Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 17G / Solution

In SI units, the source-free Maxwell equations in vacuum areUse complex amplitudes with phase $e^{i(\omega t-k\cdot x)}$ and take real parts at the end. The electric divergence equation gives $k\cdot E_0=0$. Faraday's equation gives $k\times E_0=\omega B_0$, soThe magnetic divergence equation follows because $k\cdot (k\times E_0)=0$. Ampere's equation becomes $k\times B_0=-{\mu }_0{ϵ}_0\omega E_0$. Since $k\times (k\times E_0)=-|k{|}^2{E}_0$, a nonzero wave satisfiesThus both fields are transverse, mutually perpendicular for linear polarization, and obey the vacuum plane-wave dispersion relation.

For the interface let $n=e_z$ point from the minus side to the plus side and write $[E]=E_{+}-E_{-}$, similarly for $B$. Applying the integral Maxwell equations to a thin pillbox and a thin loop gives the electromagnetic boundary conditionsHere $j$ is the tangential surface current. The jump notation fixes the signs, including the direction of the normal.

For the perfect conductor, take the time-varying interior fields to vanish. Choose positive $k$ with $\omega =ck$. The incident electric field has complex form $E_0{e}_x{e}^{i\omega t+ikz}$. To cancel its tangential value at the surface, use the reflected field $-E_0{e}_x{e}^{i\omega t-ikz}$, whose wavevector points upward. The total vacuum fields are thereforeThe sign of each magnetic contribution follows from $B_0=k\times E_0/\omega$. At $z=0$, the electric field is zero and the magnetic field is tangential. The jump conditions yieldFor a real incident amplitude this is $j=(2E_0/{\mu }_{0}c)\cos (\omega t)e_x$. It supplies the reflected wave while satisfying the normal-field and tangential-electric conditions. Any independently imposed static interior field is separate from this oscillatory response.