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 ?

  • $\begingroup$ A realistic model would require considering all reflection/absorption processes and realistic and 100% efficiency does not go well together. Maybe you can get an approximate model together from the refractive index and the absorption coefficient (refractiveindex.info/?group=METALS&material=Copper)? $\endgroup$
    – Alexander
    Commented Jun 10, 2012 at 21:13
  • $\begingroup$ Are you sure it's "50% reflection 50% transmission" not "50% reflection 50% absorption by copper"? For the latter, you would use thick copper and need to figure out the appropriate glass thickness to put on top. It seems like a more plausible and doable exam question. $\endgroup$ Commented Jun 11, 2012 at 13:08
  • $\begingroup$ I cited the examquestion. (And btw, the copper layer must be on top of the glass plate) $\endgroup$
    – tgoossens
    Commented Jun 11, 2012 at 14:46

2 Answers 2


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.

$$ T= n_s \frac{((1 + g_1)^2 + h_1^2)((1 + g_2)^2 + h_2^2)}{e^{2\alpha} + (g_1^2 + h_1^2)(g_2^2 + h_2^2)e^{-2\alpha} + Ccos(2\gamma) + Dsin(2\gamma)}$$


$$ C = 2(g_1g_2 - h_1h_2)$$ $$ D = 2(g_1h_2 + g_2h_1)$$ $$ g_1 = \frac{1 - n^2 - k^2}{(1+n)^2 + k^2} $$ $$ g_2 = \frac{n^2 - n_s^2 + k^2}{(n+n_s)^2 + k^2} $$ $$ h_1 = \frac{2k}{(1+n)^2 + k^2} $$ $$ h_2 = \frac{-2n_sk}{(n+n_s)^2 + k^2} $$ $$ a = \frac{2\pi k d}{\lambda} $$ $$ \gamma = \frac{2\pi n d}{\lambda} $$

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.

  • $\begingroup$ Thanks. I'm sure they don't expect us to do it in such detail. (It's not a specializing course). But you gave me rough ideas thanks! $\endgroup$
    – tgoossens
    Commented Jun 11, 2012 at 10:10
  • $\begingroup$ And you helped a lot of my collega students as well :) $\endgroup$
    – tgoossens
    Commented Jun 11, 2012 at 10:34
  • $\begingroup$ What is the proper calculation? The rough calculation uses currents proportional to the electric fields I presume, does the proper calculation include a phase shift and a delay, or does it do Feynman diagrams for the current carriers? $\endgroup$
    – Ron Maimon
    Commented Jun 12, 2012 at 2:38
  • $\begingroup$ The calculation I described uses refractive index and absorption co-efficients measured experimentally. It doesn't attempt to calculate the values from first principles. The complexity arises, as ptomato says, from having to consider reflections from each interface and the resulting interferance. It's really not that hard; it's just that the algebra quickly gets messy, which is why the calculation I quoted for even just an air-metal-glass-air composite looks so lengthy. $\endgroup$ Commented Jun 12, 2012 at 5:54

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.


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