Diffraction grating

CIE A-Level Physics· 35 min read

1. Structure and Working Principle★★☆☆☆⏱ 10 min

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A diffraction grating consists of many hundreds or thousands of equally spaced parallel slits cut into a transparent or reflecting material. Unlike Young's double slit (which only has 2 slits), a typical teaching grating has 300 to 1000 lines per millimetre.

📘 Definition

Diffraction grating

Specified by lines per mm ( $N$ )

A periodic optical component that diffracts light into multiple beams at different angles based on wavelength via interference

Example:

A 500 lines per mm grating has 500 evenly spaced slits across 1 mm of material

When monochromatic light passes through the grating, each slit diffracts the light, and diffracted waves from different slits interfere with each other. Bright maxima (fringes) only form where the path difference between waves from adjacent slits is an integer multiple of the wavelength, producing constructive interference.

📐 Worked Example

A diffraction grating is labelled 600 lines per mm. Calculate the grating spacing $d$ in SI units.

1
Convert lines per mm to lines per metre for SI consistency:
2
$600 \text{ lines/mm} = 600 \times 10^3 \text{ lines/m}$
3
Grating spacing is the reciprocal of lines per metre:
4
$d = \frac{1}{\text{number of lines per m}} = \frac{1}{600 \times 10^3}$
5
Calculate the result:
6
$d = 1.67 \times 10^{-6} \text{ m}$

Exam tip:

Always convert lines per mm to lines per m to get $d$ in SI units for wavelength calculations

2. The Diffraction Grating Equation★★★☆☆⏱ 15 min

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Bright maxima occur when the path difference between light from adjacent slits equals an integer multiple of the wavelength. This gives the core diffraction grating equation, which can be derived easily from basic geometry.

🔬 Derivation

Goal:

Derive the diffraction grating equation

Starting from:

Condition for constructive interference of adjacent slits

1
For two adjacent slits separated by distance $d$ , the path difference of light leaving the slits at angle $θ$ (from the central direction) is $d sin θ$ .
2
Constructive interference occurs when path difference equals $nλ$ , where $n$ is the integer order of the maximum.
3
Equate the two expressions to get the final equation.

Result:

The diffraction grating equation is $d sin θ = nλ$

📐 Worked Example

Light of wavelength 500 nm is incident normally on a 300 lines per mm diffraction grating. Calculate the angle of the first order maximum.

1
Calculate grating spacing $d$ :
2
$d = \frac{1}{300 \times 10^3} = 3.33 \times 10^{-6} \text{ m}$
3
For first order maximum, $n = 1$ . Rearrange the grating equation for $sin θ$ :
4
$\sin\theta = \frac{n\lambda}{d}$
5
Substitute values ( $λ = 500 \times 1 0^{- 9}$ m):
6
$\sin\theta = \frac{1 \times 500 \times 10^{-9}}{3.33 \times 10^{-6}} = 0.15$
7
Calculate $θ$ :
8
$\theta = \sin^{-1}(0.15) = 8.6^\circ$

3. Maximum Number of Diffracted Orders★★★★☆⏱ 12 min

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The maximum possible value of $sin θ$ is 1, which occurs when $θ = 9 0^{\circ}$ (the maximum is along the plane of the grating). Any value of $n$ that gives $sin θ > 1$ is impossible, so this limits the highest order of diffraction we can observe.

To find the maximum order, substitute $sin θ \leq 1$ into the grating equation to get $n \leq \frac{d}{λ}$ . We always round the result down to the nearest integer, because a fractional order does not exist, and rounding up would give an impossible value of $sin θ$ .

📐 Worked Example

Light of wavelength 600 nm is incident on a grating with $d = 2.0 \times 1 0^{- 6}$ m. Find the total number of maxima produced by the grating.

1
Apply the maximum order condition:
2
$n \leq \frac{d}{\lambda}$
3
Substitute values:
4
$n \leq \frac{2.0 \times 10^{-6}}{600 \times 10^{-9}} = 3.33$
5
Round down to get maximum order $n_{ma x} = 3$
6
Maxima exist for all integer $n$ from $- n_{ma x}$ to $+ n_{ma x}$ , including the central $n = 0$ maximum. Calculate total:
7
$\text{Total maxima} = 2n_{max} + 1 = (2 \times 3) + 1 = 7$

4. Advantages Over Double Slits★★☆☆☆⏱ 8 min

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Diffraction gratings are preferred over double slits for accurate measurement of wavelength for two key reasons:

Sharper, brighter fringes: Many slits contribute to each maximum, so constructive interference only occurs at very specific angles, producing narrow, bright fringes that are easy to measure accurately. Double slits produce broad, dim fringes with high uncertainty in angle measurements.
Larger angular separation between orders: The grating spacing $d$ is much smaller than typical double slit separation, so the angular separation between consecutive orders is much larger, making it easier to resolve different wavelengths and avoid overlapping fringes.

This makes diffraction gratings the standard component in optical spectrometers used to identify unknown elements from their emission spectra.

5. Common Pitfalls

Wrong move:

Forgetting to convert lines per mm to lines per m, so $d$ is 1000× wrong

Why:

Units are inconsistent, leading to wavelength values that are 1000× too big or small

Correct move:

Always multiply lines per mm by 1000 to get lines per m, then calculate $d = 1/ (lines per m)$

Wrong move:

Rounding up when calculating maximum order e.g. rounding 3.3 to 4

Why:

Rounding up gives $sin θ > 1$ , which is physically impossible

Correct move:

Always round down to the nearest integer when finding maximum $n$

Wrong move:

Only counting positive orders and forgetting the central maximum when asked for total number of maxima

Why:

Orders exist on both sides of the central maximum, which counts as an additional maximum

Correct move:

Use the formula $Total maxima = 2 n_{ma x} + 1$ , where $n_{ma x}$ is the highest order

Wrong move:

Measuring $θ$ from the first order maximum instead of the central maximum

Why:

Misinterpreting the definition of $θ$ in the grating equation

Correct move:

$θ$ is always measured from the straight-through central (n=0) maximum

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Notes
Grating spacing $d$	$d = 1/ N$	$N$ = lines per m, convert from lines per mm first
Diffraction grating equation	$d sin θ = nλ$	$θ$ from central maximum, $n$ = integer order
Maximum order	$n_{ma x} = ⌊ d / λ ⌋$	Always round down, never up
Total number of maxima	$2 n_{ma x} + 1$	+1 for the central $n = 0$ maximum
Advantage vs double slit	N/A	Brighter, sharper fringes → lower uncertainty in wavelength

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
6 mark wavelength calculation
2023 · 1
Multiple choice maximum order
2024 · 2
Derive grating equation

Going deeper

simulationDiffraction Grating Interactive SimulationTest different grating spacings and wavelengths

What's Next

Diffraction gratings are a core application of superposition and wave interference, and their ability to split light by wavelength forms the basis of all modern optical spectroscopy, used in astronomy to detect exoplanets and in chemistry to identify unknown compounds. The principles of diffraction gratings also extend to other areas of physics, including X-ray diffraction used to determine crystal structure. Extending your knowledge of superposition will prepare you for more advanced exam questions on wave phenomena and help you connect concepts across the syllabus.