# Reflected wave at complex refrective index

1. Sep 8, 2011

### superg33k

A plane wave can be described by the real part of the exponential wave equation:

$$\mathbf{E}=E_{0}e^{i(kz-wt)}$$

Adding the subscript i or r for incident or reflected waves, the ratio of the amplitude of reflected to incident wave is given by:

$$\frac{E_{r0}}{E_{i0}} = \frac{n_1-n_2}{n_1+n_2}$$

But if n2 is complex, then this leads to a complex Er0. What does this mean for the physical wave, the real part of E?

$$\mathbf{E}=E_{r0}e^{i(kz-wt)}=(Re\{E_{r0}\}+iIm\{E_{r0}\})e^{i(kz-wt)}=Re\{E_{r0}\}e^{i(kz-wt)}+Im\{E_{r0}\}e^{i(kz-wt+\pi/2)}$$

To me it looks like 2 out of phase waves are reflected. If this is right can you point me somewhere I can read up more about it? Or have I abused some notation somewhere?

Thanks for your help understanding whats going on.

2. Sep 8, 2011

### DrDu

No, as you said, you have to take the real part of E, only. A complex E_0 can be writen as $|E_0|\exp(i\phi)$. Hence the reflected wave will be out of phase with the incident one.
For non-perpendicular incidence this leads to the reflected wave becoming partly circularly polarized and is used in elipsometry.