Chapter 5: Review

The chapter problem, solved

The observatory uses radio waves, infrared radiation and visible light. Take the speed of electromagnetic waves in a vacuum as $c = 3.0 \times 10^8\,\text{m/s}$ .

1. The radio signal. The signal has a frequency of 150 MHz.

\begin{aligned} \lambda &= \frac{c}{f} \\ &= \frac{3.0 \times 10^8}{150 \times 10^6} \\ &= 2.0\,\text{m} \end{aligned}

The radio signal has a wavelength of 2.0 m.

2. The infrared radiation. The camera detects radiation with a wavelength of 10 μm.

\begin{aligned} f &= \frac{c}{\lambda} \\ &= \frac{3.0 \times 10^8}{10 \times 10^{-6}} \\ &= 3.0 \times 10^{13}\,\text{Hz} \end{aligned}

3. The visible light. The image uses light with a wavelength of 600 nm.

\begin{aligned} f &= \frac{c}{\lambda} \\ &= \frac{3.0 \times 10^8}{600 \times 10^{-9}} \\ &= 5.0 \times 10^{14}\,\text{Hz} \end{aligned}

4. The order of the three. Put them in order of increasing frequency.

Radiation	Wavelength	Frequency
Radio	2.0 m	$1.5 \times 10^8\,\text{Hz}$
Infrared	10 μm	$3.0 \times 10^{13}\,\text{Hz}$
Visible	600 nm	$5.0 \times 10^{14}\,\text{Hz}$

All three travel at the same speed, so the longest wavelength has the lowest frequency. The order is radio, then infrared, then visible light.

5. The projected picture. The picture is 2.0 cm tall, and the image is 0.30 m tall. Change 0.30 m to 30 cm.

\begin{aligned} \text{magnification} &= \frac{30\,\text{cm}}{2.0\,\text{cm}} \\ &= 15 \end{aligned}

The image is 15 times taller than the picture.

6. The white walls. A white surface reflects most of the infrared radiation that reaches it and absorbs little. Less energy is transferred to the thermal store of the building, so the inside stays cooler than it would with black walls. A white surface is also a poor emitter, so the walls radiate little energy back.

Mixed practice — choose the method first

For each question, say which relationship or idea you will use before you calculate. Keep your units with your numbers.

The observatory radio telescope receives a signal with a frequency of 150 MHz. Calculate the wavelength.
The observatory infrared camera detects radiation with a wavelength of 10 μm. Calculate its frequency.
A visible-light image uses light with a wavelength of 600 nm. Calculate its frequency.
The visitor centre lens projects a picture 2.0 cm tall so that its image is 0.30 m tall. Calculate the magnification.
Place the radio signal, the infrared radiation and the visible light from the chapter problem in order of increasing frequency. Explain your answer.
The observatory walls are painted white. Explain why white walls keep the building cooler in sunshine than black walls would.
A sound wave travels through air at 340 m/s and has a frequency of 170 Hz. Calculate its wavelength.
A sound wave has a frequency of 50 Hz. Calculate its period.
A ball of mass 0.20 kg moves at 15 m/s. Calculate the energy in its kinetic store.
A motor transfers 200 J of energy, and 150 J is usefully transferred. Calculate the efficiency of the motor.
A pupil of mass 60 kg stands on the ground. Take $g = 9.8\,\text{N/kg}$ . Calculate the pupil's weight.
A radio wave has a wavelength of 0.15 m. Calculate its frequency.
A lens has a magnification of 8.0. An object is 3.0 mm tall. Calculate the height of the image.
Ultraviolet radiation has a wavelength of 250 nm. Calculate its frequency.
A signal has a frequency of 300 MHz. Say which relationship you will use, then calculate the wavelength.
Explain why a hot object emits radiation with a shorter peak wavelength than a cool object.

Answers and concise solutions

1. Wavelength of the radio signal

Answer: $\lambda = \dfrac{c}{f} = \dfrac{3.0 \times 10^8}{1.5 \times 10^8} = 2.0\,\text{m}$ .

2. Frequency of the infrared radiation

Answer: $f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{1.0 \times 10^{-5}} = 3.0 \times 10^{13}\,\text{Hz}$ .

3. Frequency of the visible light

Answer: $f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{6.0 \times 10^{-7}} = 5.0 \times 10^{14}\,\text{Hz}$ .

4. Magnification of the projected picture

Answer: magnification $= \dfrac{30\,\text{cm}}{2.0\,\text{cm}} = 15$ , so the image is 15 times taller than the object.

5. The order of increasing frequency

Radio, then infrared, then visible light.
All electromagnetic waves travel at the same speed, so the longer the wavelength the lower the frequency. The radio wavelength of 2.0 m is the longest, then the infrared at 10 μm, then the visible light at 600 nm.

6. Why white walls keep the building cooler

A white surface reflects most of the infrared radiation that reaches it.
It absorbs little of the radiation, so less energy is transferred to the thermal store of the building.
A poor absorber is also a poor emitter, so the walls radiate little energy back.

7. Wavelength of the sound wave

Answer: $\lambda = \dfrac{v}{f} = \dfrac{340}{170} = 2.0\,\text{m}$ . The wave equation works for sound as well as for light.

8. Period of the sound wave

Answer: $T = \dfrac{1}{f} = \dfrac{1}{50} = 0.020\,\text{s}$ , so one complete wave takes 0.020 s.

9. Energy in the kinetic store

Answer: $E_k = \tfrac{1}{2}mv^2 = 0.5 \times 0.20 \times 15^2 = 22.5\,\text{J}$ , so $23\,\text{J}$ to 2 significant figures.

10. Efficiency of the motor

Answer: efficiency $= \dfrac{\text{useful output}}{\text{total input}} = \dfrac{150}{200} = 0.75$ . This is 75%.

11. Weight of the pupil

Answer: $W = mg = 60 \times 9.8 = 588\,\text{N}$ , so $590\,\text{N}$ to 2 significant figures.

12. Frequency of the radio wave

Answer: $f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{0.15} = 2.0 \times 10^9\,\text{Hz} = 2.0\,\text{GHz}$ .

13. Height of the image

Answer: image height $= 8.0 \times 3.0 = 24\,\text{mm}$ .

14. Frequency of the ultraviolet radiation

Answer: $f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{2.5 \times 10^{-7}} = 1.2 \times 10^{15}\,\text{Hz}$ .

15. Wavelength of the signal

Answer: Use the wave equation as $\lambda = \dfrac{c}{f} = \dfrac{3.0 \times 10^8}{3.0 \times 10^8} = 1.0\,\text{m}$ .

16. Why the peak wavelength is shorter

A hotter object emits more radiation and its particles have more energy.
As the temperature rises, the wavelength of the most intense radiation becomes shorter.

What you can now do

list the groups of the electromagnetic spectrum in order of wavelength and frequency;
state that all electromagnetic waves travel at the same speed in a vacuum;
use $v = f\lambda$ for any electromagnetic wave;
describe absorption, reflection, transmission and refraction;
give a use and a danger of each group of electromagnetic waves;
draw a ray diagram for a convex lens, and describe real and virtual images;
calculate the magnification of a lens;
explain the colour of an object in terms of absorption and reflection;
explain how temperature affects the radiation an object emits.

Your relevance reflection

Return to the prompt from the chapter opener: where in your own week do you use light or another kind of electromagnetic radiation? Write one more example now that you have finished the chapter. Has the chapter changed what you notice about a screen, a remote control or a warm object in the dark?