6.1 Superposition and interference — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Keep the wavelength and the path difference in the same unit, and work in metres.

Warm-up: remember

A wave has a frequency of 250 Hz. Find its period.
Write 0.60 mm in metres.
A car travels 150 m in 6.0 s. Find its average speed.

Answers

$T = \dfrac{1}{f} = \dfrac{1}{250} = 4.0 \times 10^{-3}\,\text{s}$ .
$0.60\,\text{mm} = 6.0 \times 10^{-4}\,\text{m}$ .
$v = \dfrac{s}{t} = \dfrac{150}{6.0} = 25\,\text{m}\,\text{s}^{-1}$ .

If question 1 goes wrong, remember that the period is the time for one wave, so it is one divided by the frequency. If question 2 goes wrong, the prefix milli- means $10^{-3}$ . If question 3 goes wrong, divide the distance by the time.

Practise with help

1. A sound wave has a frequency of 425 Hz and travels at $340\,\text{m}\,\text{s}^{-1}$ . Find its wavelength.

Hint

The wave equation links the speed, the frequency and the wavelength.

Next step

Rearrange $v = f\lambda$ to make the wavelength the subject, then divide 340 by 425.

Full working

\begin{aligned} v &= f\lambda \\ \lambda &= \frac{v}{f} = \frac{340}{425} = 0.80\,\text{m} \end{aligned}

The wavelength is 0.80 m.

2. A wave has a wavelength of 0.25 m and a frequency of 680 Hz. Find its speed.

Hint

The speed of a wave is the distance one wave travels each second.

Next step

Multiply the frequency by the wavelength: $v = f\lambda$ .

Full working

v = f\lambda = 680 \times 0.25 = 170\,\text{m}\,\text{s}^{-1}

The wave travels at $170\,\text{m}\,\text{s}^{-1}$ .

3. Two points on a wave are 0.80 m apart. The wavelength is 2.0 m. Find the phase difference between them, in degrees.

Hint

One whole wavelength is $360^\circ$ . The phase difference is the same fraction of a full cycle as the distance is of a wavelength.

Next step

Divide the distance by the wavelength, then multiply by $360^\circ$ .

Full working

\Delta\phi = \frac{\Delta x}{\lambda} \times 360^\circ = \frac{0.80}{2.0} \times 360^\circ = 144^\circ

The points are $144^\circ$ out of step.

4. Two loudspeakers are driven in phase at 850 Hz. The speed of sound is $340\,\text{m}\,\text{s}^{-1}$ . A listener is 4.30 m from one speaker and 3.70 m from the other. Decide whether the sound is loud or quiet, and give a reason.

Hint

Find the wavelength first, then the path difference. Compare the path difference with the wavelength, not with the frequency.

Next step

The wavelength is $\dfrac{340}{850} = 0.40\,\text{m}$ . The path difference is $4.30 - 3.70 = 0.60\,\text{m}$ . Divide the path difference by the wavelength.

Full working

\begin{aligned} \lambda &= \frac{v}{f} = \frac{340}{850} = 0.40\,\text{m} \\ \Delta x &= 4.30 - 3.70 = 0.60\,\text{m} \\ \frac{\Delta x}{\lambda} &= \frac{0.60}{0.40} = 1.5 \end{aligned}

The path difference is one and a half wavelengths, so the two waves arrive antiphase. The sound at the listener is quiet.

5. A laser of wavelength 590 nm passes through two slits 0.40 mm apart. The screen is 1.80 m away. Find the spacing of the bright fringes.

Hint

Use $w = \dfrac{\lambda D}{s}$ . Change the nanometres and the millimetres to metres before you substitute.

Next step

Write $\lambda = 5.90 \times 10^{-7}\,\text{m}$ and $s = 4.0 \times 10^{-4}\,\text{m}$ , then substitute into the fringe formula.

Full working

\begin{aligned} w &= \frac{\lambda D}{s} \\ &= \frac{5.90 \times 10^{-7} \times 1.80}{4.0 \times 10^{-4}} \\ &= 2.7 \times 10^{-3}\,\text{m} \end{aligned}

The fringes are $2.7\,\text{mm}$ apart, to two significant figures.

6. Explain what is meant by coherent sources, and explain why two separate lamps cannot produce a steady interference pattern, even when they look the same colour.

Hint

Coherence is about the phase between the two waves, and about the frequency. Think about how an ordinary lamp emits its light.

Next step

Say that coherent sources have the same frequency and a constant phase difference. Then say what an ordinary lamp does instead.

Full working

Coherent sources have the same frequency and a constant phase difference, so the position of the bright and dark fringes stays fixed. Two separate lamps emit light in short, unrelated bursts. The phase difference between them changes many times each second, so the pattern shifts from moment to moment. The eye sees only the average, which has no visible fringes. A single source feeding both slits, or one laser, is needed because the light is then coherent.