A radio transmitter emits electromagnetic waves of frequency 1.0 × 10^8 Hz, which travel through free space at speed c = 3.00 × 10^8 m s^-1. (a) State the region of the electromagnetic spectrum in which this wave lies. [1] (b) Calculate the wavelength of this wave in free space.…

Answer: ((a)) radio waves ((b)) = 3.00 m ((c)) e.g. all electromagnetic waves travel at the same speed, c = 3.00 × 10^8 m s^-1, in free space (a vacuum), and all transfer energy without transferring matter ((d)) electromagnetic waves are transverse, so their oscillations are perpendicular to the direction of travel and can be restricted to a single plane by a polarising filter; sound waves are longitudinal, so their oscillations are already parallel to the direction of travel and there…. Part (a): Region of the electromagnetic spectrum A frequency of 1.0 × 10^8 Hz (100 MHz) lies in the radio wave region of the electromagnetic spectrum. This is a typical frequency for FM radio broadcasting. Part (b): Wavelength in free spac…

AS Level Physics 9702: EM Spectrum, Wavelength and Why Only Transverse Waves Polarise

Part (a): Region of the electromagnetic spectrum

A frequency of $1.0 \times 10^{8}\text{ Hz}$ ( $100\text{ MHz}$ ) lies in the radio wave region of the electromagnetic spectrum. This is a typical frequency for FM radio broadcasting.

Part (b): Wavelength in free space

All electromagnetic waves travel at speed $c$ in free space, so $c = f\lambda$ gives: $\lambda = \frac{c}{f} = \frac{3.00 \times 10^{8}}{1.0 \times 10^{8}} = 3.00\text{ m}$

Check (rearranging the other way): $f\lambda = 1.0 \times 10^{8} \times 3.00 = 3.00 \times 10^{8}\text{ m s}^{-1}$ , which reproduces $c$ , confirming $\lambda = \boxed{3.00}\text{ m}$ .

Part (c): A property common to the whole spectrum

Besides all being transverse waves, every electromagnetic wave, from radio waves to gamma rays, travels at the same speed, $c = 3.00 \times 10^{8}\text{ m s}^{-1}$ , in free space (a vacuum). (Equivalently: they all transfer energy without transferring matter, and can all travel through a vacuum.)

Part (d): Why electromagnetic waves can be polarised but sound cannot

Polarisation restricts the oscillations of a wave to a single plane. This is only possible for a wave whose oscillations are perpendicular to its direction of travel, because there is a whole range of possible perpendicular directions (a full circle of planes) that can be filtered down to one.

Electromagnetic waves are transverse, so their (electric and magnetic field) oscillations are perpendicular to the direction of travel, and a polarising filter can select oscillations in just one of these perpendicular planes.

Sound waves are longitudinal: the oscillations of air particles are already parallel to (along) the direction of travel, so there is no perpendicular plane to restrict, a longitudinal wave therefore cannot be polarised.

Final answers

(a) Radio waves
(b) $\lambda = \boxed{3.00}\text{ m}$
(c) All electromagnetic waves travel at speed $c = 3.00 \times 10^{8}\text{ m s}^{-1}$ in free space (accept: all transfer energy without transferring matter)
(d) Electromagnetic waves are transverse (oscillations perpendicular to travel, so can be restricted to one plane); sound is longitudinal (oscillations already parallel to travel, so cannot be restricted this way)