Diffraction Gratings

Spec mapping (OCR H556 Module 4.5): Derive and apply the diffraction-grating equation $d\sin\theta = n\lambda$dsinθ=nλ for normal incidence. Distinguish $N$N-slit grating physics from two-slit Young's interference. Use the exact (non-small-angle) form to predict maximum order, angular separation, and to measure wavelength in spectroscopy applications.

A diffraction grating is the natural extension of the double-slit experiment to many slits. Instead of two slits, we use hundreds or thousands of closely-spaced parallel slits — typically several hundred per millimetre. The result is a much sharper, more intense and more precisely-located fringe pattern that allows extremely accurate measurement of wavelengths. Diffraction gratings are the standard tool of spectroscopy, used to analyse the light from stars, the composition of flames and the chemistry of laboratory samples.

OCR A-Level Physics A requires you to derive and apply the grating equation:

$d \sin\theta = n\lambda$dsinθ=nλ

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Physical Principle

Like Young's experiment, a diffraction grating works by superposition of coherent waves from multiple sources. Each slit in the grating produces its own secondary wave (Huygens). Because the same primary wavefront reaches all slits at the same time (for normal incidence), the secondary waves are all coherent with each other.

Consider a parallel beam of monochromatic light incident normally on a grating with spacing $d$d between adjacent slits. At an angle $\theta$θ from the straight-through direction, the wave from slit $k+1$k+1 has travelled an extra distance $d\sin\theta$dsinθ compared with the wave from slit $k$k. The cumulative path difference between slit $1$1 and slit $N$N is $(N-1)\,d\sin\theta$(N−1)dsinθ.

For a principal maximum, all $N$N waves must arrive at the detector in phase. This requires the path difference between adjacent slits to be a whole number of wavelengths:

$d \sin\theta = n \lambda$dsinθ=nλ

where $n$n is an integer ($0, \pm 1, \pm 2, \pm 3, \ldots$0,±1,±2,±3,…). This is the diffraction-grating equation.

The path difference between slits $k+1$k+1 and $k$k is the same as that between slits $k+2$k+2 and $k+1$k+1, so if it is a whole number of wavelengths for one pair, it is for every pair. All $N$N waves automatically arrive in phase at the same angle.

What the Variables Mean

Symbol	Meaning
$d$d	Slit spacing — distance between adjacent slit centres (m)
$\theta$θ	Angle of the maximum, measured from the straight-through direction
$n$n	Order of the maximum: integer $0, \pm 1, \pm 2, \ldots$0,±1,±2,…
$\lambda$λ	Wavelength of the light

Why Gratings Are Sharper Than Double-Slits

With only two slits, the transition from "in phase" to "out of phase" as you move off a principal maximum is gradual; the fringes are broad sinusoidal $\cos^2$cos2 profiles. With $N$N slits, all $N$N waves must be simultaneously in phase to produce a principal maximum — a much more restrictive condition. The consequence is that principal maxima become very narrow and very intense, separated by wide regions of near-zero intensity. The full intensity pattern of an $N$N-slit grating is:

$I(\theta) = I_0 \left[\frac{\sin(N\pi d \sin\theta / \lambda)}{\sin(\pi d \sin\theta / \lambda)}\right]^2$I(θ)=I0​[sin(πdsinθ/λ)sin(Nπdsinθ/λ)​]2

You will not need this at A-Level, but its shape — sharp peaks of width $\propto 1/N$∝1/N — is what makes gratings useful for spectroscopy: they can resolve wavelengths that differ by less than a nanometre, which a double-slit cannot.

