Solution

The method of the fictitious wave for a moving reflector requires adapting the boundary condition. For a perfect metal at rest, the fictitious wave is chosen so that the total tangential electric field vanishes at the surface. If the metal moves with velocity v , the perfect conductor condition must be satisfied in the reference frame where the metal is at rest. In that frame, the tangential electric field vanishes:

$\mathbf{E}'_{\text{tan}} = 0$

By transforming this condition to the laboratory system (CGS) using the field transformation laws to first order in $v/c$ one obtains:

$\mathbf{E}_{\text{tan}} + \frac{\mathbf{v}}{c} \times \mathbf{B} = \mathbf{0}$

For a plane wave incident normally, $\mathbf{v}, \mathbf{E}, and \mathbf{B}$ are mutually perpendicular. At the surface, the superposition of the incident wave and the fictitious wave (which will become the real reflected wave upon exiting the metal) must satisfy the above relation. From it, it follows that the amplitudes of the combined fields satisfy:

$E = \frac{v}{c}\,B$

This is precisely the condition mentioned: the intensity of the resulting electric field is$v/c$times smaller than the magnetic induction.