 Solution

The dispersion relation is $\omega =ck$. At a perfect conductor the tangential electric field and normal magnetic field vanish. Thus $E_{ref}=-E_0\hat{x}{e}^{i(-kz+\omega t)}$ and $B_{ref}=-E_0\hat{y}{e}^{i(-kz+\omega t)}/c$. At the surface the magnetic field is $-2E_0\hat{y}/c$, so with outward normal $\hat{z}$ the surface current is $K=2E_0\hat{x}/({\mu }_{0}c)$ (real parts understood).