Electromagnetic Waves

Maxwell's contribution was a single term. Ampère's law as inherited was $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$, which is inconsistent with charge conservation. Take the divergence and you get $\nabla \cdot \mathbf{J}=0$, which is false whenever charge accumulates anywhere. Adding the displacement current fixes it:

$$\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{\epsilon }_0\frac{\partial \mathbf{E}}{\partial t}$$

Light, unasked for

In vacuum, take the curl of Faraday's law and substitute:

$${\nabla }^2\mathbf{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\mathbf{E}}{\partial t^2}$$

A wave equation, with speed

$$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$$

Both constants came from benchtop experiments on capacitors and current-carrying wires. Neither had anything to do with optics. The number that came out was the measured speed of light, and there was no longer any question about what light was.

Polarisation and energy

The fields are transverse and mutually perpendicular, with $\mathbf{E}$, $\mathbf{B}$, $\hat{\mathbf{k}}$ forming a right-handed set and $|\mathbf{E}|=c|\mathbf{B}|$. Energy flux is the Poynting vector:

$$\mathbf{S}=\frac{1}{{\mu }_0}\mathbf{E}\times \mathbf{B}$$

Dispersion

In a medium the speed becomes $c/n$, and $n$ depends on frequency because the bound electrons are driven oscillators with their own resonances. Model them as damped harmonic oscillators and you get

$$n^2(\omega )=1+\frac{N{e}^2}{{\epsilon }_{0}m}\sum _j\frac{f_j}{{\omega }_j^2-{\omega }^2-i{\gamma }_j\omega }$$

Away from resonance $n$ rises with frequency, so blue refracts more than red. Prisms, rainbows and chromatic aberration are all this formula.

Near resonance the medium absorbs strongly and $n$ can fall below one, so the phase velocity exceeds $c$. Nothing is violated: information travels at the group velocity, which does not.

Boundaries

Matching $\mathbf{E}$ and $\mathbf{B}$ across an interface gives the Fresnel coefficients, and with them essentially all of practical optics.

Past the critical angle for total internal reflection the transmitted wave does not vanish; it becomes evanescent, decaying exponentially into the second medium while carrying no net energy across. Bring a third surface close enough and it tunnels, the optical analogue of quantum tunnelling, and the mechanism behind fibre couplers.

Radiation needs acceleration

A static charge has a field but radiates nothing. A charge in uniform motion is a static charge in another frame, so it radiates nothing either. Only acceleration radiates, and the power follows Larmor's formula:

$$P=\frac{q^2{a}^2}{6\pi {\epsilon }_0{c}^3}$$

Dipole radiation goes as ${\sin }^2\theta$, with nothing emitted along the dipole axis. That single angular factor explains an unreasonable amount: why the sky is polarised at ninety degrees to the Sun, why Brewster's angle exists at all, and why an antenna is oriented the way it is.

Why the sky is blue, quantitatively

Scattering off particles much smaller than the wavelength scales as ${\omega }^4$. Blue light at 450 nm scatters roughly $(700/450{)}^4\approx 5.9$ times more strongly than red at 700 nm.

It is the same ${\omega }^4$ that appears in the dispersion formula above: a driven oscillator far below resonance, radiating according to Larmor. Blue sky, red sunsets and the polarisation of both follow from that one calculation.