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Diffraction gratings

Replace two slits with thousands and the fringes sharpen into thin bright lines at precisely predictable angles. One equation, d sin θ = nλ, locates every line, and because a sine cannot exceed 1 only a limited number of orders can exist.

Year 12AQA 3.3.2.2 CIE 8.4.1, 8.4.2 OCR A 5.5.2(g), 5.5.2(h)EDEXCEL 84

Builds on Interference and Young's double slit.

IN THIS TOPIC

Explain why many slits give sharper and brighter maxima than two.
Use $d \sin\theta = n\lambda$ , getting d from the number of lines per millimetre.
Work out the highest order that can exist, and give uses of gratings.

COMMON MISCONCEPTION

Adding thousands more slits should smear the pattern into a blur.

From two slits to thousands

A diffraction grating is a plate ruled with hundreds of slits per millimetre. Light leaving all of them overlaps, and the many-slit sum is far stricter than the two-slit one. At most angles, the thousands of contributions arrive with a scatter of phases and cancel almost perfectly. Only at a few special angles does every slit's light arrive exactly in step, and there the maxima are brighter, fed by every slit at once, and much sharper, because even a tiny step away from the exact angle restores the cancellation.

The grating equation

The special angles come from the path difference between neighbouring slits, whose centres sit a distance d apart, the grating spacing. Light leaving adjacent slits at angle $\theta$ to the normal differs in path by $d\sin\theta$ . Every slit stays in step with every other only when that difference is a whole number of wavelengths.

d \sin\theta = n\lambda

ON THE AQA DATA SHEET

FIG. 1Adjacent slits, and the extra distance

d\sin\theta

the lower ray travels. When it equals

n\lambda

, light from every slit on the grating arrives in phase.

The whole number n is the order of the maximum. Zero order is the straight-through beam, first order the pair either side of it, and so on outwards. Gratings are labelled in lines per millimetre, so d is one millimetre shared out between that many lines.

d = \frac{1}{N}

NOT ON THE AQA DATA SHEET: LEARN IT

Feed N in lines per metre and d comes out in metres. A grating of 600 lines per mm is 6.00 × 10⁵ lines per metre, giving $d = 1.67 \times 10^{− 6}$ m.

Orders, and the one that cannot exist

Rearranged, $\sin\theta = n\lambda/d$ , and a sine can never exceed 1. That alone caps the pattern: orders exist only while $n\lambda/d \le 1$ , so the highest order is the whole-number part of $d/\lambda$ . Beyond it, the geometry simply has no angle to offer.

FIG. 2The full pattern for d = 2.4

\lambda

: a zero order and two orders either side, each a sharp beam. A third order would need sin

\theta

= 1.25, which no angle can supply.

WORKED EXAMPLE

Monochromatic light of wavelength 550 nm falls on a grating with 600 lines per mm. Find the angle of the first-order maximum, and the highest order visible.

$d = 1/N$ = (1.0 × 10^{− 3}) / 600 = 1.67 × 10^{− 6} m.

For the first order, $\sin\theta = \lambda/d$ = (550 × 10^{− 9}) / (1.67 × 10^{− 6}) = 0.330, so $\theta$ = 19.3°.

Now the cap. $d/\lambda$ = 3.03, so n = 3 works, needing $\sin\theta$ = 0.990, while n = 4 would need 1.32. The third order is the last one visible, out at a steep 81.9°.

GUIDED PRACTICE

A wavelength from the second order

Light through a 600 lines-per-millimetre grating puts its second-order maximum at 42.5°. Find d from the line count, then the wavelength.

Show the working

d = 1/600 mm = 1.67 × 10⁻⁶ m.

With n = 2, λ = d sin θ/n = 1.67 × 10⁻⁶ × 0.676/2 = 5.6 × 10⁻⁷ m, so 560 nm, green. Forget to divide by the order and you land on a wavelength no visible light has.

INDEPENDENT PRACTICE

Why gratings make rainbows

White light passes through a 400 lines-per-millimetre grating. Find the first-order angles for violet (400 nm) and red (700 nm), and the angular width of the first-order spectrum.

Show the working

d = 2.5 × 10⁻⁶ m. Violet: sin θ = 0.16, θ = 9.2°. Red: sin θ = 0.28, θ = 16.3°.

The spectrum spans about 7°, violet innermost. Longer wavelengths diffract to larger angles, which is the reverse of a prism's ordering and a favourite comparison question.

Because the maxima are so sharp, their angles can be measured precisely, and the grating equation then delivers $\lambda$ to matching precision. That precision is the instrument's real job. A grating splits light into line spectra, which lets a chemist identify an element from the wavelengths it emits and lets an astronomer read the composition of a star nobody will ever visit.

ASSESSMENT FOCUS

Finding d is where marks leak. Convert the millimetre to metres first, then divide by the number of lines. For 600 lines per mm, d = (1.0 × 10^{− 3})/600, and the answer should land near 10^{− 6} m.
For the highest order, work out $d/\lambda$ and take the whole-number part. Writing n = 3.03, or rounding it up to 4, both lose the mark. When $\sin\theta$ comes out above 1 the order simply does not exist, so say so instead of forcing an angle out of the calculator.
Grating maxima are sharper and brighter than double-slit fringes. The sharpness is the reason a grating measures wavelength well.
Check the mode. An angle of 19.3° arriving as 0.337 means the calculator is in radians.

CHECK YOURSELF

Light of wavelength 550 nm falls on a grating with 300 lines per mm. Find (a) the angle of the first-order maximum and (b) the highest order that can be seen.

Show a hint

Find d first, and remember that sin $\theta$ can never pass 1.

Show the answer

(a) $d = 1/N$ = (1.0 × 10^{− 3}) / 300 = 3.33 × 10^{− 6} m. Then $\sin\theta = \lambda/d$ = (550 × 10^{− 9}) / (3.33 × 10^{− 6}) = 0.165, so $\theta$ = 9.5°.

(b) $d/\lambda$ = 6.06, so the largest whole n with $\sin\theta \le 1$ is n = 6. That sixth order sits far out at 81.9°, almost along the grating itself.

d sin θ = nλ.

sin θ can never pass 1, and that caps n.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the diffraction gratings questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 2: interference: double slit and diffraction grating puts this topic in the lab, and the written papers ask about it.
Trigonometry and resolving vectors is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Explain why many slits give sharper and brighter maxima than two.
Use $d \sin\theta = n\lambda$ , getting d from the number of lines per millimetre.
Work out the highest order that can exist, and give uses of gratings.

Open the full revision checklist to track your progress across the whole unit.

Diffraction gratings

From two slits to thousands

The grating equation

Orders, and the one that cannot exist

Question workbook

Question workbook

Answer book