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From Lines-per-Millimetre to $d$d

The slit spacing $d$d and the manufacturer's specification "lines per unit length" are reciprocals. A grating labelled "$300$300 lines per mm" has:

$d = \frac{1\text{ mm}}{300} = \frac{1}{300}\text{ mm} = 3.33 \times 10^{-6}\text{ m}$d=3001 mm​=3001​ mm=3.33×10−6 m

Grating spec	Slit spacing $d$d
$100$100 lines/mm	$1.00 \times 10^{-5}\text{ m}$1.00×10−5 m
$300$300 lines/mm	$3.33 \times 10^{-6}\text{ m}$3.33×10−6 m
$600$600 lines/mm	$1.67 \times 10^{-6}\text{ m}$1.67×10−6 m
$1\,200$1200 lines/mm	$8.33 \times 10^{-7}\text{ m}$8.33×10−7 m
$2\,400$2400 lines/mm	$4.17 \times 10^{-7}\text{ m}$4.17×10−7 m

Very fine gratings (with $d$d close to the wavelength of light) produce only a few orders of maximum, at wide angles; coarser gratings produce many orders at small angles. Fluency in this reciprocal conversion is essential.

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Worked Example 1 — First-Order Angle

Q. Light of wavelength $600\text{ nm}$600 nm is incident normally on a diffraction grating with $300$300 lines per mm. Calculate the angle of the first-order maximum.

A. Slit spacing: $d = 1/300\text{ mm} = 3.33\times 10^{-6}\text{ m}$d=1/300 mm=3.33×10−6 m.

For $n = 1$n=1:

$\sin\theta = \frac{n\lambda}{d} = \frac{(1)(6.00\times 10^{-7})}{3.33\times 10^{-6}} = 0.180$sinθ=dnλ​=3.33×10−6(1)(6.00×10−7)​=0.180

$\theta = \arcsin(0.180) = 10.4°$θ=arcsin(0.180)=10.4°

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Worked Example 2 — Maximum Order

Q. For the same grating and wavelength as above, how many orders can be seen?

A. The highest order corresponds to $\sin\theta \le 1$sinθ≤1. At the limit $\sin\theta = 1$sinθ=1, $\theta = 90°$θ=90° (the maximum would emerge along the plane of the grating):

$n_{\max} \le \frac{d}{\lambda} = \frac{3.33\times 10^{-6}}{6.00\times 10^{-7}} = 5.55$nmax​≤λd​=6.00×10−73.33×10−6​=5.55

Since $n$n must be a positive integer, $n_{\max} = 5$nmax​=5. You see the central maximum ($n = 0$n=0) plus five orders on each side, for a total of eleven bright fringes.

In practice the $n = 5$n=5 orders would be very close to grazing emergence ($\theta \approx 64°$θ≈64°) and may be dim or distorted, but they are geometrically present.

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Worked Example 3 — Measuring Wavelength

Q. When illuminated with monochromatic light, a grating with $500$500 lines per mm shows its first-order maximum at $\theta = 18.2°$θ=18.2°. Calculate the wavelength.

A. $d = 1/500\text{ mm} = 2.00\times 10^{-6}\text{ m}$d=1/500 mm=2.00×10−6 m.

$\lambda = \frac{d\sin\theta}{n} = \frac{(2.00\times 10^{-6})(\sin 18.2°)}{1} = (2.00\times 10^{-6})(0.3123)$λ=ndsinθ​=1(2.00×10−6)(sin18.2°)​=(2.00×10−6)(0.3123)

$\lambda = 6.25\times 10^{-7}\text{ m} = 625\text{ nm}$λ=6.25×10−7 m=625 nm

Orange-red, in the visible spectrum.

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Worked Example 4 — Resolving Two Wavelengths (Sodium Doublet)

Q. A grating with $600$600 lines per mm is illuminated with a sodium lamp emitting two close wavelengths, $589.0\text{ nm}$589.0 nm and $589.6\text{ nm}$589.6 nm. Calculate the angular separation of the two first-order maxima.

