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Plane Wave at Barrier

  • #1

Homework Statement


Sorry for the dull question. Problem is as shown/attached
Screen Shot 2016-05-28 at 12.10.47.png


Homework Equations


The waves in part ii) are travelling 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?
 

Answers and Replies

  • #2
RUber
Homework Helper
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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.
 

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