Intensity in Single-Slit and Double-Slit Diffraction · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

What sets the brightness across a slit pattern

Science · Waves, Light & Sound · ages 20-21
Name ______________________   Date ____________

  1. For a uniformly illuminated slit at normal incidence in the far field, theta is measured from the beam direction. Which condition gives a single-slit minimum, provided its angle exists?

    • a sin(theta) = p lambda for nonzero integer p
    • a sin(theta) = 0 at the central direction
    • a sin(theta) = (p + 1/2) lambda for integer p

  2. For nonzero beta in radians, continuous slit phasors trace an arc of radius r > 0 in field-amplitude units. Its traced length is E0 = 2r|beta| and chord magnitude is E = 2r|sin(beta)|. What is E/E0?

    • |beta/sin(beta)|
    • |sin(beta)/beta|
    • |sin(beta)/(2beta)|

  3. In the uniform-slit intensity formula, beta = 0 gives zero central intensity because sin(beta) is zero.

    Circle one:   True   False

  4. Two identical uniformly illuminated slits have a = 0.030 mm and centre spacing d = 0.24 mm. Use coherent 600 nm light at normal incidence, common polarisation and far-field observation. How many non-missing integer interference orders lie strictly between the first envelope minima?

    Answer: ______________

  5. Two identical uniformly illuminated slits have width a = 3lambda and centre separation d = 12lambda. Light is coherent, monochromatic and normally incident with common polarisation; observation is far-field. Which listed interference order is envelope-suppressed?

    • Order 2
    • Order 3
    • Order 4

  6. For a uniformly illuminated slit in the far field, evaluate I/I0 at beta = 5pi/2 radians using [sin(beta)/beta]^2. Use sin(5pi/2) = 1. Which expression gives the relative intensity?

    • 4/(25pi²), the squared amplitude ratio
    • 2/(5pi), the unsquared amplitude ratio
    • 25pi²/4, the inverse squared amplitude ratio

  7. In the identical, uniformly illuminated two-slit far-field model with coherent normal illumination and common polarisation, a = 5lambda and centre spacing d = 50lambda. Which listed pair is suppressed at the second envelope minima?

    • Orders −5 and +5
    • Orders −11 and +11
    • Orders −20 and +20

  8. In small-angle far-field diffraction with a much larger than lambda, a learner doubles slit width at fixed wavelength and predicts twice the central angular width. Which correction follows from 2lambda/a?

    • The central width doubles because a is in the numerator
    • The central width halves because a is in the denominator
    • The central width stays fixed because wavelength is fixed
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Answer key

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What sets the brightness across a slit pattern W1-mt_HURhzZdIkM-s1

  1. a sin(theta) = p lambda for nonzero integer p · Minima occur at beta = p pi for nonzero integers p. Beta zero instead gives central intensity I0 through the limiting ratio.
  2. |sin(beta)/beta| · Divide 2r|sin(beta)| by 2r|beta|. Cancelling the positive factor 2r gives |sin(beta)/beta|, a nonnegative amplitude-magnitude ratio.
  3. False · At beta zero the ratio cannot be evaluated by direct substitution. Its limit is 1, so the central intensity is I0.
  4. 15 · d/a = 8, so orders −7 through +7 lie strictly inside. Count seven negative orders, zero once and seven positive orders: 15.
  5. Order 4 · d/a = 12/3 = 4. The first envelope minimum suppresses order 4, whose required sine is 4/12 = 1/3.
  6. 4/(25pi²), the squared amplitude ratio · The amplitude-magnitude ratio is 1/(5pi/2) = 2/(5pi). Squaring gives I/I0 = 4/(25pi²), approximately 0.0162.
  7. Orders −20 and +20 · d/a = 10 and the second envelope minimum has p = ±2, giving m = ±20. Their required sine magnitude is 20/50 = 0.4, so the angles exist.
  8. The central width halves because a is in the denominator · The small-angle width is approximately 2lambda/a radians. Replacing a by 2a halves it.
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