (C) different frequencies and a constant phase difference
- (D) the same frequency and a phase difference that changes randomly
Show Answer
Answer: A – Coherent sources have the same frequency (or wavelength) and maintain a constant phase difference over time. Only then do the maxima and minima of the interference pattern stay in fixed positions.
Q3. Two separate filament lamps placed side by side do not produce a stable interference pattern on a screen. The main reason is that
- (A) the light from the lamps is not monochromatic
- (B) the lamps produce light of different intensities
- (C) each lamp emits light in short bursts, so the phase difference between the two sources changes randomly many times per second
- (D) light from the two lamps travels at different speeds
Show Answer
Answer: C – Light is emitted by individual atoms in random, short-lived bursts. The phase difference between two independent lamps keeps changing randomly, so any interference pattern shifts around far too quickly to be observed and the illumination averages out to be uniform.
Q4. Two identical waves arrive at a point exactly in phase with each other. The result at that point is
- (A) a displacement of zero at all times
- (B) a wave with twice the frequency
- (C) a wave with the same amplitude as one wave alone
- (D) constructive interference, giving a wave whose amplitude equals the sum of the two amplitudes
Show Answer
Answer: D – Waves arriving in phase reinforce each other: crests meet crests and troughs meet troughs. The resultant amplitude is the sum of the individual amplitudes – for two identical waves, double the amplitude.
Q5. In a two-source interference pattern, a maximum (bright fringe) is produced at points where the path difference between the two waves is
- (A) an odd number of wavelengths
- (B) a whole number of wavelengths, nλ
- (C) always exactly half a wavelength
- (D) an odd number of half wavelengths
Show Answer
Answer: B – A path difference of nλ (n = 0, 1, 2, …) means the waves arrive in phase, so they interfere constructively and produce a maximum.
Q6. At a point in an interference pattern the path difference is 2.5λ. The interference at this point is
- (A) destructive, producing a minimum
- (B) constructive, producing a maximum
- (C) neither constructive nor destructive
- (D) constructive, because 2.5 is greater than 2
Show Answer
Answer: A – 2.5λ = (2 + ½)λ, an odd number of half wavelengths. The waves arrive in antiphase (180° out of phase), so they cancel and a minimum is produced.
Q7. Two waves meet at a point with a path difference of 0.75λ. The phase difference between the waves at that point is
- (A) 90°
- (B) 135°
- (C) 270°
- (D) 360°
Show Answer
Answer: C – Phase difference = (path difference / λ) × 360° = 0.75 × 360° = 270°. One full wavelength of path difference corresponds to a phase difference of 360°.
Q8. Light of wavelength 600 nm from two coherent sources reaches a point with a path difference of 300 nm. The phase difference and the type of interference are
- (A) 90°, constructive
- (B) 360°, constructive
- (C) 180°, constructive
- (D) 180°, destructive
Show Answer
Answer: D – Path difference / λ = 300 / 600 = 0.5, so the phase difference is 0.5 × 360° = 180°. The waves arrive in antiphase and cancel, giving destructive interference.
Q9. Complete destructive interference between two waves of equal amplitude occurs when the waves arrive with a phase difference of
- (A) 180°, so a crest of one meets a trough of the other
- (B) 0°, so the waves are exactly in phase
- (C) 360°, so the waves have completed a full cycle
- (D) 90°, so the waves are a quarter of a cycle apart
Show Answer
Answer: A – A phase difference of 180° means the waves are in antiphase: the positive displacement of one wave is exactly cancelled by the equal negative displacement of the other, at every instant.
Q10. Coherent sources are essential for an observable interference pattern because, without coherence,
- (A) the waves would not overlap in the same region
- (B) the waves would travel too slowly to meet
- (C) the positions of the maxima and minima keep changing, so the pattern averages out to uniform illumination
- (D) the waves would have different amplitudes
Show Answer
Answer: C – If the phase difference between the sources changes randomly, each point on the screen is a maximum at one instant and a minimum a fraction of a second later. The eye or detector averages this out and no fringes are seen.
Young’s Double Slit Experiment
Q11. In Young’s double slit experiment, light from a single source first passes through a single slit before reaching the double slits. The purpose of the single slit is to
- (A) make the light monochromatic
- (B) ensure the light arriving at the two slits comes from the same wavefront, so the two slits act as coherent sources
- (C) increase the intensity of the light
- (D) polarise the light before it reaches the double slits
Show Answer
Answer: B – Both slits are illuminated by the same original wavefront, so any phase change at the source affects both slits equally. The phase difference between the two slits stays constant, making them coherent sources.
Q12. The interference pattern observed on the screen in Young’s experiment consists of
- (A) a single bright spot at the centre of the screen
- (B) two bright bands, one behind each slit
- (C) equally spaced bright and dark fringes parallel to the slits, with the central bright fringe the brightest
- (D) coloured fringes that get wider further from the centre
Show Answer
Answer: C – The diffracted waves from the two slits overlap and interfere, producing a series of parallel, equally spaced bright and dark fringes. The central maximum (zero path difference) is the brightest.
