Use diffraction gratings to determine wavelength
Apply d sin θ = nλ and describe how a grating pattern can be used to estimate light wavelength.
Section 8 · Superposition
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 8. These original models and calculations are study aids; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use wavelength, frequency, phase difference, trigonometry and consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In d sin θ = nλ, what does n represent?
What you will learn
- Recall and use d sin θ = nλ for a diffraction grating
- Relate grating spacing, diffraction angle and spectral order
- Describe a grating-based method for determining wavelength without requiring spectrometer construction
A grating creates maxima at specific angles
A diffraction grating has many equally spaced slits. For a principal maximum at angle θ to the normal, d sin θ = nλ, where d is slit spacing and n is an integer order number. The zero order has n = 0; first order has n = 1.
If a grating has N lines per metre, spacing d = 1/N. Use the measured angle from the normal, not from the grating surface. Check that the calculated sine is physically possible and that the chosen order is consistent with the observed maximum.
Describe a wavelength measurement
Direct a known monochromatic light beam normally at a grating and observe the central and first-order maxima on a distant screen. Measure the perpendicular grating-to-screen distance L and the lateral displacement y of a first-order maximum from the central one.
Find θ from tan θ = y/L, then use d sin θ = nλ with the known grating spacing and measured order. Repeat on both sides and average consistent estimates. A spectrometer's structure and operation are outside this syllabus outcome; this screen method explains the physical measurement without claiming instrument training.
Pause and explain
A grating has 2.0 × 10⁵ lines per metre. What is d?
Worked example
A grating has 500 lines per mm. A first-order maximum appears at 30°. Find wavelength.
Show the worked solution
- Convert line density: 500 lines mm⁻¹ = 5.0 × 10⁵ lines m⁻¹.
- d = 1/N = 2.0 × 10⁻⁶ m.
- For n = 1, λ = d sin 30° = 1.0 × 10⁻⁶ m.
Answer λ = 1.0 × 10⁻⁶ m.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A grating has 200 lines per mm. Find spacing d in metres.
Give me a hint
Convert lines per millimetre to lines per metre, then take the reciprocal.
Compare my reasoning
- N = 200 × 1000 = 2.0 × 10⁵ lines m⁻¹.
- d = 1/N.
- State metres per line as the slit spacing.
d = 5.0 × 10⁻⁶ m.
Look for these in your work
- Density converted
- Reciprocal taken
- Spacing unit stated
A grating spacing is 1.0 × 10⁻⁶ m. A second-order maximum occurs at 30°. Find wavelength.
Give me a hint
Rearrange to λ = d sin θ/n.
Compare my reasoning
- Use n = 2 and sin 30° = 0.5.
- λ = (1.0 × 10⁻⁶ × 0.5)/2.
- Evaluate in metres.
λ = 2.5 × 10⁻⁷ m.
Look for these in your work
- Order selected
- Equation rearranged
- Value and unit correct
A first-order maximum is 0.40 m from the centre on a screen 1.2 m from the grating. The grating has spacing 2.0 × 10⁻⁶ m. Estimate wavelength.
Give me a hint
Find θ from the screen triangle before applying the grating equation.
Compare my reasoning
- tan θ = 0.40/1.2, so θ ≈ 18.4°.
- Use λ = d sin θ/n with n = 1.
- λ ≈ 2.0 × 10⁻⁶ sin 18.4°.
λ ≈ 6.3 × 10⁻⁷ m.
Look for these in your work
- Angle determined from geometry
- First order used
- Wavelength estimated
Common mistakes
- Using line density as d instead of taking its reciprocal
- Measuring θ from the grating plane rather than the normal
- Using degrees and radians inconsistently in a calculator
What can you explain now?
In d sin θ = nλ, what does n represent?
Compare with the explanation
The spectral order number
It identifies the principal maximum order.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Why must grating line density be converted to slit spacing before using d sin θ = nλ?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.