Optical mode leakage through a layer of gold

Question

The geometry of my semiconductor device is given below. The blue regions are gold, the grey ones - gallium arsenide (n-doped to $2.9 \times 10^{15} \mathrm{cm^{-3}}$). The dimensions are μm, i.e. it is 500μm wide, the bottom GaAs is 400μm thick, then a layer of gold follows, then a 15μm layer of GaAs, and a 5μm layer of Au on the top:

The thinner strip of GaAs is a laser medium where radiation of 3THz is emitted. I would like to investigate the effect of the thickness of the thin layer of gold on the optical mode profile. I simulated the profile for two thicknesses: 20nm and 200nm. Below are the plots of the normalised E field, first 2D and then a cross section along x=0 line. Arc length is simply the y axis of the first plot.

20nm

200nm

In the first (20nm) case, |E| decreases from 450V/m to 90V/m (5x). In the second (200nm) case, |E| decreases from 2800V/m to 10V/m (280x).

The properties of the materials are as follows (Drude model for GaAs, literature for Au):

+----------+--------+---------+
| Material | Im(ε)  | Re(ε)   |
+----------+--------+---------+
| Au       | 1.43e5 | -5.11e4 |
| GaAs     | -0.17  | 12.98   |
+----------+--------+---------+

The field should decrease as $\exp(-\alpha z)$ (Beer's law), where $\frac{1}{\alpha} = \delta \approx 50\mathrm{nm}$ is the skin depth for 3THz in gold - this is calculated from: $$\alpha = \frac{1}{\delta} = \frac{\kappa \omega}{c}$$ where $\kappa$ is the known value for $Im(\tilde{n}) = 319$ (Ordal et al., http://dx.doi.org/10.1364/AO.26.000744). Hence: $$E_2(20nm) = 450 \exp(-20/50) \approx 300$$ $$E_2(200nm) = 2800 \exp(-200/50) \approx 50$$

The first result is 3x higher and the second 5x higher than I see in the simulation.

To check this, I simulated the system for different thicknesses of the gold layer, and fitted the data to the equation: $$\frac{|E_2|}{|E_1|} = A \exp \left(-\frac{z}{\delta} \right)$$

I got the following plot:

And the fit yields these parameters: $$\frac{|E_2|}{|E_1|} = 0.32 \exp \left(-\frac{z}{41\mathrm{nm}} \right)$$

So $\delta=41\mathrm{nm}$ is not too far from the calculated value. What I am trying to understand now is:

Where does the factor of 0.32 come from? It obviously matters much more for lower thickness of gold, which I am interested in. Does it have something to do with the normalisation? Boundary conditions?

I have realised that I should be looking at $\mathrm{E_y}$ rather than $\mathrm{|E|}$, but I checked it and the values look the same.

COMSOL FEM software was used to visualise the problem below. The question is related - albeit different - to this one: The skin effect and the reflectivity of gold

Answer 1

I think your refractive index for gold is not correct. I just did a quick look-up at refractiveindex.info for 100 µm, using published data by Ordal et al (which I usually use for a first guess).

Seems, your real part is almost there, but you're missing the imaginary part completely. Yet, this is significant, since it determines the losses of your gold layer. From the numbers, I would guess that the gold absorbs 2/3 of the light, which leaves you with the pre-factor of 0.32 to Beer's law.

Answer 2

I think the assumption that I should see decay according to Beer's law is wrong. It would apply if I sent a wave through a thin layer of gold and looked at the transmission. In my original model though, I was dealing with a wave confined in a sub-wavelength cavity, so the electric field norm was a sum of reflections between the two layers of metal.

This is not to say there is nothing spurious with the results I got, but it is probably not the right way of checking them.