A. $d = 1/600\text{ mm} = 1.667\times 10^{-6}\text{ m}$d=1/600 mm=1.667×10−6 m.

For $\lambda_1 = 589.0\text{ nm}$λ1​=589.0 nm:

$\sin\theta_1 = \frac{5.890\times 10^{-7}}{1.667\times 10^{-6}} = 0.3534 \Rightarrow \theta_1 = 20.690°$sinθ1​=1.667×10−65.890×10−7​=0.3534⇒θ1​=20.690°

For $\lambda_2 = 589.6\text{ nm}$λ2​=589.6 nm:

$\sin\theta_2 = \frac{5.896\times 10^{-7}}{1.667\times 10^{-6}} = 0.3538 \Rightarrow \theta_2 = 20.714°$sinθ2​=1.667×10−65.896×10−7​=0.3538⇒θ2​=20.714°

Angular separation: $\Delta\theta = 0.024° \approx 0.42\text{ mrad}$Δθ=0.024°≈0.42 mrad.

Although small, this is easily resolved by a grating with a few thousand illuminated slits; a double-slit could not distinguish them.

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Worked Example 5 — Beyond the Maximum Order

Q. A grating has $1\,200$1200 lines per mm. Light of $\lambda = 700\text{ nm}$λ=700 nm is incident. Is a second-order maximum visible?

A. $d = 1/1\,200\text{ mm} = 8.33\times 10^{-7}\text{ m}$d=1/1200 mm=8.33×10−7 m. For $n = 2$n=2:

$\sin\theta = \frac{(2)(7.00\times 10^{-7})}{8.33\times 10^{-7}} = 1.68$sinθ=8.33×10−7(2)(7.00×10−7)​=1.68

This is greater than $1$1, so the $n = 2$n=2 order is not physically realisable — there is no second-order maximum for this combination. Only the central and first-order ($\sin\theta = 0.840$sinθ=0.840, $\theta = 57°$θ=57°) are visible. This is a frequent OCR question style: compute, get $\sin\theta > 1$sinθ>1, report "no such order exists".

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Worked Example 6 — Linear Position on a Screen

Q. A diffraction grating of $300$300 lines per mm is illuminated normally by a $\lambda = 590\text{ nm}$λ=590 nm sodium lamp. A screen is placed $1.50\text{ m}$1.50 m from the grating, parallel to it. Calculate the distance on the screen from the central fringe to the first-order fringe.

A. $d = 1/300\text{ mm} = 3.33\times 10^{-6}\text{ m}$d=1/300 mm=3.33×10−6 m. For $n = 1$n=1:

$\sin\theta = \frac{(1)(5.90\times 10^{-7})}{3.33\times 10^{-6}} = 0.177$sinθ=3.33×10−6(1)(5.90×10−7)​=0.177

$\theta = \arcsin(0.177) = 10.2°$θ=arcsin(0.177)=10.2°

Linear distance on the screen (small enough angle that $\tan\theta \approx \sin\theta/\cos\theta$tanθ≈sinθ/cosθ):

$y = D\tan\theta = 1.50 \times \tan 10.2° = 1.50 \times 0.180 = 0.270\text{ m} = 27.0\text{ cm}$y=Dtanθ=1.50×tan10.2°=1.50×0.180=0.270 m=27.0 cm

Each side of the central, so the first-order fringes are $54.0\text{ cm}$54.0 cm apart on the screen. A useful intuition: with this geometry ($\theta \approx 10°$θ≈10°), $y \approx D\tan\theta$y≈Dtanθ, but for the more accurate exam answer use $\tan\theta$tanθ, not $\sin\theta$sinθ (the latter is the small-angle approximation, which is not what the grating equation uses).

Watch out. The grating equation gives $\sin\theta$sinθ, not $\tan\theta$tanθ. To find the screen-plane linear distance, you must compute $\theta$θ first, then take $\tan\theta$tanθ. Many students slip by writing $y \approx D\sin\theta$y≈Dsinθ, which is only valid in the small-angle limit.

