OxfordAQA International AS and A-level Physics: Interference and diffraction: double slits and gratings (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.5.6 (Interference) and 3.5.7 (Diffraction) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. The content is shared International AS and A-level. Use these data where needed: speed of electromagnetic waves in air 3.00 × 10⁸ m s⁻¹; speed of sound in air 340 m s⁻¹.

Read the study guide first if needed, and use the revision notes for recall. Course pages: physics hub and checklist. For a quick check of where you stand, try the free 10-minute diagnostics.

Questions

1.

(a) State what is meant by coherent sources. [2] (b) State the path difference, in terms of wavelength, that gives destructive interference between two sources that are in phase. [1]

2. Two loudspeakers are connected to one signal generator set to 850 Hz.

(a) Calculate the wavelength of the sound. [1] (b) A microphone is 4.30 m from one loudspeaker and 5.10 m from the other. State and explain whether the sound is loud or quiet there. [2] (c) At a second position the distances are 3.00 m and 3.60 m. State and explain whether the sound is loud or quiet. [2]

3. A laser of wavelength 610 nm illuminates two slits 0.25 mm apart. The screen is 3.0 m from the slits.

(a) Calculate the fringe spacing. [2] (b) Calculate the distance from the central fringe to the fourth bright fringe. [2]

4. A student uses a laser to demonstrate Young’s slits.

(a) Give two safety precautions. [2] (b) Give two reasons a laser is more suitable than a filament lamp. [2]

5. A microwave transmitter has frequency 10.7 GHz and is placed in front of a metal plate with two slits.

(a) Calculate the wavelength. [2] (b) A detector is 0.620 m from one slit and 0.676 m from the other. Show that it records a maximum. [2] (c) State the smallest path difference that would give a minimum. [1]

6. In a Young’s slits experiment the slit separation is 0.45 mm and the screen is 2.10 m away. The distance across ten fringe spacings is 31.5 mm.

(a) Calculate the wavelength of the light. [3] (b) Explain why ten spacings are measured rather than one. [1] (c) The ruler reading is uncertain by ±1 mm. Calculate the percentage uncertainty in the ten-spacing distance and in a single spacing, and comment. [2]

7. Slits 0.60 mm apart are lit with white light. The screen is 1.20 m away. Red light has wavelength 680 nm and blue light 450 nm.

(a) Describe the pattern seen. [2] (b) Calculate the fringe spacing for red and for blue. [2] (c) Calculate the separation of the second-order red and blue maxima. [2]

8. A single narrow slit is placed in a light beam, with a screen behind it.

(a) Describe the pattern for monochromatic light. [2] (b) State how the width of the central maximum changes when the slit is made narrower. [1] (c) State how it changes when red light is replaced with blue light. [1] (d) Describe the pattern for white light. [2]

9. Derive d sin θ = nλ for a plane transmission grating at normal incidence. [4]

10. Violet light of wavelength 405 nm is incident normally on a grating of 800 lines per mm.

(a) Calculate the grating spacing. [1] (b) Find the angle of each of the first two maxima either side of the centre. [3] (c) Find the highest order and the total number of maxima. [2] (d) A screen is 1.50 m from the grating. Calculate the distance of the first order maximum from the centre. [2]

11. A grating has 3000 lines per cm. Light of unknown wavelength gives a third-order maximum at 31.4°.

(a) Calculate the wavelength. [3] (b) State the highest order observed. [2] (c) Calculate the angle of the fourth-order maximum. [1]

12. White light (400 nm to 700 nm) is incident normally on a grating of 450 lines per mm.

(a) Calculate the grating spacing. [1] (b) Calculate the angles of the first-order maxima for 400 nm and 700 nm. [2] (c) Show that the third-order violet overlaps the second-order spectrum. [3] (d) State one application of a diffraction grating. [1]

Answers

1. (a) Same frequency [1] and a constant phase difference [1]. (b) An odd number of half wavelengths, (n + ½)λ [1]. [3] Examiner insight: “Same wavelength” alone is incomplete; the answer must say that the phase difference stays constant.

2. (a) λ = v ÷ f = 340 ÷ 850 = 0.40 m [1] (b) Path difference = 5.10 − 4.30 = 0.80 m = 2λ [1]. The waves arrive in phase, so the sound is loud [1]. (c) Path difference = 0.60 m = 1.5λ [1]. The waves arrive in antiphase, so the sound is quiet [1]. [5] Examiner insight: A loud or quiet answer gains little without the path difference expressed as a number of wavelengths.

3. (a) w = λD ÷ s = 610 × 10⁻⁹ × 3.0 ÷ 0.25 × 10⁻³ [1] = 7.3 × 10⁻³ m [1] (b) 4w = 4 × 7.32 mm [1] = 29 mm [1]. [4] Examiner insight: Carry the unrounded spacing (7.32 mm) into part (b) and round only the final answer.

4. (a) Any two from: never look into the beam or at reflections; use a low-power laser; stop the beam with a matt block; keep the beam away from eye level [1] [1]. (b) It is monochromatic, so the fringes are distinct and the spacing relates to one wavelength [1]. It is bright, narrow and coherent across its width, so a clear pattern forms without extra slits or filters [1]. [4] Examiner insight: “Be careful” is too vague for a safety mark; name the specific action.

