When aluminum foil is placed in a microwave, I see sparks generated by what I assume is dielectric breakdown. However, if I put aluminum foil in visible light (assuming the same intensity), there are no sparks generated even though visible light has a higher frequency and therefore, higher energy. Why is that?
[Physics] Why doesn’t visible light generate sparks in aluminum foil when microwaves do
microwavesvisible-light
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Mostly, you see things because they reflect light. They absorb some of it, which gives them their color, but you will also see them, if you shine infrared or ultraviolett light at them. So: Whatever light you shine at them, a large part of this light will be reflected and you can detect this light to "see" matter.
Your argumentation therefore seems backward. The most plausible idea is that most light on earth is of a given wavelength and therefore, most animals' eyes adapted to this wavelength.
More precisely, have a look at the sun's spectrum: As you can see (yellow part), the radiation is most intense in the area of the visible light. This is due to the fact that the sun is a near ideal black body of the temperature of its surface. Now, the light that reaches the surface is not all of the light of the sun, since some wavelengths are blocked by the atmosphere (red remains), which is due to the fact that elements absorb certain levels of radiation.
Now, light detection is more difficult, if there is less light (you can't see very well in the dark), hence it's easier to detect intensive light - thus it's a good idea to adjust your eyes to the area where light is most intensive.
There are a few other aspects worth mentioning though:
Note also that higher energy "light" can create other difficulties. Much of organic matter becomes transparent for gamma radiation (some even for x-rays - that's why tomographys works), which also means that it is much harder to detect x-rays with organic material, so it would be even harder to build an organic eye to "see" and make use of low intensities of gamma radiation. Still: with a good detector and enough intensive x-rays, I could probably also see a good picture of my surroundings.
The same holds in the other direction: radio waves have very long wavelengths. A simple eye is not big enough to see them.
The upshot of all of this is:
Seeing the whole spectrum requires a much larger variety of detectors, one type of "eye" will simply not be enough.
Light on earth is most abundant in a narrow band of the electromagnetic spectrum
This does not explain why we only see a certain band of the electromagnetic spectrum, unless you want biological economy.
EDIT: So why do some animals see UV and none see IR light? Unlike I previously claimed, this seems to be more a biological problem: you'd probably need a very different "eye", similarly to what I hinted at when saying we need a larger variety of detectors: The only animals with really confirmed IR vision are snakes, who don't use their eyes to "see" IR light. On the other hand, all animals with confirmed UV senses use their eyes, they have just a slightly different window of visibility shifted to the ultraviolett, or simply another type of receptors (some birds apparently have up to five different color receptors, which also spread a larger band of wavelenghts).
I did not include a more complete survey of the biology - this is, after all, a question about the physics. See also Thomas' answer for a more complete argument of some biological arguments showing that it is probably not beneficial to have multiple eyes.
EDIT 2: There were some questions added for clarification, so let me try to answer those:
Do most energy level transitions in matter of every day objects correspond quite precisely to the wavelenghts of visible light?
Answer: No they don't. Let's have a look at the emission spectrum of hydrogen, the most abundant element in the universe and also very present on earth (albeit normally bound): Hydrogen spectrum and in particular this Wikipedia picture. We can see many lines, only a few of which are visible (four lines in the Balmer series). The NIST has a database of spectral lines for every element (see http://physics.nist.gov/PhysRefData/ASD/lines_form.html), where you can see that there is an abundancy of lines that are not visible. However, I don't know how probable all those transitions are. The Balmer lines for hydrogen are of course very probable transition.
If no electronic transitions happened in the band of visible light, would we still be able to use this band to see? If no, what would be the most efficient ways to see?
Assuming that we had a device to actually detect the light in these frequencies without using electron transitions (this is more a biophysical question and beyond my capabilities): We would be able to use this band, precisely because of what I said in my original answer: Most of the light we see is reflected sunlight, not absorbed and reemitted or just emitted light. Since sunlight is abundand precisely in the visible spectrum (and this has nothing to do with the emission spectra of atoms), we would see very well. However, colours will be problematic: Sunlight is white and the colours result from an absorption of certain parts of this light, while others are simply reflected.
The absorption process is linked to the spectral lines, but I don't feel that I know enough to make this connection more precise. So it might be that the lack of any absorption in this part of the spectrum will make our world rather colourless - we'd see black and white.
The problem is that the shorter wavelengths get absorbed by anything in the middle, but micro and radiowaves (depending on the freqency), are either transparent or bounce back from the upper athmosphere. The highest frequencies (such as x rays) can pass trough objects, but they are harmful to us.
Best Answer
Most microwave ovens have a wavelength of about 12 cm. This means that you get strong electric fields over the scale of cm, giving you inductive heating and sizable potential differences over macroscopic scales. Visible light, with a wavelength of on the order of 0.00005 cm, will also move electrons around in the foil, but the resistance will be five orders of magnitude less (resistance is proportional to length in a material of constant resistivity), so you won't see such effects.