· H2 Physics — Ten-Year Series, worked ← All 7 questions

2022 A-Level · Paper 2 · Question 5

Why the quiet spots in front of a metal plate are never silent

Superposition · stationary waves 10 marks Microwaves Wavelength from a graph

A point source of microwaves faces a metal plate, and a detector slides between them. Ten marks: what makes a wave stationary, how this one forms, why the readings dip but never reach zero, and the wavelength and frequency read off the intensity graph. Below is the full solution, and beside it the two kinds of wave and the detector's view.

The model, live

Progressive / intensityStationaryMinima

Part (a)

2 marks

State two features of a stationary wave that distinguish it from a progressive wave.

1. The amplitude of a stationary wave varies with position, from zero at the nodes to a maximum at the antinodes; in a progressive wave every point has the same amplitude.

2. A stationary wave does not transfer energy along the wave; a progressive wave does. (Also accepted: all points between adjacent nodes oscillate in phase, whereas in a progressive wave the phase changes continuously with distance.)

Part (b)

2 marks

Explain how the stationary wave between the source and the plate is formed.

Microwaves from the source are reflected by the metal plate. The incident and reflected waves have the same frequency and speed and travel in opposite directions, so they overlap and superpose, giving nodes and antinodes at fixed positions.

Part (c)(i)

3 marks

A detector is moved from the source towards the plate. Explain why the intensity at a minimum is never zero.

Approach

Complete cancellation needs two waves of equal amplitude in antiphase. Here the two waves at the detector have different amplitudes. Say why (three marks, three reasons): the source is a point source, the waves have travelled different distances, and the plate does not reflect everything.

The source is a point source, so the waves spread out and their amplitude decreases with distance travelled. At any point between the source and the plate the reflected wave has travelled further than the incident wave (and some energy is lost at the plate), so the two waves have different amplitudes. When they meet in antiphase they cannot cancel completely, so the intensity at a minimum is not zero.

Note

The graph shows the effect too: the maxima fall and the minima rise as the detector moves away from the source. Make the amplitudes equal in the model and the minima drop to zero.

Part (c)(ii)

1 mark

Use the intensity graph to find the wavelength.

Approach

Adjacent minima are adjacent nodes, half a wavelength apart. Span as many as possible: the first minimum is at about 2.3 cm2.3\,\text{cm} and the third at about 7.5 cm7.5\,\text{cm}, two half-wavelengths apart.

λ2=7.5−2.32=2.6 cmλ=5.2 cm\frac{\lambda}{2} = \frac{7.5 - 2.3}{2} = 2.6\,\text{cm} \qquad \lambda = 5.2\,\text{cm}

wavelength =5.2 cm= 5.2\,\text{cm}

Watch

Minimum to minimum is λ/2\lambda/2, not λ\lambda. A wavelength of 2.6 cm2.6\,\text{cm} would give a frequency twice as large.

Part (c)(iii)

2 marks

Calculate the frequency of the microwaves.

f=cλ=3.00×1085.2×10−2=5.8×109 Hzf = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{5.2\times10^{-2}} = 5.8\times10^{9}\,\text{Hz}

frequency =5.8×109 Hz= 5.8\times10^{9}\,\text{Hz}

What this question was really testing

  1. Stationary vs progressive. Amplitude varies with position; no energy transfer.
  2. Formation by reflection. Same frequency and speed, opposite directions, superposition.
  3. Why cancellation is incomplete. Unequal amplitudes, because of spreading and imperfect reflection.
  4. Reading a wavelength off a graph. Adjacent minima are λ/2\lambda/2 apart; measure over several.

This is how we mark, and how we teach.

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