Phasor Addition and the Mathematics of Interference · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Adding waves with arrows

Science · Waves, Light & Sound · ages 19-21
Name ______________________   Date ____________

  1. A light wave's resultant amplitude doubles in the same medium. What happens to its intensity?

    • It quadruples
    • It doubles
    • It is unchanged

  2. Two coherent waves have matching polarisation, amplitude 1 unit each and phase 0°. What do their phasors do?

    • They oppose and give a resultant of length 0 units
    • They align and give a resultant of length 2 units
    • They align and give a resultant of length 1 unit

  3. Two narrow coherent slits start in phase with matching polarisation. Their separation is d. In the far-screen approximation, which condition gives a bright interference maximum at angle theta?

    • d sin(theta) = (m + 1/2) lambda, for integer m
    • Any angle works when the screen distance is halved
    • d sin(theta) = m lambda, for integer m

  4. Two narrow coherent slits start in phase with matching polarisation. At a far-screen point, each slit alone gives 5 W/m² and the path difference is lambda/6. What is the combined intensity? Use cos²(pi/6) = 3/4.

    • 10 W/m²
    • 15 W/m²
    • 20 W/m²

  5. Three coherent waves have matching polarisation, amplitude 3 units each and phases 0°, 60° and 120°. Add their phasors. What is the resultant amplitude in units?

    Answer: ______________

  6. More equally spaced slits are illuminated coherently, with equal contributions from each slit. Wavelength and slit spacing stay fixed. Why do the principal maxima become narrower?

    • Each slit shortens the wavelength, contracting fringes
    • The principal maxima shift halfway between their old angles
    • The nearest minima move closer to each principal maximum

  7. Maya says a dark interference fringe proves that light energy has been destroyed. In a lossless arrangement, which reply explains the energy accounting?

    • Interference changes where energy arrives; it does not destroy it
    • Dark fringes absorb light and convert it to heat
    • Dark fringes turn light into shorter wavelengths outside the visible band

  8. Three coherent waves have equal amplitudes, matching polarisation and phases 0°, 120° and 240°. Their resultant amplitude is zero.

    Circle one:   True   False

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Answer key

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

Adding waves with arrows W1-mt_dxh7HHbIUw-s1

  1. It quadruples · Intensity is proportional to amplitude squared. The intensity factor is 2² = 4.
  2. They align and give a resultant of length 2 units · Both phasors point in the same direction. Their lengths add: 1 + 1 = 2 units.
  3. d sin(theta) = m lambda, for integer m · An integer-wavelength path difference adds whole phase cycles, so the phasors align.
  4. 15 W/m² · The path difference gives phi = pi/3. Thus I = 4I1 cos²(phi/2) = 4 × 5 × 3/4 = 15 W/m².
  5. 6 · The component sums are X = 3 and Y = 3√3 units. The resultant amplitude is √(3² + (3√3)²) = √36 = 6 units.
  6. The nearest minima move closer to each principal maximum · The nearest minima occur at adjacent phase gaps ±360/N degrees from alignment. Increasing N reduces those gaps.
  7. Interference changes where energy arrives; it does not destroy it · Destructive interference reduces intensity locally. Energy is conserved when the full flow is counted, with enhanced intensity elsewhere.
  8. True · The equal phasors close a triangle when placed head to tail. The final head meets the first tail, so the resultant has zero length.
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