A Level Physics 8.4 Diffraction Grating MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice questions for Cambridge International AS Level Physics (9702), Section 8.4, The Diffraction Grating. All 30 questions are original practice questions with worked answers, covering the grating equation d sin θ = nλ, the orders of the pattern, and how a grating is used with white light and in a spectrometer.
Formula recap. Slit separation: d = 1 / N, where N is the number of lines per metre. Grating equation: d sin θ = nλ, where n is the order (0, 1, 2, …) and θ is measured from the normal (the zero-order direction). The highest possible order is the largest whole number with nλ ≤ d, because sin θ cannot exceed 1. See the full A Level Physics Formula Sheet.
The Grating Equation
Q1. What is a diffraction grating?
- (A) A single very narrow slit cut in an opaque screen
- (B) A plate carrying a very large number of equally spaced parallel slits (or lines) close together
- (C) Two slits separated by a distance of about 1 mm
- (D) A glass prism used to split white light into colours
Show Answer
Answer: B – A grating has hundreds of lines per millimetre, i.e. thousands to millions of slits in total, all equally spaced. Light passing through all of them interferes to give sharp bright maxima at angles given by d sin θ = nλ.
Q2. A diffraction grating has 300 lines per millimetre. What is the slit separation d?
- (A) 3.33 × 10-3 m
- (B) 3.33 × 10-9 m
- (C) 3.33 × 10-6 m
- (D) 3.0 × 10-5 m
Show Answer
Answer: C – 300 lines/mm = 300 × 103 = 3.00 × 105 lines per metre. d = 1 / N = 1 / (3.00 × 105) = 3.33 × 10-6 m. Option A forgets to convert millimetres to metres.
Q3. A grating is marked “600 lines / mm”. What is the slit separation d?
- (A) 1.67 × 10-6 m
- (B) 1.67 × 10-3 m
- (C) 6.00 × 10-6 m
- (D) 1.67 × 10-9 m
Show Answer
Answer: A – N = 600 × 103 = 6.00 × 105 lines per metre, so d = 1 / (6.00 × 105) = 1.67 × 10-6 m. More lines per millimetre always means a smaller slit separation.
Q4. In the grating equation d sin θ = nλ, the angle θ is the angle between the diffracted beam and
- (A) the surface of the grating
- (B) the normal to the grating, which is the direction of the zero-order beam
- (C) the first-order maximum
- (D) the edge of the grating
Show Answer
Answer: B – Angles in the grating equation are always measured from the normal to the grating (the straight-through, zero-order direction for light arriving at normal incidence), never from the grating surface.
Q5. White light is shone through a diffraction grating. The zero-order maximum is white because
- (A) the grating only works with monochromatic light, so white light is not split up at all
- (B) the central maximum contains no red or violet light
- (C) the colours in the central maximum are too close together to be seen separately
- (D) for the zero order, θ = 0 for every wavelength, so all colours fall at the same position and add up to white
Show Answer
Answer: D – With n = 0, d sin θ = 0 whatever the wavelength, so every colour is diffracted to θ = 0. All colours overlap exactly, giving a white central maximum. Splitting into spectra happens only in orders 1, 2, 3, …
Q6. Light of wavelength 500 nm passes through a grating with 400 lines per millimetre (d = 2.5 × 10-6 m). At what angle is the first-order maximum?
- (A) 11.5°
- (B) 0.20°
- (C) 23.6°
- (D) 5.7°
Show Answer
Answer: A – sin θ = nλ / d = (1 × 500 × 10-9) / (2.5 × 10-6) = 0.200, so θ = sin-1(0.200) = 11.5°. Option B is the value of sin θ itself, not the angle; option C is the second-order angle.
Q7. At a third-order maximum of a grating pattern, the path difference between light coming from two adjacent slits is
- (A) λ / 3
- (B) λ
- (C) 3λ
- (D) 9λ
Show Answer
Answer: C – The path difference between adjacent slits is d sin θ, and at a maximum this equals a whole number of wavelengths: d sin θ = nλ. For the third order (n = 3) the path difference is 3λ, which is why the waves arrive in phase and reinforce.
Q8. For light of a given wavelength, which change increases the angle of the first-order maximum?
