When, for instance, a strip of zinc is placed in a copper sulfate solution, a thin layer of copper is formed upon the surface of the zinc. At first, this layer is black, and fairly non reflective. Once this layer becomes thicker, it becomes more reflective, taking on the characteristic coppery sheen. Do any Dopers know what causes this effect? I mean, why does adding a thin layer of reflective metal reduce the reflectivity of the metal, if that makes any sense.
-Oli
I may actually be talking right out of my butt here, but I seem to remember someone saying that (possibly only some) metals are not fully opaque to visible light. If this is true then it could be destructive interference between the top and bottom layer reflections.
I wish I had a good answer to this, but I can’t remember having seen the effect you describe. Shoulda looked closer in chemistry class.
In any event, the optics of thin films is complex, because you have to deal with the rays of light that pass through the sample to the underlying layer and reflect from what’s underneath, then come back. Yes, even metals are, in thin enough layers, transmissive. It’s just that with metals and visible light, those layers have to be very thin. Your replacement chemistry, however, is doing exactly that – making thin layers.
A thin uniform layer applied above a substrate can generate interference effects, if the path length difference is less than the coherence length of the light (which is really short for ordinary white light), and destructive interference between the front and back layers can make the part look dark. Thin film optics is an interesting and commercially important field. Usually one deals with dielectrics deposited on dielectrics. If you start using metals, then the equations become much more complicated, because your index of refraction becomes a complex number, with real and imaginary parts. It wouldn’y surprise me that it got darker.
If the above is true (thin film interference with a material having a complex index), then you’d expect the effect to go away as the copper got thicker – then virtually all of it gets either reflected or absorbed.
All of this is set out with more mathematical detail than you’d ever want in chaptert 13 of Born and Wolf’s Principles of Optics. Thet note that a discussion of the optics of thin metal films is in JOSA 37 451 and 38 483 (1947 and 1948, respectively). B&W, unfortunately, don’t give a word description of what this all looks like. rats.
Of course, I might be completely wrong. It might be that you’re putting down a layer of some zinc-copper alloy or weird compound at first that looks black, and that it all looks coppery when the pure copper starts coming out.
That would make sense, given the extremely thin nature of the layer of copper. However, even with such a thin layer, I should think the vast majority of the light should be reflected by the copper, and therefore the second reflection would be extremely faint, and the destructive inteference effect would be minimised. I must add, I know precisely nothing about this topic. It’s all speculation.
-Oli
Uh, that was in reply to the post under the OP.
-Oli
BTW, CalMeacham, really informative post. Do you have any links to info pages on refractive indices of metals etc.?
(OT: This forum is great. Its a heck of a lot more informative than other forums such as sciforums.)
-Oli
starman – not websites. I’m old school – I know which books you can find this in. The aforementioned Born and Wolf gives the real and imaginary parts for the refractive indices of 22 metals – but all based on data that’s typically about 100 years old. For copper, n = 0.62, and nk = 2.57 . For more up=to-date info, look up Palik’s three books Optical Properties of Solids.
You’re right about the high absorption quelling most interference. For most metals, say B&W, the penetration depth is less than 1/4 wave in the visible, and you want a 1/4 wave coating for destructive interference.
CalMeacham is right on to things.
Another source that is available at any Barnes and Nobel is QED by Richard Feynman. It would be a very good book to look at to understand exactly the mechanisms involved in this instance.
I don’t buy the interference theory. If it were interference, I’d expect only some colors to be absorbed, and others for which 1/4 wavelength isn’t the thickness of the copper to be reflected. Hence, I’d expect the strip to look colored, and further, change colors as the copper thickness changed. Similar to a sheen of oil on a puddle of water in a parking lot.
My WAG is that the Copper atoms right at the junction have to space themselves at the Zinc atom spacing, and that this spacing doesn’t match the spacing in bulk copper. This could make it difficult for current to flow at the surface, leading to wideband absorption of light.
The coefficients of reflection and transmission vary with frequency, but I think materials can reflect or transmit electromagnetic waves over almost the entire visible spectrum i.e. glass and mirrors. (The equation’s a bear and I’m too lazy to actually figure this out.)
If this true then when the thickness of the copper film is less than the skin depth you could very well get destructive interference, but as soon as the build up exceeds the skin depth you would then get reflection.
Conductors due to their free electrons quickly attenuate EM waves. For example the skin depth in copper for a microwave is about 10[sup]-4[/sup] mm.
Skin depth = the distance at which the amplitude of an em wave drops by e[sup]-1[/sup] of its value at the surface.
I wonder what the hell I’m talking about. Destructive interference only depends on the length of the light path through the material…and yet it still seems this effect must be caused by it. I’ll have to think about this.