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Worked Example 7 — Identifying Spectral Lines

Q. A grating with $600$600 lines per mm is illuminated normally by a discharge lamp containing a mixture of mercury and helium. The student sees first-order maxima at $\theta_1 = 14.1°$θ1​=14.1°, $\theta_2 = 17.1°$θ2​=17.1°, $\theta_3 = 21.4°$θ3​=21.4°. Identify the three wavelengths and suggest which spectral lines they correspond to (one is the He yellow line at $587.6\text{ nm}$587.6 nm).

A. $d = 1/600\text{ mm} = 1.667\times 10^{-6}\text{ m}$d=1/600 mm=1.667×10−6 m. Using $\lambda = d\sin\theta$λ=dsinθ at $n = 1$n=1:

$\lambda_1 = (1.667\times 10^{-6})\sin 14.1° = (1.667\times 10^{-6})(0.2436) = 406\text{ nm} \text{ (violet)}$λ1​=(1.667×10−6)sin14.1°=(1.667×10−6)(0.2436)=406 nm (violet) $\lambda_2 = (1.667\times 10^{-6})\sin 17.1° = (1.667\times 10^{-6})(0.2940) = 490\text{ nm} \text{ (blue–green)}$λ2​=(1.667×10−6)sin17.1°=(1.667×10−6)(0.2940)=490 nm (blue–green) $\lambda_3 = (1.667\times 10^{-6})\sin 21.4° = (1.667\times 10^{-6})(0.3649) = 608\text{ nm} \text{ (orange)}$λ3​=(1.667×10−6)sin21.4°=(1.667×10−6)(0.3649)=608 nm (orange)

None of the three is at the known He $587.6\text{ nm}$587.6 nm line — but $\lambda_1 \approx 406\text{ nm}$λ1​≈406 nm is close to a known Hg violet line ($404.7\text{ nm}$404.7 nm), $\lambda_2 \approx 490\text{ nm}$λ2​≈490 nm close to Hg blue–green ($491.6\text{ nm}$491.6 nm), and $\lambda_3 \approx 608\text{ nm}$λ3​≈608 nm close to Hg orange ($607.3\text{ nm}$607.3 nm). This is a credible Hg-Hg-Hg identification. The He line would be at $\sin\theta = 587.6/1667 = 0.3525$sinθ=587.6/1667=0.3525, $\theta = 20.6°$θ=20.6° — present somewhere in the pattern but apparently not reported by the student.

This is a typical spectroscopy-style question — measure angles, convert to wavelengths, compare with a known atomic spectrum table.

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Order of Maxima

The order $n$n labels the principal maxima outward from the central direction:

• $n = 0$n=0: central (straight-through) maximum. All wavelengths superpose here, so in white light the central fringe is white.
• $n = \pm 1$n=±1: first-order maxima, one on each side. Closest coloured fringes to the centre.
• $n = \pm 2$n=±2: second-order, further out, at wider angles.
• $n = \pm 3, \pm 4, \ldots$n=±3,±4,…: higher orders.

The highest visible order is limited by the geometric condition $\sin\theta \le 1$sinθ≤1, giving:

$n_{\max} = \left\lfloor \frac{d}{\lambda} \right\rfloor$nmax​=⌊λd​⌋

Dispersion in Higher Orders

For a given grating and a given wavelength, higher orders are at larger angles. For a polychromatic source (e.g. white light), higher orders are more spread out — the second-order spectrum is about twice as wide as the first, the third-order about three times as wide. This is useful for spectroscopy but also produces order overlap: the red end of the third order ($\lambda \approx 700\text{ nm}$λ≈700 nm) can overlap the violet end of the fourth order ($\lambda \approx 525\text{ nm}$λ≈525 nm for $n = 4$n=4), because $3 \times 700 = 4 \times 525 = 2\,100\text{ nm}$3×700=4×525=2100 nm of path difference per slit.

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Gratings vs Young's Double Slit