- (A) Using light of a shorter wavelength instead
- (B) Using a grating with fewer lines per millimetre
- (C) Using a grating with a larger slit separation d
- (D) Using a grating with more lines per millimetre, so that d is smaller
Show Answer
Answer: D – sin θ = λ / d for the first order. A smaller d gives a larger sin θ and so a larger angle. More lines per millimetre means the lines are closer together, i.e. d is smaller – this spreads the pattern out further.
Q9. A grating is ruled with 5.0 × 105 lines per metre. What is the slit separation d?
- (A) 5.0 × 10-6 m
- (B) 2.0 × 10-6 m
- (C) 2.0 × 10-3 m
- (D) 5.0 × 105 m
Show Answer
Answer: B – d = 1 / N = 1 / (5.0 × 105) = 2.0 × 10-6 m. (This is a 500 lines/mm grating.)
Q10. Monochromatic light falls normally on a grating of slit separation d. For light diffracted at angle θ, the path difference between rays from two adjacent slits is
- (A) d / sin θ
- (B) d cos θ
- (C) d sin θ
- (D) d tan θ
Show Answer
Answer: C – From the geometry of the two parallel rays leaving adjacent slits at angle θ, the extra path travelled by the ray from the further slit is d sin θ. A bright maximum occurs when this equals a whole number of wavelengths, giving the grating equation d sin θ = nλ.
Orders & Angle Calculations
Q11. A grating with 500 lines per millimetre (d = 2.0 × 10-6 m) is used with light of wavelength 600 nm. The first-order maximum is at angle
- (A) 8.6°
- (B) 64.2°
- (C) 17.5°
- (D) 36.9°
Show Answer
Answer: C – sin θ = λ / d = (600 × 10-9) / (2.0 × 10-6) = 0.300, so θ = sin-1(0.300) = 17.5°. Option D is the second-order angle and option B the third-order angle.
Q12. For the same grating and light as Q11 (500 lines/mm, λ = 600 nm), the second-order maximum is at angle
- (A) 17.5°
- (B) 36.9°
- (C) 64.2°
- (D) 73.7°
Show Answer
Answer: B – sin θ = 2λ / d = 2 × 0.300 = 0.600, so θ = sin-1(0.600) = 36.9°. Note that the angle is not simply double the first-order angle (2 × 17.5° = 35.0°), because it is sin θ, not θ, that is proportional to n.
Q13. For the grating and light of Q11 and Q12 (d = 2.0 × 10-6 m, λ = 600 nm), no fourth-order maximum is observed. This is because
- (A) the grating is not wide enough to produce a fourth order
- (B) the fourth order is too faint to be seen
- (C) the fourth order falls at exactly the same angle as the third order
- (D) the fourth order would require sin θ = 1.20, which is impossible because sin θ can never exceed 1
Show Answer
Answer: D – For n = 4, sin θ = nλ / d = (4 × 600 × 10-9) / (2.0 × 10-6) = 1.20. No angle has a sine greater than 1, so this order simply does not exist. The third order (sin θ = 0.900, θ = 64.2°) is the last one.
Q14. A grating with 600 lines per millimetre (d = 1.67 × 10-6 m) is illuminated with light of wavelength 450 nm. What is the highest order of maximum that can be seen?
- (A) 3
- (B) 4
- (C) 7
- (D) 2
Show Answer
Answer: A – The largest possible order has nλ ≤ d, so nmax = d / λ = (1.67 × 10-6) / (450 × 10-9) = 3.7. The highest whole order is therefore 3 (the 4th order would need sin θ = 4 / 3.7 > 1).
Q15. A grating with 200 lines per millimetre (d = 5.0 × 10-6 m) is used with light of wavelength 600 nm. How many bright maxima are there in the whole pattern?
- (A) 8
- (B) 9
- (C) 17
- (D) 16
Show Answer
Answer: C – nmax = d / λ = (5.0 × 10-6) / (600 × 10-9) = 8.33, so the highest order is 8. Maxima appear at orders 0, ±1, ±2, …, ±8, giving 2 × 8 + 1 = 17 maxima in total.
Q16. Sodium light of wavelength 589 nm is passed through a grating with 300 lines per millimetre (d = 3.33 × 10-6 m). The first-order maximum is at angle
- (A) 10.2°
- (B) 20.7°
- (C) 5.1°
- (D) 0.18°
Show Answer
Answer: A – sin θ = λ / d = (589 × 10-9) / (3.33 × 10-6) = 0.177, so θ = sin-1(0.177) = 10.2°. Option B is the second-order angle (sin θ = 0.353).