Q13. The fringe width x (separation of adjacent bright fringes) in Young’s experiment is given by
- (A) x = λD / a
- (B) x = λa / D
- (C) x = aD / λ
- (D) x = D / (λa)
Show Answer
Answer: A – x = λD / a, where λ is the wavelength, D is the distance from the slits to the screen and a is the separation of the two slits. Fringes are wider for longer wavelengths and larger D, and narrower for wider slit separations.
Q14. In Young’s experiment the screen is moved further away from the double slits, increasing D. The fringes on the screen become
- (A) closer together
- (B) unchanged in separation
- (C) fewer in number but the same width
- (D) further apart (wider fringe width)
Show Answer
Answer: D – x = λD / a, so the fringe width is directly proportional to D. A larger slit-to-screen distance spreads the fringes further apart (although each fringe becomes less intense).
Q15. The separation a of the two slits is reduced, with λ and D unchanged. The fringe width
- (A) decreases
- (B) increases
- (C) stays the same
- (D) becomes zero
Show Answer
Answer: B – x = λD / a: the fringe width is inversely proportional to the slit separation. Bringing the slits closer together makes the pattern spread out, which is why the slits must be close (typically less than 1 mm apart).
Q16. A Young’s experiment is first performed with blue light and then repeated with red light of longer wavelength, all else unchanged. With red light the fringes are
- (A) further apart
- (B) closer together
- (C) the same width
- (D) no longer visible
Show Answer
Answer: A – x = λD / a is proportional to λ. Red light has a longer wavelength than blue light, so the red fringes are more widely spaced.
Q17. The two slits in Young’s experiment must be narrow so that
- (A) only one wavelength of light can pass through each slit
- (B) the light is not absorbed by the slit material
- (C) the light diffracts strongly at each slit, so the two diffracted beams spread out and overlap over a wide region
- (D) the two beams have exactly equal path lengths to the screen
Show Answer
Answer: C – A narrow slit causes significant diffraction, spreading the light from each slit over a wide angle. The two spread-out beams then overlap across the screen, which is the region where interference fringes can form.
Q18. A student measures the wavelength of laser light using a double slit. To reduce the percentage uncertainty in the fringe width, the best method is to
- (A) measure the width of a single fringe with a ruler
- (B) measure across as many fringes as possible and divide the total width by the number of fringe separations counted
- (C) move the screen as close to the slits as possible
- (D) use a brighter laser
Show Answer
Answer: B – Measuring across, say, 10 fringes and dividing by 10 spreads the fixed ruler uncertainty (about ±1 mm) over a much larger total width, reducing the percentage uncertainty in x – and therefore in λ = ax / D.
Q19. If the slit separation a in Young’s experiment is made much larger, the interference pattern becomes difficult to observe because
- (A) the fringes disappear completely
- (B) the fringe width becomes very small, so the fringes are too close together to resolve
- (C) the light from the two slits is no longer coherent
- (D) the central maximum moves off the screen
Show Answer
Answer: B – x = λD / a: a large a gives a very small fringe width. The fringes are still formed, but they are packed too closely together for the eye to separate them.
Q20. At the central maximum of a Young’s pattern, the path difference between the two waves is zero. This point is
- (A) always a bright fringe, for any wavelength
- (B) a dark fringe, because the waves arrive in antiphase
- (C) bright for red light but dark for blue light
- (D) only bright if the two slits have the same width
Show Answer
Answer: A – Zero path difference means the waves always arrive exactly in phase, whatever the wavelength, so the centre of the pattern is always a bright maximum (the zeroth-order maximum).
Interference Calculations
Q21. Monochromatic light of wavelength 600 nm is incident on two slits separated by 0.50 mm. The screen is 1.5 m from the slits. The fringe width is
- (A) 0.18 mm
- (B) 1.8 mm
- (C) 18 mm
- (D) 0.020 mm
Show Answer
Answer: B – x = λD / a = (600 × 10−9 × 1.5) / (0.50 × 10−3) = (9.0 × 10−7) / (5.0 × 10−4) = 1.8 × 10−3 m = 1.8 mm. (Option A comes from forgetting to convert mm to m for a.)
Q22. Light of wavelength 500 nm passes through slits 0.25 mm apart, and fringes are viewed on a screen 2.0 m away. The separation of adjacent bright fringes is
- (A) 0.40 mm
- (B) 1.0 mm
- (C) 2.0 mm
- (D) 4.0 mm
Show Answer
Answer: D – x = λD / a = (500 × 10−9 × 2.0) / (0.25 × 10−3) = (1.0 × 10−6) / (2.5 × 10−4) = 4.0 × 10−3 m = 4.0 mm.
