Reflected wave at complex refrective index

AI Thread Summary
A plane wave is represented by the exponential wave equation, with the amplitude ratio of reflected to incident waves defined by the refractive indices. When the second index is complex, the reflected wave's amplitude becomes complex, indicating that the physical wave consists of two out-of-phase components. This results in the reflected wave being partly circularly polarized, particularly in non-perpendicular incidence scenarios. The discussion highlights the importance of considering only the real part of the wave equation for physical interpretations. Understanding these concepts is crucial for applications like ellipsometry.
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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 what's going on.
 
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No, as you said, you have to take the real part of E, only. A complex E_0 can be written 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.
 
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