Q17. In a spectrometer experiment, the first-order maximum for a spectral line is found at 15.0° to the zero order, using a grating with 400 lines per millimetre (d = 2.5 × 10-6 m). The wavelength of the line is
- (A) 259 nm
- (B) 647 nm
- (C) 517 nm
- (D) 1290 nm
Show Answer
Answer: B – λ = d sin θ / n = (2.5 × 10-6 × sin 15.0°) / 1 = 2.5 × 10-6 × 0.259 = 6.47 × 10-7 m = 647 nm.
Q18. Light of wavelength 450 nm passes through a grating with 600 lines per millimetre (d = 1.67 × 10-6 m). The second-order maximum is at angle
- (A) 15.7°
- (B) 0.54°
- (C) 65.4°
- (D) 32.7°
Show Answer
Answer: D – sin θ = 2λ / d = (2 × 450 × 10-9) / (1.67 × 10-6) = 0.540, so θ = sin-1(0.540) = 32.7°. Option A is the first-order angle (sin θ = 0.270).
Q19. A grating has a slit separation d = 1.2 × 10-6 m. When it is used with red light of wavelength 650 nm, the first order is seen but no second-order maximum exists. This is because
- (A) the second order falls outside the range of the screen
- (B) red light cannot be diffracted more than once
- (C) the grating has too few lines ruled on it
- (D) d is less than 2λ, so the second order would need sin θ = 2λ / d = 1.08, which is greater than 1
Show Answer
Answer: D – sin θ for n = 2 is (2 × 650 × 10-9) / (1.2 × 10-6) = 1.08 > 1, so no second order can be formed. In general, whenever d < 2λ only the zero and first orders exist. (The first order itself is at sin θ = 0.542, θ = 32.8°.)
Q20. A grating with 100 lines per millimetre (d = 1.0 × 10-5 m) is used with light of wavelength 500 nm. The total number of bright maxima in the pattern is
- (A) 20
- (B) 41
- (C) 21
- (D) 40
Show Answer
Answer: B – nmax = d / λ = (1.0 × 10-5) / (500 × 10-9) = 20. Counting both sides plus the zero order: 2 × 20 + 1 = 41 maxima. A coarse grating (large d) produces many orders close together.
Using a Grating & White Light
Q21. White light is shone through a diffraction grating. The appearance of the first-order maximum is
- (A) a single white band, just like the zero order
- (B) a spectrum, with violet nearest to the zero order and red furthest from it
- (C) a spectrum, with red nearest to the zero order and violet furthest from it
- (D) a series of dark fringes with no colour
Show Answer
Answer: B – For n = 1, sin θ = λ / d, so the shortest wavelength (violet, about 400 nm) is diffracted through the smallest angle and the longest (red, about 700 nm) through the largest angle. Each order (except the zero order) is therefore a full spectrum, violet on the inside, red on the outside.
Q22. Within any one order of a white-light grating spectrum, the colour deviated (diffracted) through the largest angle is
- (A) violet
- (B) green
- (C) yellow
- (D) red
Show Answer
Answer: D – sin θ = nλ / d. In a given order, n and d are fixed, so the largest θ belongs to the largest λ. Red light has the longest wavelength in the visible spectrum (about 700 nm), so it is deviated most.
Q23. With white light, the red part of the second-order spectrum (λ = 700 nm) appears at a larger angle than the violet part of the third-order spectrum (λ = 400 nm), so the two orders overlap. This happens because
- (A) the value of nλ for second-order red (2 × 700 = 1400 nm) is greater than for third-order violet (3 × 400 = 1200 nm)
- (B) red light travels faster than violet light through the air
- (C) the slit separation d is different for different colours
- (D) violet light is partly absorbed by the grating
Show Answer
Answer: A – The angle depends on the product nλ through sin θ = nλ / d. Since 1400 nm > 1200 nm, second-order red has the larger sin θ and so the larger angle. At high orders the spectra of neighbouring orders increasingly overlap in this way.
Q24. Compared with a double slit having the same slit separation, a diffraction grating produces maxima that are
- (A) wider and fainter
- (B) at much smaller angles
- (C) much sharper (narrower) and brighter, so the angle, and hence the wavelength, can be measured more accurately
- (D) in completely different positions, because the value of d is different
Show Answer
Answer: C – The maxima are in the same positions (same d), but with thousands of slits interfering they become very narrow and intense. A sharp maximum can be located far more precisely than a broad double-slit fringe, which is why a grating measures wavelengths much more accurately.