Q23. In a Young’s experiment the fringe width is 2.0 mm. Approximately how many bright fringes appear across a 20 mm wide region centred on the central maximum?
- (A) 10
- (B) 20
- (C) 40
- (D) 5
Show Answer
Answer: A – Number of fringe separations = 20 mm / 2.0 mm = 10, so about 10 bright fringes (one at each multiple of the fringe width, including the central maximum) lie within the 20 mm region.
Q24. At a point P on the screen in a double slit experiment, the path difference between the two waves is 3.5λ. Point P is
- (A) a bright fringe (maximum)
- (B) halfway between a bright and a dark fringe
- (C) a dark fringe (minimum)
- (D) the central maximum
Show Answer
Answer: C – 3.5λ = (3 + ½)λ, an odd number of half wavelengths, so the waves arrive in antiphase and cancel: P is a dark fringe.
Q25. Two loudspeakers connected to the same signal generator emit sound of wavelength 0.80 m. A listener stands at a point where the sound from one speaker travels 3.6 m and from the other 2.4 m. The listener hears
- (A) a loud sound, because the path difference is a whole number of wavelengths
- (B) a quiet sound, because the path difference is 1.5λ
- (C) a loud sound, because both speakers have the same frequency
- (D) no sound at all from either speaker
Show Answer
Answer: B – Path difference = 3.6 − 2.4 = 1.2 m. 1.2 / 0.80 = 1.5, so the path difference is 1.5λ = (1 + ½)λ. The waves arrive in antiphase and interfere destructively, so the sound is quiet (a minimum).
Q26. In a microwave interference experiment, two coherent sources emit microwaves of wavelength 3.0 cm. A receiver is placed where the path difference from the two sources is 4.5 cm. The detected signal is
- (A) a minimum, because the path difference is 1.5λ
- (B) a maximum, because the path difference is 1.5λ
- (C) a maximum, because 4.5 cm is a whole number of centimetres
- (D) zero path difference, so no interference occurs
Show Answer
Answer: A – 4.5 / 3.0 = 1.5, so the path difference is 1.5λ, an odd number of half wavelengths. The microwaves arrive in antiphase, giving a minimum (weak) signal.
Q27. Each of two coherent waves alone produces an intensity I at a point. At a maximum of their interference pattern, the total intensity is
- (A) I
- (B) 2I
- (C) zero
- (D) 4I
Show Answer
Answer: D – Intensity is proportional to (amplitude)2. At a maximum the amplitudes add: amplitude = A + A = 2A, so the intensity is (2A)2 / A2 = 4 times the intensity of one wave alone, i.e. 4I. (Energy is conserved overall because the minima have zero intensity.)
Q28. In a double slit experiment using light of wavelength 590 nm, 10 fringe separations span 18 mm on a screen 1.2 m from the slits. The slit separation is
- (A) 0.039 mm
- (B) 3.9 mm
- (C) 0.39 mm
- (D) 39 mm
Show Answer
Answer: C – Fringe width x = 18 mm / 10 = 1.8 mm = 1.8 × 10−3 m. Rearranging x = λD / a gives a = λD / x = (590 × 10−9 × 1.2) / (1.8 × 10−3) = (7.08 × 10−7) / (1.8 × 10−3) = 3.9 × 10−4 m = 0.39 mm.
Q29. Two loudspeakers, driven in phase by one generator, emit sound of frequency 340 Hz. The speed of sound is 340 m s−1. A listener at a point where the path difference is 2.5 m hears
- (A) a maximum, because the path difference is 2.5 wavelengths
- (B) a maximum, because the path difference is 5 wavelengths
- (C) a minimum, because the path difference is 2.5 wavelengths
- (D) a minimum, because the path difference is 1.5 wavelengths
Show Answer
Answer: C – λ = v / f = 340 / 340 = 1.0 m. Path difference = 2.5 m = 2.5λ = (2 + ½)λ, an odd number of half wavelengths, so the waves arrive in antiphase and the listener hears a minimum (quiet point).
Q30. In a ripple tank, two dippers connected to the same vibrator produce water waves of wavelength 2.0 cm. At a point where the distances from the two dippers differ by 5.0 cm, the water surface
- (A) has double the usual amplitude
- (B) is a point of maximum disturbance
- (C) shows waves of wavelength 5.0 cm
- (D) is on a nodal line, with almost no disturbance
Show Answer
Answer: D – Path difference = 5.0 cm = 5.0 / 2.0 = 2.5λ, an odd number of half wavelengths. The waves arrive in antiphase and cancel, so the point lies on a nodal line where the water stays almost still.
Keep Practising
Work through more A Level topics and every other board from the Physics MCQs hub, and keep the key interference equations beside you in the A Level Physics Formula Sheet.
All questions are original practice questions written for QuaidianPoint, in the style of the Cambridge 9702 syllabus – they are not Cambridge past-paper questions. QuaidianPoint is not affiliated with Cambridge Assessment International Education.