Electromagnetic Spectrum Worked ExamplesIGCSE Physics 0625

(a) In order of increasing frequency the regions are: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. So P is microwaves and Q is ultraviolet.

(b) All electromagnetic waves have the same speed in a vacuum, so from $v = f\lambda$ the lowest frequency has the longest wavelength: radio waves, at the left-hand end.

(c) They all travel at the same speed in a vacuum, $3.0 \times 10^{8}$ m/s. (Other accepted answers: they are all transverse waves; they can all travel through a vacuum.)

A is fully correct: ultraviolet makes security ink fluoresce (detecting fake notes, security marking), and too much ultraviolet damages surface cells and eyes, leading to skin cancer and eye conditions.

B: sterilising water is an ultraviolet use, and mutation is caused by X-rays and gamma rays. The infrared hazard is skin burns.

C: intruder alarms use infrared, and skin burns are the infrared hazard. The microwave hazard is internal heating of body cells.

D: thermal imaging uses infrared, and internal heating is the microwave hazard. Gamma rays are used to sterilise food and medical equipment and cause mutation or damage to cells.

Identify each region: 1 remote control — infrared; 2 security scanner — X-rays; 3 satellite television — microwaves; 4 sterilising water — ultraviolet.

Wavelength increases from gamma towards radio: X-rays, ultraviolet, (visible), infrared, microwaves. So the order is 2, 4, 1, 3: B.

A is the same list in order of decreasing wavelength — the trap for anyone who ordered by frequency. C swaps ultraviolet and infrared, and D puts ultraviolet before X-rays.

(a) Electromagnetic waves travel at $3.0 \times 10^{8}$ m/s in a vacuum, and at approximately the same speed in air.

(b) Convert the wavelength to metres first: nano means $10^{-9}$.

Write the wave equation and rearrange for $f$.

Evaluate, to 2 significant figures to match the data.

(c) Monochromatic light has a single frequency.

(a) Convert the frequency to hertz: giga means $10^{9}$.

Rearrange $v = f\lambda$ for the wavelength and substitute.

Evaluate.

(b) Microwaves can penetrate some walls, so the signal reaches other rooms, and they only need a short aerial for transmission and reception, so the aerial fits inside a phone or router.

(a) Communication with artificial satellites is mainly by microwaves.

(b) Direct broadcast satellite television uses the geostationary satellite, G.

(c) A geostationary satellite orbits in the same direction as the Earth rotates and takes the same time (24 hours) to go round, so it stays above the same point on the Earth's surface. The dish never has to track it.

(d) Some satellite phones use low-orbit satellites. (Some satellite phones use geostationary satellites too.)

(a) X is analogue; Y is digital.

(b) An analogue signal can take any value within a range and varies continuously. A digital signal has only a few set values — here two, high (1) and low (0).

(c) The noise on Z is small compared with the gap between the two levels, so every part of Z is still clearly high or clearly low. A regenerator reads each level and sends out a fresh, clean pulse, removing the noise completely and restoring Y.

An analogue signal can have any value, so the circuit cannot tell which variations are noise and which are the real signal. Amplifying it amplifies the noise too.

(d) Digital signalling gives an increased rate of transmission of data, and an increased range, because the signal can be regenerated accurately again and again along the way.

All electromagnetic waves travel at the same speed in a vacuum, $3.0 \times 10^{8}$ m/s, whatever their frequency or wavelength. The pulses cover the same distance at the same speed, so they arrive together: C.

A and B assume that frequency or wavelength affects the speed. In $v = f\lambda$ a higher frequency simply goes with a shorter wavelength; the product $v$ stays the same. D invents a different speed for infrared.

(a) Bluetooth uses radio waves.

(b) Radio waves pass through walls, but the signal is weakened each time it does so. After two walls (and the extra distance) the signal is too weak for the headphones to receive.

(c) Radio waves are electromagnetic, so they travel at $3.0 \times 10^{8}$ m/s in air (not the speed of sound).

(d) Glass is transparent to visible light and some infrared, so little is absorbed along the cable, and visible light and short-wavelength infrared can carry high rates of data.

(a) Measure across as many whole waves as possible to reduce reading error. The wave passes through zero going upwards at $0$ cm and again at $5.6$ cm, which is two complete wavelengths.

Check with crests: $0.7$ cm to $3.5$ cm is also $2.8$ cm.

(b) Use $v = f\lambda$ with the wavelength in metres.

(a) Total distance: up and down, converted to metres.

The signal is a microwave, so it travels at $3.0 \times 10^{8}$ m/s. Use time $=$ distance ÷ speed.

(b) Repeat for the low-orbit satellite.

(c) The delay is about $65$ times shorter, so there is no noticeable pause in a conversation (a quarter-second pause each way via a geostationary satellite is easy to hear). Being much closer, the satellite also receives a stronger signal from a small handset.

(a) Each small tick is $10$ nm. P is $4$ ticks past $400$ nm and Q is $1$ tick past $650$ nm.

(b) P is blue: blue light has a shorter wavelength than red light, and P has the shorter wavelength. It is the difference in wavelength (and frequency) that makes the colours different.

(c) Convert to metres and use $f = v / \lambda$.

Repeat for Q.

(d) Use the unrounded values. The frequency ratio equals the inverse of the wavelength ratio, because $f \propto 1/\lambda$ when the speed is fixed.