Q25. The principal maxima of a diffraction grating are very narrow. The best explanation is that
- (A) moving only slightly away from a maximum angle puts the waves from the very many slits out of step with each other, so they cancel and the intensity falls almost to zero within a very small angle
- (B) each slit is extremely narrow, so each slit spreads the light out only a little
- (C) the grating reflects most of the light back towards the source
- (D) only one single wavelength can pass through the grating at each angle
Show Answer
Answer: A – At an exact maximum, waves from all the slits arrive in phase. A tiny change of angle destroys this agreement across thousands of slits, giving near-complete cancellation. The result is an intense maximum confined to a very narrow range of angles.
Q26. Grating X has 300 lines per millimetre and grating Y has 600 lines per millimetre. The same light is used with both. For the first order,
- (A) Y gives the larger angle because its slit separation is larger
- (B) Y gives the larger angle because its slit separation is smaller
- (C) both give the same angle because the same light is used
- (D) X gives the larger angle because it has fewer lines
Show Answer
Answer: B – 600 lines/mm means d = 1.67 × 10-6 m, half the d of the 300 lines/mm grating. Since sin θ = λ / d, halving d doubles sin θ, so Y spreads the spectrum out more (greater dispersion). More lines per millimetre = smaller d = bigger angles.
Q27. A grating spectrometer is used to examine the light from a sodium discharge lamp. What is observed, and why is it useful?
- (A) A continuous rainbow, showing that the lamp emits every wavelength
- (B) Nothing at all, because a discharge lamp emits only one colour
- (C) The light appears brighter, which makes the lamp easier to study
- (D) Bright lines at specific angles in each order; measuring the angles gives wavelengths that are characteristic of the element in the lamp
Show Answer
Answer: D – A discharge lamp emits only certain wavelengths, so instead of continuous spectra the grating gives sharp bright lines (images of the slit) in each order. From d sin θ = nλ the wavelengths follow – for sodium, a close pair of yellow lines near 589 nm – and the pattern of lines identifies the element.
Q28. The rainbow colours seen on the surface of a CD are mainly caused by
- (A) interference in a thin transparent film on the surface
- (B) coloured pigments in the printed label
- (C) the very closely spaced spiral track of the disc acting as a reflection grating, diffracting each wavelength through a different angle
- (D) refraction of light inside the plastic of the disc
Show Answer
Answer: C – The CD track is a spiral with a spacing of about 1.6 × 10-6 m, comparable to a grating’s slit separation. Reflected white light is diffracted by this regular structure, so different colours are reflected strongly at different angles, producing the rainbow effect (a DVD, with even closer tracks, behaves similarly).
Q29. A student wants the most accurate possible value of the wavelength from a grating experiment. Which approach is best?
- (A) Measure the angle of the zero order very carefully
- (B) Use only the first-order maximum because it is the brightest
- (C) Measure the angle of the highest-order maximum visible, because the larger angle can be measured with a smaller percentage uncertainty
- (D) Use a grating with fewer lines per millimetre so that the pattern is smaller
Show Answer
Answer: C – Since λ = d sin θ / n, the accuracy of λ follows the accuracy of θ. The measuring instrument has a fixed absolute uncertainty, so a larger angle (from a higher order) gives a smaller percentage uncertainty. Wrong options: the zero order contains no wavelength information, and fewer lines/mm makes the angles smaller, not larger.
Q30. The principal maxima of a diffraction grating are much brighter than the fringes from a double slit. This is because
- (A) at a maximum, the waves from a huge number of slits all arrive in phase and reinforce each other, and far more light in total passes through the many slits of the grating
- (B) each slit of the grating is much wider than the slits of a double slit
- (C) the grating reflects the light back through the slits a second time
- (D) the maxima are closer together, so the same light is squeezed into a smaller space
Show Answer
Answer: A – With N slits, the amplitudes from all N slits add up in phase at a principal maximum, and the grating also lets far more light through than just two slits. The maxima therefore stand out as intense sharp lines against an almost dark background.
Keep practising: browse all boards and topics on the Physics MCQs hub, and keep the A Level Physics Formula Sheet open while you revise – it collects the grating equation together with every other 9702 equation you need.
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