A-Level — physicsHard4 min read

Diffraction and Diffraction Gratings

Single slit diffraction; d sin θ = nλ; using gratings to measure wavelength

Tutor AI Team
Tutor AI Team

# Diffraction and Diffraction Gratings — A-Level Physics

Diffraction is the spreading of waves when they pass through a gap or around an obstacle. Diffraction gratings exploit this phenomenon to produce precise measurements of wavelength — essential in spectroscopy.

1. Diffraction

Diffraction occurs when a wave passes through a gap or past an edge and spreads out.

Key Facts

  • Maximum diffraction when gap width ≈ wavelength
  • Wide gap → little diffraction; narrow gap → lots of spreading
  • Diffraction provides evidence that light is a wave

Single Slit Diffraction Pattern

  • Central maximum is bright and wide
  • Side maxima are dimmer and narrower
  • Narrower slit → wider central maximum (more spreading)
  • Central maximum width: w≈2λD/aw \approx 2\lambda D/a (where aa = slit width)

Minima at: asin⁡θ=nλa\sin\theta = n\lambda (where n=1,2,3,...n = 1, 2, 3, ...)

2. Diffraction Gratings

A diffraction grating has many equally spaced slits (typically hundreds per mm).

The Grating Equation

dsin⁡θ=nλ\boxed{d\sin\theta = n\lambda}

Where:

  • dd = slit spacing (m) — distance between adjacent slits
  • θ\theta = angle of diffraction from the central axis
  • nn = order number (0, 1, 2, 3, ...)
  • λ\lambda = wavelength

If the grating has NN slits per metre: d=1/Nd = 1/N

If NN slits per mm: d=10−3/Nd = 10^{-3}/N

Finding Maximum Order

The maximum possible order is when sin⁡θ=1\sin\theta = 1 (i.e., θ=90°\theta = 90°): nmax=floor(dλ)n_{\text{max}} = \text{floor}\left(\frac{d}{\lambda}\right)

3. Grating vs Double Slit

Property Double Slit Diffraction Grating
Number of slits 2 Hundreds/thousands
Maxima Broad, not very bright Sharp, bright
Pattern Gradually fading fringes Distinct sharp lines
Precision Low High (better for measurement)

More slits → sharper, brighter maxima.

4. Applications

Spectroscopy

  • Gratings separate white light into its component wavelengths
  • Each order shows a spectrum
  • Used to identify elements (atomic emission spectra)
  • Used in astronomy to analyse starlight

Measuring Wavelength

  • Shine light through grating
  • Measure angle θ\theta for each order
  • Use dsin⁡θ=nλd\sin\theta = n\lambda to calculate λ\lambda

Worked Example: Basic Grating Calculation

Problem

A grating has 500 lines per mm. Light of wavelength 589 nm is used. Find the angle of the 2nd order maximum.

d=10−3/500=2×10−6d = 10^{-3}/500 = 2 \times 10^{-6} m dsin⁡θ=nλd\sin\theta = n\lambda 2×10−6×sin⁡θ=2×589×10−92 \times 10^{-6} \times \sin\theta = 2 \times 589 \times 10^{-9} sin⁡θ=1178×10−9/2×10−6=0.589\sin\theta = 1178 \times 10^{-9}/2 \times 10^{-6} = 0.589 θ=36.1°\theta = 36.1°

Solution

Worked Example: Maximum Order

Problem

Same grating and wavelength. Find the maximum order visible.

nmax⁡=d/λ=2×10−6/589×10−9=3.40n_{\max} = d/\lambda = 2 \times 10^{-6}/589 \times 10^{-9} = 3.40

So nmax⁡=3n_{\max} = 3 (must be a whole number).

Solution

Worked Example: Finding Slit Spacing

Problem

The first-order maximum for 650 nm light occurs at 19.5°. Calculate the slit spacing.

d=nλ/sin⁡θ=1×650×10−9/sin⁡19.5°=650×10−9/0.334=1.95×10−6d = n\lambda/\sin\theta = 1 \times 650 \times 10^{-9}/\sin19.5° = 650 \times 10^{-9}/0.334 = 1.95 \times 10^{-6} m

Number of lines per mm = 10−3/1.95×10−6=51310^{-3}/1.95 \times 10^{-6} = 513 lines/mm.

Solution

Worked Example: White Light Through a Grating

Problem

With white light (400–700 nm), the first-order spectrum spans: θmin⁡=sin⁡−1(400×10−9/2×10−6)=sin⁡−1(0.2)=11.5°\theta_{\min} = \sin^{-1}(400 \times 10^{-9}/2 \times 10^{-6}) = \sin^{-1}(0.2) = 11.5° θmax⁡=sin⁡−1(700×10−9/2×10−6)=sin⁡−1(0.35)=20.5°\theta_{\max} = \sin^{-1}(700 \times 10^{-9}/2 \times 10^{-6}) = \sin^{-1}(0.35) = 20.5°

The spectrum spans from 11.5° (violet) to 20.5° (red).

Solution

6. Practice Questions

    1. A diffraction grating has 300 lines per mm. Calculate the slit spacing. (1 mark)
    1. Light of wavelength 550 nm passes through a grating (d=1.67×10−6d = 1.67 \times 10^{-6} m). Calculate the angle for the 1st and 2nd order maxima. (4 marks)
    1. How many orders are visible for 400 nm light with a grating of 600 lines/mm? (3 marks)
    1. Explain why a diffraction grating produces sharper maxima than a double slit. (2 marks)

    Answers

    1. d=10−3/300=3.33×10−6d = 10^{-3}/300 = 3.33 \times 10^{-6} m.

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Summary

  • Diffraction: spreading of waves through gaps; most when gap ≈ wavelength
  • Grating equation: dsin⁡θ=nλd\sin\theta = n\lambda
  • d=1/Nd = 1/N where NN = slits per metre
  • Maximum order: nmax⁡=floor(d/λ)n_{\max} = \text{floor}(d/\lambda)
  • More slits → sharper maxima
  • Used in spectroscopy to measure wavelengths and identify elements
Tags:
#a-level#physics#waves#diffraction#diffraction-grating#wavelength

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