# Interaction of EM waves with media at boundaries

1. Dec 27, 2009

### barnflakes

Please see the diagram attached, I don't understand how my lecturer is resolving the E-field in the perpendicular case.

He obtains:

$$E_i = \bold{y} e^{-i\omega(t - \frac{n_1}{c} (-xsin \theta_i - z cos \theta_i))}$$

How has he obtained the $$(-xsin \theta_i - z cos \theta_i)$$ part from that diagram? What are x and z in that expression? As far as I can see, the E field is perpendicular to the plane of incidence so why does it have those components? Just to reiterate I'm talking about the E field being polarised perpendicular to the plane, so the right hand diagram.

Cheers guys.

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2. Dec 27, 2009

### Born2bwire

The general wave equation of a plane wave is of the form:

$$e^{-i\omega t + i\mathbf{k}\cdot\mathbf{r}}$$

The direction of propagation is the unit vector of \mathbf{k} and the phase velocity and frequency is related to the magnitude of \mathbf{k}. We can easily see that the wave number in your case is

$$k = \frac{\omega n_1}{c}$$

as we would expect. The rest of the equation is just \hat{k} dotted with the position vector which is described in reference to the incident angle.

Polarization has to do with the direction that the electric and magnetic field vectors point. In this case, we see that the TE case, the electric field must point along the \hat{y} direction, as it does in your given equation.

3. Dec 28, 2009

### barnflakes

Thank you born2bwire, that's explained it very well and I understand it perfectly, I think. I tried to do the same thing with the parallel case from the diagram he has provided.

Can you confirm the following equations I got:

$$E_i = (-\hat{x} cos\theta_i + \hat{z}sin \theta_i)E_{i0} e^{-i\omega(t - \frac{n_1}{c} (-xsin \theta_i - z cos \theta_i))} H_i = \hat{y}\frac{E_{i0}}{Z1} e^{-i\omega(t - \frac{n_1}{c} (-xsin \theta_i - z cos \theta_i))}$$

Similarly,

$$E_r = (-\hat{x} cos\theta_r - \hat{z}sin \theta_r)E_{r0} e^{-i\omega(t - \frac{n_1}{c} (-xsin \theta_r + z cos \theta_r))} H_r = \hat{y}\frac{E_{r0}}{Z1} e^{-i\omega(t - \frac{n_1}{c} (-xsin \theta_r + z cos \theta_r))}$$

and

$$E_t = (-\hat{x} cos\theta_t + \hat{z}sin \theta_t)E_{t0} e^{-i\omega(t - \frac{n_2}{c} (-xsin \theta_t - z cos \theta_t))} H_i = \hat{y}\frac{E_{t0}}{Z2} e^{-i\omega(t - \frac{n_2}{c} (-xsin \theta_t - z cos \theta_t))}$$

So using the fact that the tangential components of E and H are continuous across the boundary at z = 0 and t = 0, it should give

$$\frac{E_{i0}}{Z1} + \frac{E_{r0}}{Z1} = \frac{E_{t0}}{Z2}$$

and $$E_{i0}cos\theta_i + E_{r0}cos\theta_r = E_{t0}cos\theta_t$$

Have I done that correctly? I only ask because my lecturer has something different and I'm not sure if he's made the mistake or I have. Thank you.