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9702 · 8.4

The diffraction grating — practice questions

Practice and worked examples for 9702 The diffraction grating. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A diffraction grating has 500 lines per millimetre. Monochromatic light of wavelength 600 nm is incident normally on the grating. Calculate the angle of the first-order maximum.

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  1. First, determine the grating spacing, dd. Given 500 lines per mm, d=1 mm/500=1×10−3 m/500=2.0×10−6 md = 1 \text{ mm} / 500 = 1 \times 10^{-3} \text{ m} / 500 = 2.0 \times 10^{-6} \text{ m}.
  2. Identify the known values for the formula: d=2.0×10−6 md = 2.0 \times 10^{-6} \text{ m} λ=600 nm=600×10−9 m\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m} n=1n = 1 (for the first-order maximum)
  3. Apply the diffraction grating formula: dsin⁡θ=nλd \sin \theta = n\lambda 2.0×10−6×sin⁡θ=1×600×10−92.0 \times 10^{-6} \times \sin \theta = 1 \times 600 \times 10^{-9}
  4. Solve for sin⁡θ\sin \theta: sin⁡θ=(600×10−9)/(2.0×10−6)=0.3\sin \theta = (600 \times 10^{-9}) / (2.0 \times 10^{-6}) = 0.3
  5. Calculate θ\theta: θ=sin⁡−1(0.3)=17.5∘\theta = \sin^{-1}(0.3) = 17.5^\circ (to 3 significant figures)

Worked example 2

A diffraction grating has 600 lines per millimetre. It is illuminated with monochromatic light of wavelength 550 nm. What is the maximum number of orders of diffraction that can be observed?

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  1. Calculate the grating spacing, dd. The grating has 600 lines per mm, so the spacing between lines is: d=1 mm600=1×10−3 m600≈1.667×10−6 md = \frac{1 \text{ mm}}{600} = \frac{1 \times 10^{-3} \text{ m}}{600} \approx 1.667 \times 10^{-6} \text{ m}.
  2. State the condition for the maximum order. The maximum possible angle of diffraction is θ=90∘\theta = 90^\circ, for which sin⁡θ=1\sin \theta = 1. The diffraction grating equation is dsin⁡θ=nλd \sin \theta = n\lambda.
  3. Substitute the maximum condition into the formula to find the maximum value of nn. dsin⁡(90∘)=nmaxλd \sin(90^\circ) = n_{max} \lambda d=nmaxλd = n_{max} \lambda
  4. Solve for nmaxn_{max}: nmax=dλn_{max} = \frac{d}{\lambda}
  5. Substitute the numerical values: d=1.667×10−6 md = 1.667 \times 10^{-6} \text{ m} λ=550 nm=550×10−9 m\lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m} nmax=1.667×10−6550×10−9≈3.03n_{max} = \frac{1.667 \times 10^{-6}}{550 \times 10^{-9}} \approx 3.03
  6. Determine the highest integer order. Since the order nn must be an integer, the highest observable order is the largest integer less than or equal to 3.03. Therefore, the maximum number of orders is 3. (This means we can see n=1, n=2, and n=3 on each side of the central maximum, plus the central n=0 maximum).
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