6.1 Superposition and interference — Practise

Work through these on your own. Change every distance to metres before you substitute, and write the unit of every answer. In the questions that ask for a reason, write your answer in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A sound wave has a frequency of 1.5 kHz and a speed of 330 m/s. Find its wavelength.
A wave has a wavelength of 2.5 m and a speed of 340 m/s. Find its frequency.
Two points on a wave are 0.15 m apart. The wavelength is 0.60 m. Find the phase difference between them, in degrees.
Two loudspeakers are driven in phase at 425 Hz. The speed of sound is 340 m/s. A point is 2.60 m from one speaker and 1.40 m from the other. State whether the sound there is loud or quiet, and give a reason.
A laser of wavelength 640 nm passes through two slits 0.30 mm apart. The screen is 2.0 m away. Find the fringe spacing.
In a two-slit experiment the slits are 0.20 mm apart and the screen is 1.5 m away. The fringe spacing is 2.4 mm. Find the wavelength of the light.
Two loudspeakers are placed side by side. Describe how to make them act as coherent sources, and explain why this matters.
Two loudspeakers 2.0 m apart are driven in phase at 850 Hz. The speed of sound is 340 m/s. Find the angle of the first loud band to the side of the centre.
You must find the wavelength of a laser. Describe a method that uses two slits, and state the three measurements you need.
Explain why the dark fringes in an interference pattern do not mean that energy has been destroyed.
A sound wave has a frequency of 340 Hz with a percentage uncertainty of 0.5%, and a wavelength of 1.00 m with a percentage uncertainty of 2.0%. Find its speed and its absolute uncertainty.
A wavelength is measured as 6.328e-7 m. Write it in nanometres, and round it to three significant figures.

Answers and concise solutions

1. The wavelength

Answer: $\lambda = \dfrac{v}{f} = \dfrac{330}{1500} = 0.22\,\text{m}$ .

2. The frequency

Answer: $f = \dfrac{v}{\lambda} = \dfrac{340}{2.5} = 136\,\text{Hz}$ .

3. The phase difference

Answer: $\Delta\phi = \dfrac{0.15}{0.60} \times 360^\circ = 90^\circ$ .

4. The sound at the point

Answer: The wavelength is $\dfrac{340}{425} = 0.80\,\text{m}$ . The path difference is $2.60 - 1.40 = 1.20\,\text{m}$ , which is one and a half wavelengths. The waves arrive antiphase, so the sound is quiet.

5. The fringe spacing

Answer: $w = \dfrac{\lambda D}{s} = \dfrac{640 \times 10^{-9} \times 2.0}{0.30 \times 10^{-3}} = 4.3 \times 10^{-3}\,\text{m}$ , or $4.3\,\text{mm}$ to two significant figures.

6. The wavelength

Answer: $\lambda = \dfrac{ws}{D} = \dfrac{2.4 \times 10^{-3} \times 0.20 \times 10^{-3}}{1.5} = 3.2 \times 10^{-7}\,\text{m}$ , or $320\,\text{nm}$ .

7. Two loudspeakers are placed side by side. Describe how to make them act as coherent sources, and explain why this matters.

Connect both loudspeakers to the same signal generator, so one signal drives both of them.
They then emit sound of the same frequency.
They also keep a constant phase difference between them, because they started together and nothing changes that.
This matters because only coherent sources give a steady interference pattern. With unconnected speakers the phase difference keeps changing and the loud and quiet places move.

8. The angle of the first loud band

Answer: The wavelength is $\dfrac{340}{850} = 0.40\,\text{m}$ . Then $\sin\theta = \dfrac{\lambda}{s} = \dfrac{0.40}{2.0} = 0.20$ , so $\theta = 11.5^\circ$ .

9. You must find the wavelength of a laser. Describe a method that uses two slits, and state the three measurements you need.

Measure the distance across several bright fringes and divide by the number of spaces, to get the fringe spacing w.
Find the separation s of the two slits, which is printed on the slide.
Measure the distance D from the slits to the screen.
Use $w = \\dfrac{\\lambda D}{s}$ , rearranged to give $\\lambda = \\dfrac{ws}{D}$ .

Answer: Measure the fringe spacing across several fringes, the slit separation and the screen distance. Then use $\lambda = \dfrac{ws}{D}$ .

10. Explain why the dark fringes in an interference pattern do not mean that energy has been destroyed.

At a dark fringe the two waves arrive antiphase and their displacements add to zero, because of the principle of superposition.
The energy is not lost at that point. It is moved to the places where the waves arrive in phase.
The energy is redistributed across the pattern, so the total energy is unchanged.
The bright fringes are brighter than either wave alone, which is where the missing energy has gone.

11. The speed and its uncertainty

Answer: The speed is $v = f\lambda = 340 \times 1.00 = 340\,\text{m}\,\text{s}^{-1}$ . The percentage uncertainties add: $0.5\% + 2.0\% = 2.5\%$ . The absolute uncertainty is $340 \times \dfrac{2.5}{100} = 8.5\,\text{m}\,\text{s}^{-1}$ .

12. The wavelength in nanometres and to three significant figures

Answer: $6.328 \times 10^{-7}\,\text{m} = 632.8\,\text{nm}$ , and to three significant figures the wavelength is $6.33 \times 10^{-7}\,\text{m}$ , or $633\,\text{nm}$ .