The question:
How would you make a semi transparent mirror (50% reflection, 50% transmission) with glass with a layer of copper. For light $\lambda$ = 500nm Try to be as realistic as possible
What I've done so far.
The reflection coefficient of copper for 500nm light is 50%. So that's great. But I'm not sure whether that's enough. (It is an exam question, so I guess not). I'm guessing it'll have to do something with skin depth, but I'm not sure.
Any suggestions on how to continue ?
To a very good approximation the transmission of a metal film falls exponentially with thickness i.e.:
$$ T = e^{-\alpha t}$$
where $\alpha$ is the absorption coefficient given on the web site Alexander mentioned, http://refractiveindex.info/?group=METALS&material=Copper, and at 500nm wavelength this gives $\alpha = 6.4297\times 10^5/cm$. So you just have to solve for $T = 0.5$.
If you want to do the calculation properly it turns into a bit of a nightmare. By one of those strange co-incidences I did exactly this calculation as part of my PhD, and even more amazingly I have my thesis to hand. The reference I used for the calculation was O. S. Heavens, Optical Properties of Thin Solid Films, Butterworths Scientific Publications, London 1955. It's on Google Books here, but annoyingly hasn't been scanned so you can't see the contents.
I'll copy the equation for the optical transmission from my thesis, but I imagine one look will make you run for cover. I compared the full calculation to the simple exponential formula and agreement was basically perfect except at very small film thicknesses (below about 5nm) but in any case the metal films break up into islands at these thicknesses so the equation doesn't really apply.
$$ T= n_s \frac{((1 + g_1)^2 + h_1^2)((1 + g_2)^2 + h_2^2)}{e^{2\alpha} + (g_1^2 + h_1^2)(g_2^2 + h_2^2)e^{-2\alpha} + Ccos(2\gamma) + Dsin(2\gamma)}$$
where:
$$ C = 2(g_1g_2 - h_1h_2)$$ $$ D = 2(g_1h_2 + g_2h_1)$$ $$ g_1 = \frac{1 - n^2 - k^2}{(1+n)^2 + k^2} $$ $$ g_2 = \frac{n^2 - n_s^2 + k^2}{(n+n_s)^2 + k^2} $$ $$ h_1 = \frac{2k}{(1+n)^2 + k^2} $$ $$ h_2 = \frac{-2n_sk}{(n+n_s)^2 + k^2} $$ $$ a = \frac{2\pi k d}{\lambda} $$ $$ \gamma = \frac{2\pi n d}{\lambda} $$
where $k$ and $n$ are the extinction co-efficient and refractive index of the metal film and $n_s$ is the refractive index of the glass substrate. The film thickness is $d$ and the light wavelength is $\lambda$. If you do a sample calculation for some test film thickness you'll probably find most of the terms are approximately zero or unity, which is why it approximates to an exponential equation in the film thickness.
Bear in mind that I've hand copied this from my thesis, so there may be transcription errors lurking as traps for the unwary.
There's a reflection on the front interface, then absorption within the copper, then reflection on the back interface. Some light survives this and is transmitted through the copper. But the light reflected on the back interface travels backwards through the copper, is absorbed some, reflects on the front interface, is absorbed more, and what doesn't reflect on the back interface is also transmitted. This contribution interferes with the directly transmitted light, possibly constructively. What's more, this process happens an infinite number of times.
I expect your exam question wants you to solve this problem.
