Properties of Electromagnetic Waves

Accelerating charges create ripples in the electromagnetic field

Switch on a mobile phone and an alternating current drives electrons back and forth along the antenna at hundreds of millions of times per second. Each time those electrons reverse direction they accelerate — and that acceleration sends electromagnetic waves rippling outward through empty space at $c = 3.00 \times 10^8$ m/s, reaching a base station kilometres away without any physical medium connecting them. Electromagnetic waves are produced whenever electric charges accelerate: the oscillating electron creates a changing electric field, which in turn creates a changing magnetic field, forming a self-sustaining wave.

The key features of an electromagnetic wave are:

• The electric field ($\vec{E}$) and magnetic field ($\vec{B}$) oscillate perpendicular to each other.
• Both fields are perpendicular to the direction of propagation — making EM waves transverse.
• The ratio of field amplitudes is fixed: $E/B = c$.

Different regions of the spectrum are produced by different mechanisms:

• Radio waves: Accelerated electrons in antenna circuits.
• Microwaves: Electron transitions in molecules (rotational) and special vacuum tubes (magnetrons).
• Infrared: Molecular vibrations and thermal motion of atoms.
• Visible light and UV: Electron transitions between energy levels in atoms.
• X-rays: Rapid deceleration of high-energy electrons (bremsstrahlung) or electron transitions in inner atomic shells.
• Gamma rays: Nuclear decay and transitions within the nucleus.

Electromagnetic waves are produced by accelerating (oscillating) charges. The electric field $\vec{E}$ and magnetic field $\vec{B}$ oscillate perpendicular to each other and to the direction of propagation — making EM waves transverse. Each changing field sustains the other ($E/B = c$), so no medium is required.

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Why signals fade with distance

We just saw that EM waves are produced by accelerating charges and require no medium. That raises a question: as those waves travel outward in all directions, how does their strength change with distance? This card answers it → the inverse square law $I = P/(4\pi r^2)$.

As an electromagnetic wave spreads out from a point source, its energy is distributed over an increasingly large area. The intensity $I$ at distance $r$ from a source of power $P$ is:

This is the inverse square law: intensity is inversely proportional to the square of the distance from the source. Doubling the distance reduces intensity to one-quarter. Tripling the distance reduces it to one-ninth.

Reflection occurs when an EM wave strikes a boundary and bounces back. The angle of incidence equals the angle of reflection. Smooth surfaces (mirrors) produce specular reflection; rough surfaces produce diffuse reflection.

Refraction occurs when an EM wave crosses a boundary between media with different refractive indices, changing direction and speed. The frequency stays constant; wavelength and speed change.

Absorption occurs when the wave's energy is transferred to the medium. Different materials absorb different wavelengths — this is why glass is transparent to visible light but opaque to UV, and why skin absorbs UV and gets sunburned.

Intensity from a point source: $I = P/(4\pi r^2)$ (W/m²). Doubling the distance reduces intensity to $1/4$ — the inverse square law. Ratio form: $I_1/I_2 = (r_2/r_1)^2$. At boundaries: reflection (angle in = angle out), refraction (frequency unchanged, speed and wavelength change), absorption (energy transfers to medium).

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Calculating intensity and safe distances

We just saw that intensity follows $I = P/(4\pi r^2)$. That raises a question: how do you apply this formula in multi-step problems — finding intensity at a given distance, solving for distance, or finding a safe distance? This card answers it → worked examples using direct substitution and the ratio method.

Three-step inverse square law method: (1) direct calculation $I = P/(4\pi r^2)$; (2) ratio form $I_1/I_2 = (r_2/r_1)^2$ when power is not given; (3) rearrange for distance $r = \sqrt{P/(4\pi I)}$. HSC trap: square the distance ratio and then invert it to get the intensity ratio.

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Avoid the most common exam errors

We just saw how to apply $I = P/(4\pi r^2)$ in worked examples. That raises a question: what are the most common mistakes students make in exam questions about the inverse square law? This card answers it → the key conceptual traps and a reminder that only point sources obey this law.

The most common error with the inverse square law is confusing the factor by which distance changes with the factor by which intensity changes.

• If distance doubles ($\times 2$), intensity becomes $1/2^2 = 1/4$.
• If distance triples ($\times 3$), intensity becomes $1/9$.
• Square the distance ratio, then invert.

Also, the inverse square law only applies to point sources radiating uniformly in all directions. Lasers and focused beams do not obey this law because their energy is collimated.

The inverse square law $I \propto 1/r^2$ applies only to point sources radiating uniformly in all directions (area $= 4\pi r^2$). Lasers and collimated beams do NOT obey it. Common trap: doubling distance reduces intensity by factor 4, not 2 — always square the distance ratio and invert.

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