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9702 · 7.1

Progressive waves — practice questions

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

Worked example 1

Unpolarised light of intensity 8.0 W m−28.0 \text{ W m}^{-2} is incident on a polarising filter. The light that passes through this filter then passes through a second polarising filter. The axis of the second filter is at an angle of 60∘60^{\circ} to the axis of the first. Calculate the intensity of the light emerging from the second filter.

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  1. First Filter: When unpolarised light passes through the first polariser, its intensity is halved. The intensity after the first filter, I1I_1, is I1=12×8.0=4.0 W m−2I_1 = \frac{1}{2} \times 8.0 = 4.0 \text{ W m}^{-2}. This light is now polarised.
  2. Recall Malus' Law: For the second filter, the formula is I2=I1cos⁡2θI_2 = I_1\cos^2\theta.
  3. Identify Given Values: Intensity incident on the second filter I1=4.0 W m−2I_1 = 4.0 \text{ W m}^{-2}. Angle between filter axes θ=60∘\theta = 60^{\circ}.
  4. Calculate cos⁡(60∘)\cos(60^{\circ}): cos⁡(60∘)=0.5\cos(60^{\circ}) = 0.5.
  5. Calculate (cos⁡(60∘))2(\cos(60^{\circ}))^2: (0.5)2=0.25(0.5)^2 = 0.25.
  6. Substitute into Malus' Law: I2=4.0 W m−2×0.25I_2 = 4.0 \text{ W m}^{-2} \times 0.25.
  7. Calculate Final Intensity: I2=1.0 W m−2I_2 = 1.0 \text{ W m}^{-2}.
  8. State the Answer: The intensity of the transmitted light is 1.0 W m−21.0 \text{ W m}^{-2}.

Worked example 2

An ambulance emits a siren with a frequency of 550 Hz. It travels towards a stationary observer at a speed of 30 m/s. If the speed of sound in the air is 340 m/s, what is the frequency heard by the observer?

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  1. Identify the formula: For a source moving towards a stationary observer, the Doppler effect formula for sound is fo=fs(vv−vs)f_o = f_s \left( \frac{v}{v - v_s} \right), where fof_o is the observed frequency, fsf_s is the source frequency, vv is the wave speed, and vsv_s is the source speed.
  2. List the known values:
    • Source frequency, fs=550 Hzf_s = 550 \text{ Hz}
    • Speed of the source, vs=30 m s−1v_s = 30 \text{ m s}^{-1}
    • Speed of sound, v=340 m s−1v = 340 \text{ m s}^{-1}
  3. Substitute the values into the formula: fo=550×(340340−30)f_o = 550 \times \left( \frac{340}{340 - 30} \right)
  4. Calculate the denominator: 340−30=310 m s−1340 - 30 = 310 \text{ m s}^{-1}
  5. Calculate the fraction: 340310≈1.0968\frac{340}{310} \approx 1.0968
  6. Calculate the observed frequency: fo=550×1.0968≈603.24 Hzf_o = 550 \times 1.0968 \approx 603.24 \text{ Hz}
  7. State the final answer: The frequency heard by the observer is approximately 603 Hz (to 3 significant figures).
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