The Diffraction Grating and Measuring Wavelength · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Sharp lines measure light

Science · Waves, Light & Sound · ages 17-18
Name ______________________   Date ____________

  1. For normal incidence in the far field, d is slit spacing, lambda wavelength, n order and theta the angle from the grating normal. Which equation gives principal reinforcement directions?

    • d cos theta equals n lambda
    • d sin theta equals n lambda
    • d tan theta equals n lambda

  2. A grating has 500 lines per millimetre. What is its centre-to-centre slit spacing d?

    • 2 × 10^−6 metres
    • 5 × 10^−1 metres
    • 6 × 10^−7 metres

  3. For normal incidence, an order requiring sin(theta) greater than 1 can occur at an angle beyond 90 degrees, with theta measured from the grating normal.

    Circle one:   True   False

  4. For normal incidence, d = 2 × 10^−6 m and lambda = 600 nm. What is the largest positive order permitted by the sine bound? Include a grazing direction if the ratio is exactly an integer.

    Answer: ______________

  5. For normal incidence in the far field, d = 3.0 × 10^−6 m and lambda = 450 nm. For order n = 2, find sin(theta), with theta measured from the grating normal.

    Answer: ______________

  6. With spacing and wavelength fixed, why does increasing the number of illuminated grating slits narrow the principal maxima?

    • More slits reduce the wavelength passing through the grating
    • More slits increase the spacing between neighbouring slits
    • More slits increase cancellation away from principal maxima

  7. Ava calculates 1/250 = 0.004 for a grating with 250 lines per millimetre and writes d = 0.004 m. Which step needs correction?

    • Replace reciprocal density with diffraction order
    • Replace reciprocal density with wavelength squared
    • Convert the reciprocal spacing from mm to metres

  8. For normal incidence in the far field, order n = 2 has sin(theta) = 0.625, with theta measured from the grating normal. The spacing is d = 1.6 × 10^−6 m. Find lambda in metres.

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Sharp lines measure light W1-mt_yIjB-uJuFt-s1

  1. d sin theta equals n lambda · The neighbouring path difference is d sin(theta), and full reinforcement requires an integer n times the wavelength.
  2. 2 × 10^−6 metres · d = (1/500) mm = 0.002 mm = 2 × 10^−6 m.
  3. False · The sine of a real angle never exceeds 1. Going beyond 90 degrees does not provide a larger sine.
  4. 3 · d/lambda = 2 × 10^−6 / (6 × 10^−7) = 3.333..., so the largest permitted integer order is 3. Visibility is not guaranteed.
  5. 0.3 · sin(theta) = 2 × 450 × 10^−9 / (3.0 × 10^−6) = 0.3.
  6. More slits increase cancellation away from principal maxima · More contributions give stronger phase cancellation away from principal reinforcement directions. Finite-width peaks and weaker secondary maxima still remain.
  7. Convert the reciprocal spacing from mm to metres · The reciprocal gives 0.004 mm, not 0.004 m. Multiplying by 0.001 gives 0.000004 m.
  8. 5e-7 · lambda = d sin(theta)/n = 1.6 × 10^−6 × 0.625 / 2 = 5 × 10^−7 m.
Worksheet · LightMySky