Plane Wave at Barrier

[bananabandana]

Homework Statement

Sorry for the dull question. Problem is as shown/attached

Homework Equations

The waves in part ii) are traveling in a HIL dielectric of permittivity $\epsilon_{r}$ from $0 <z<d$ and then hit an ideal metal boundary at $z=d$.

The Attempt at a Solution

I figure this should be quite obvious! The total field in the dielectric $\vec{E_{D}}$ is the superposition of the reflected and incident waves. At $z=d$ the field parallel to the boundary must be 0 -ideal metal is ideal conductor and so electrons can move to precisely counteract the field. We also know (from applying Faraday's law to a small loop around the boundary) that the parallel electric field must be continuous across the boundary. i.e

$$ \vec{E_{d}} =E_{D}\hat{\mathbf{x}}= \bigg(E_{xi}exp[i(kz-\omega t)] + E_{xr} exp[i(-kz-\omega t) ]\bigg)\hat{\mathbf{x}} $$
[n.b - the frequency of the incident and reflected waves has to be the same, else we couldn't satisfy the boundary condition for all t,] So:
$$ E_{D}(z=d) =0 \implies E_{xi} \big[cos(kd) +isin(kd)\big] =-E_{xr} \big[cos(kd)-isin(kd)\big] $$
$$ E_{xr} = -\frac{ cos(kd)+isin(kd)}{cos(kd)-isin(kd)} E_{xi} $$
Which is obviously not the result I'm meant to get! What has gone wrong?

 

[RUber]

It looks to me that the given formula for the plane wave is for free space for z<0. So, you have the first get the expression for the wave inside the dielectric. That is where you will get the $\epsilon_r$ and $k'$ terms.