5. (a) λ = c ÷ f = 3.00 × 10⁸ ÷ 10.7 × 10⁹ [1] = 0.0280 m [1] (b) Path difference = 0.676 − 0.620 = 0.056 m [1] = 2λ, a whole number of wavelengths, so a maximum [1]. (c) λ ÷ 2 = 0.014 m [1] [5] Examiner insight: A “show that” answer needs the path difference divided by the wavelength, giving the whole number.

6. (a) w = 31.5 ÷ 10 = 3.15 mm [1]. λ = ws ÷ D = 3.15 × 10⁻³ × 0.45 × 10⁻³ ÷ 2.10 [1] = 6.75 × 10⁻⁷ m [1] (b) The ruler uncertainty is shared over ten spacings, so the percentage uncertainty is smaller [1]. (c) Ten spacings: 1 ÷ 31.5 = 3.2% [1]. One spacing: 1 ÷ 3.15 = 32%, ten times larger, so measuring across ten spacings is far better [1]. [6] Examiner insight: Comment questions need a comparison of the two figures, not just two numbers.

7. (a) The central fringe is white, because all colours have zero path difference there [1]. On either side the fringes are coloured, with blue nearer the centre and red further out [1]. (b) Red: 680 × 10⁻⁹ × 1.20 ÷ 0.60 × 10⁻³ = 1.36 mm [1]. Blue: 450 × 10⁻⁹ × 1.20 ÷ 0.60 × 10⁻³ = 0.90 mm [1]. (c) Second-order separation = 2 × (1.36 − 0.90) [1] = 0.92 mm [1]. [6] Examiner insight: Explain why the centre is white (zero path difference for every colour), not just that it is white.

8. (a) A bright, wide central maximum [1], with narrower, dimmer maxima either side separated by dark minima [1]. (b) It becomes wider [1]. (c) It becomes narrower [1]. (d) A white central maximum [1] with coloured side maxima, blue at the inner edge and red at the outer edge [1]. [6] Examiner insight: The direction of the change matters; “changes” earns nothing.

9. A plane wave arrives normally, so the light is in phase at every slit [1]. For waves from adjacent slits (spacing d) leaving at angle θ, the path difference is d sin θ [1]. Maxima need this to be a whole number of wavelengths, nλ [1], so that waves from all slits reinforce: d sin θ = nλ [1]. [4] Examiner insight: Draw the right-angled triangle with hypotenuse d and mark θ; the sin θ step is where credit is gained.

10. (a) d = 1 ÷ 800 000 = 1.25 × 10⁻⁶ m [1] (b) n = 1: sin θ = 405 × 10⁻⁹ ÷ 1.25 × 10⁻⁶ = 0.324 [1], θ = 18.9° [1]. n = 2: sin θ = 0.648, θ = 40.4° [1]. (c) d ÷ λ = 3.09, so highest order is 3 [1]. Maxima at 0, ±1, ±2, ±3: 7 [1]. (d) x = 1.50 × tan 18.9° [1] = 0.51 m [1]. [8] Examiner insight: Use sin in the grating equation but tan for position on a screen; mixing them is a frequent error.

11. (a) d = 1 ÷ 300 000 = 3.33 × 10⁻⁶ m [1]. λ = d sin θ ÷ n = 3.33 × 10⁻⁶ × sin 31.4° ÷ 3 [1] = 5.79 × 10⁻⁷ m [1] (b) d ÷ λ = 5.76 [1], so the highest order is 5 [1]. (c) sin θ = 4 × 579 × 10⁻⁹ ÷ 3.33 × 10⁻⁶ = 0.695, θ = 44.0° [1] [6] Examiner insight: Convert lines per cm to metres carefully; a factor of 100 slip changes every later answer.

12. (a) d = 1 ÷ 450 000 = 2.22 × 10⁻⁶ m [1] (b) 400 nm: sin θ = 0.18, θ = 10.4°. 700 nm: sin θ = 0.315, θ = 18.4° [1] [1] (c) Third-order violet: sin θ = 3 × 400 × 10⁻⁹ ÷ 2.22 × 10⁻⁶ = 0.54, θ = 32.7° [1]. Second-order spectrum runs from 21.1° (violet) [1] to 39.1° (red). 32.7° lies within this range, so the orders overlap [1]. (d) For example, measuring wavelengths of light from a lamp or star to identify elements [1]. [7] Examiner insight: “Show that” overlap answers need the angle ranges of both spectra, not only one value.

Where marks are usually lost

• Not stating the path difference as a number of wavelengths when explaining loud or quiet regions.
• Leaving out “constant phase difference” in a definition of coherence.
• Using lines per millimetre or per centimetre as d without conversion.
• Dividing by the number of fringes instead of the number of spacings.
• Using tan in the grating equation or sin for the screen position.
• Getting the direction wrong for slit width or wavelength and central maximum width.
• Giving vague laser safety such as “be careful”.
• Writing a non-whole highest order, or counting maxima on one side only.

Next steps

• Revisit the revision notes and the study guide.
• Course pages: physics hub and checklist.
• Take the free 10-minute diagnostics to find weak topics.
• Book a free trial class for help with anything unclear.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), sections 3.5.6 and 3.5.7, Version 4.4 (for exams May/June 2020 onwards), published by OxfordAQA.