What is a diffraction grating and how does it work in the context of superposition?

A diffraction grating is an optical component with a regular pattern of lines or grooves that diffract light into several beams. It works on the principle of superposition, where the overlapping of light waves from different slits creates an interference pattern. This pattern results in the separation of light into its constituent wavelengths, producing a spectrum. In the context of Cambridge International A Levels Physics, understanding this principle is crucial for analyzing how light behaves when it encounters obstacles.

How is the diffraction grating formula derived and what does it signify?

The diffraction grating formula is derived from the condition for constructive interference, which occurs when the path difference between light from adjacent slits is an integer multiple of the wavelength. The formula is given by d sin θ = nλ, where d is the grating spacing, θ is the angle of diffraction, n is the order of the spectrum, and λ is the wavelength of light. This formula is significant as it allows students to calculate the angles at which different wavelengths will be diffracted, a key concept in the Cambridge International A Levels Physics curriculum.

What are the applications of diffraction gratings in real-world scenarios?

Diffraction gratings have numerous applications in various fields. They are used in spectrometers to analyze the spectral composition of light, which is crucial in fields like chemistry and astronomy. In telecommunications, they help in wavelength division multiplexing, allowing multiple signals to be sent over a single optical fiber. Understanding these applications is important for Cambridge International A Levels students to appreciate the practical implications of theoretical concepts in Physics.

How does the number of lines per millimeter on a diffraction grating affect the diffraction pattern?

The number of lines per millimeter on a diffraction grating determines the grating spacing, d, which inversely affects the diffraction pattern. A higher number of lines per millimeter results in a smaller grating spacing, leading to a greater angular separation of the diffracted beams. This means that the spectral lines are more spread out, allowing for better resolution of closely spaced wavelengths. This concept is essential for students studying Cambridge International A Levels Physics, as it highlights the relationship between grating characteristics and the resulting diffraction pattern.

Why is it important to understand the concept of order in diffraction gratings?

Understanding the concept of order in diffraction gratings is important because it helps in identifying the different maxima in the diffraction pattern. The order, denoted by n, indicates the number of wavelengths by which paths differ between adjacent slits. Higher orders correspond to higher angles of diffraction and can provide more detailed spectral information. For Cambridge International A Levels Physics students, grasping this concept is crucial for accurately analyzing and interpreting diffraction patterns in experiments and real-world applications.

Why is "Confusing lines/mm with spacing" a common mistake in The diffraction grating?

Why it happens: Units. How to avoid it: If grating has N lines per mm, d = 1/N mm = 1/(N × 10^3) m. Source: 9702 Examiner Reports 2022-2024

Why is "Allowing $n$ exceeding maximum" a common mistake in The diffraction grating?

Why it happens: Forgetting ≤ 1. How to avoid it: n_ ≤ d/. Round DOWN to integer. Source: 9702 Examiner Reports 2022-2024

SuperpositionCambridge International A Level Physics 9702

The diffraction grating

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Diffraction Grating Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Many slits give sharp maxima at $d\sin\theta = n\lambda$ . Used for spectroscopy.

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

8.4.1 — Apply grating equation.
8.4.2 — Find maximum order.
8.4.3 — Determine $\lambda$ from angle.

Grating equation

$d\sin\theta = n\lambda$ .

Equation. $d\sin\theta = n\lambda$ for $n = 0, 1, 2, \ldots$

$d$ = grating spacing = distance between adjacent slits/lines.

If $N$ lines per mm, $d = 1/N$ mm = $1/(N \times 10^3)$ m.

$n$ = order (central maximum is $n = 0$ ).

Sharper than Young's slits. Many slits → constructive interference at exact angles is much sharper.

Maximum order. $\sin\theta \leq 1 \Rightarrow n_{\max} = d/\lambda$ (round down to integer).

Example. 500 lines/mm, 600 nm light:

$d = 2 \times 10^{-6}$ m.
First order: $\sin\theta = 600 \times 10^{-9}/2 \times 10^{-6} = 0.3$ , $\theta \approx 17.5°$ .
Max order: $d/\lambda = 3.33$ → $n_{\max} = 3$ .

Cambridge tip. Always convert lines/mm to $d$ in metres first.

A diffraction grating produces sharp maxima at angles where d sinθ = nλ; longer wavelengths diffract through larger angles.

$d\sin\theta = n\lambda$ .
Sharp, bright maxima.
$n_{\max} = d/\lambda$ .

