Electromagnetic spectrum and light speed

Electromagnetic waves are produced when electric charges accelerate, for example in the alternating currents of generators and motors; the changing electric field creates a changing magnetic field, and the two fields propagate together as a transverse wave.

For every electromagnetic wave in vacuum, the speed is the same constant: $c=299\,792\,458\ \mathrm{m/s}\approx 3.00\times10^8\ \mathrm{m/s}$. This single speed links wavelength and frequency through the relation $c=\lambda f$, so a higher frequency means a shorter wavelength and vice versa.

The electromagnetic spectrum is the ordered family of these waves: radio, microwave, infrared, visible, ultraviolet, X-ray, and gamma; they differ only by frequency and wavelength, not by their speed in vacuum. The visible range is a narrow band roughly from $4.0\times10^{14}$ to $7.9\times10^{14}\ \mathrm{Hz}$; beyond it, frequency and photon energy increase while wavelength decreases.

A radio station emits an electromagnetic wave with frequency $f=100\ \mathrm{MHz}$ in vacuum. Find the wavelength.

A green laser pointer has wavelength $\lambda=532\ \mathrm{nm}$ in vacuum. Calculate its frequency and the energy of one photon using $E=hf$, with $h=6.63\times10^{-34}\ \mathrm{J\cdot s}$. Identify the spectral region.

An X-ray photon has wavelength $\lambda_X=0.100\ \mathrm{nm}$, and a visible photon has wavelength $\lambda_V=500\ \mathrm{nm}$. In vacuum, which has the greater frequency and greater photon energy? Calculate the ratio $f_X/f_V$. Explain why the answer does not depend on intensity.

Safe self-study activity: open the PhET simulation “Molecules and Light” or “Radiating Charge” (or a similar interactive model of electromagnetic waves). Do not point a laser or your eyes at the Sun; make all observations in the model. In the simulation, set frequencies in the radio, microwave, infrared, visible, ultraviolet, and X-ray ranges; for each, record the frequency $f$ and the measured wavelength $\lambda$. Calculate $c=\lambda f$ and compare it with $3.00\times10^8\ \mathrm{m/s}$. Then note how the classification of the wave changes as you move between ranges.

To show that all electromagnetic waves form one continuous spectrum and travel in vacuum at the same speed $c$, and to highlight the inverse relationship between frequency and wavelength.

A horizontal logarithmic frequency axis: radio on the left, then microwave, infrared, visible, ultraviolet, X-ray, and gamma; below the axis is a wavelength scale decreasing to the right. Above the visible range is a color strip from red to violet. An inset shows a transverse wave with mutually perpendicular $\vec E$ and $\vec B$ fields and the relation $c=\lambda f$.

You tap call and your phone transmits radio waves at $1800\ \text{MHz}$ to a base station $2.0\ \text{km}$ away. The signal arrives after $\Delta t=\frac{d}{c}=\frac{2000\ \text{m}}{3.0\times10^{8}\ \text{m/s}}\approx 6.7\ \mu\text{s}.$ In air the wavelength is $\lambda=c/f=(3.0\times10^{8}\ \text{m/s})/(1.8\times10^{9}\ \text{Hz})\approx0.17\ \text{m}$, about $17\ \text{cm}$.

This exercises two key claims of the theory: every electromagnetic wave in vacuum (and practically in air) travels at the same speed $c$ whatever its frequency, and its wavelength follows from $\lambda=c/f$. The travel time does not change if the operator moves the call to another radio band.

In a thunderstorm you see the flash and hear the thunder $6.0\ \text{s}$ later. The light covers $2.0\ \text{km}$ in $t=\frac{d}{c}=\frac{2000\ \text{m}}{3.0\times10^{8}\ \text{m/s}}\approx6.7\ \mu\text{s},$ which is effectively instantaneous, while sound at $340\ \text{m/s}$ needs $2000\ \text{m}/340\ \text{m/s}\approx5.9\ \text{s}$. So a delay of $6.0\ \text{s}$ corresponds to a distance $d\approx340\ \text{m/s}\times6.0\ \text{s}\approx2.0\ \text{km}$.

It checks the light-speed value $c\approx3.0\times10^{8}\ \text{m/s}$ for the visible part of the spectrum in air: the whole measured $\approx6\ \text{s}$ gap is produced by sound, while the electromagnetic flash time is shorter than the eye's response and is treated as zero.

A kitchen microwave oven runs at $2.45\ \text{GHz}$. In air $\lambda=c/f=(3.0\times10^{8}\ \text{m/s})/(2.45\times10^{9}\ \text{Hz})\approx0.12\ \text{m}=12\ \text{cm}$. The turntable of radius $\approx0.13\ \text{m}$ carries the food through the standing-wave pattern whose hot spots are $\lambda/2\approx6\ \text{cm}$ apart. The door mesh has holes around $2\ \text{mm}=2\times10^{-3}\ \text{m}$, which is $\lambda/60$, so the waves cannot get out and are reflected back into the cavity.

This example places the microwave band inside the electromagnetic spectrum and applies $\lambda=c/f$ to a wave whose length is comparable to kitchen-sized objects; the same wavelength fixes both the spacing of the hot spots and why a $2\ \text{mm}$ mesh works as a shield.