See the full worked example for the diffraction grating →

Spectroscopy

Different wavelengths diffract differently.

Different $\lambda$ diffract at different angles for same $n$ .

White light spreads into spectrum at each order (red diffracts MORE than blue — longer wavelength).

Application. Measure $\lambda$ by measuring $\theta$ for known $d$ and $n$ . Rearrange grating equation: $\lambda = d\sin\theta / n$ .

Example. 300 lines/mm, $\theta = 22°$ second order: $\lambda = (3.33 \times 10^{-6})(0.375)/2 \approx 624$ nm.

Cambridge tip. Mark scheme rewards explicit calculation steps.

Different $\lambda$ → different $\theta$ .
White light spreads into spectrum.
$\lambda = d\sin\theta / n$ for measurement.

See the full worked example for the diffraction grating →

How it’s examined

Grating on every Paper 1. Most-tested: angle calculation (6-7 marks), max order (5 marks), wavelength determination (7-8 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — The diffraction grating

Step-by-step solutions to past-paper-style questions on the diffraction grating, written exactly the way a tutor would explain them at the board.

1Grating angle (6 marks)
Extended• Adapted from 9702/22 May/Jun 2024• grating
Question
Light wavelength 600 nm hits a grating with 500 lines/mm. Find the angle of the first-order maximum. (6 marks)
Step-by-step solution
1. Step 1
  $d$ = grating spacing = $1/(500 \times 10^3)$ = $2.0 \times 10^{-6}$ m.
2. Step 2
  $d\sin\theta = n\lambda$ , $n = 1$ .
  $\sin\theta = \dfrac{\lambda}{d} = \dfrac{600 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.30$
3. Step 3
  $\theta = \sin^{-1}(0.30) \approx 17.5°$ .
Answer
$\theta \approx 17.5°$ .
2Maximum order (5 marks)
Extended• max order
Question
For the above grating (500/mm) with 600 nm light, find the maximum number of orders observable. (5 marks)
Step-by-step solution
1. Step 1
  Maximum $\sin\theta = 1$ .
  $n_{\max} = \dfrac{d}{\lambda} = \dfrac{2.0 \times 10^{-6}}{600 \times 10^{-9}} = 3.33$
2. Step 2
  Round down. Maximum order = 3.
Answer
$n_{\max} = 3$ .
3Wavelength from angle (7 marks)
Extended• spectroscopy
Question
A grating with 300 lines/mm produces a second-order maximum at $\theta = 22°$ for an unknown wavelength. Find the wavelength. (7 marks)
Step-by-step solution
1. Step 1
  $d = 1/(300 \times 10^3) = 3.33 \times 10^{-6}$ m.
2. Step 2
  $d\sin\theta = n\lambda$ with $n = 2$ .
  $\lambda = \dfrac{d\sin\theta}{n} = \dfrac{3.33 \times 10^{-6} \times \sin 22°}{2}$
3. Step 3
  Evaluate.
  $\lambda = \dfrac{3.33 \times 10^{-6} \times 0.375}{2} \approx 6.24 \times 10^{-7} \text{ m} = 624 \text{ nm}$
Answer
$\lambda \approx 624$ nm.

Key Formulae — The diffraction grating

The formulae you need to memorise for the diffraction grating on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Grating equation
$d\sin\theta = n\lambda$
$d$
grating spacing (distance between adjacent slits)
$\theta$
diffraction angle to maximum
$n$
order of maximum (0, 1, 2, ...)
When to use
Position of bright maxima in grating diffraction.

Key Definitions and Keywords — The diffraction grating

Definitions to memorise and the exact keywords mark schemes credit for the diffraction grating answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Diffraction grating
Examiner keyword
Many parallel equally-spaced slits/lines. Produces sharp bright maxima at specific angles.
Grating spacing
Examiner keyword
Distance $d$ between adjacent slits/lines. Reciprocal of lines per metre.

Common Mistakes and Misconceptions — The diffraction grating

The traps other students keep falling into on the diffraction grating questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Confusing lines/mm with spacing
9702 Examiner Reports 2022-2024
Why it happens
Units.
How to avoid it
If grating has $N$ lines per mm, $d = 1/N$ mm $= 1/(N \times 10^3)$ m.
✕Allowing $n$ exceeding maximum
9702 Examiner Reports 2022-2024
Why it happens
Forgetting $\sin\theta \leq 1$ .
How to avoid it
$n_{\max} \leq d/\lambda$ . Round DOWN to integer.

Practice questions

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