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?

  • $\begingroup$ ""(assuming the same intensity"" That is the problem, how do you define that "intensity"? You can have sparks from/to an aluminium foil with DC high voltage too. What about the "intensity" of the latter? $\endgroup$
    – Georg
    Commented Aug 25, 2013 at 19:04
  • 4
    $\begingroup$ I'm not sure how to respond to this. I typed "same intensity" to try to make the situations as equal as possible. I just kind of assumed that the frequency was the dominate factor, which I didn't understand. That's why I asked the question. $\endgroup$ Commented Aug 25, 2013 at 19:12
  • $\begingroup$ Light intensity is a well defined concept. Anyway, I don't think the intensity is the key to this conundrum. $\endgroup$
    – Ali
    Commented Aug 25, 2013 at 19:20
  • $\begingroup$ I am guessing whether it is related to the surface plasmon exictation. $\endgroup$
    – unsym
    Commented Aug 26, 2013 at 3:44

3 Answers 3


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.


I had to think about this for a while but I think I have a rough idea of why these two forms of EM energy are so different when it comes to metal.

Electromagnetic radiation "experiences matter" differently depending on the frequency. For example some EM energy can pass right through your body, whereas others, like light can not. How matter reacts to the energy is a subject of much study.

In general as you move higher in energy (or frequency) electrons start to get ejected this is known as ionized radiation. As you move lower in frequency the electrons are not ejected and this is called non-ionized radiation. Visible light and microwaves fall into the non-ionizing category.

In the case of the microwave the surface of metal couples directly and strongly to the EM energy and acts like an antenna building up charge and inducing currents -- absorbing much of the EM energy. This charge builds up until it gets so high in potential that it discharges (sparks).

Light when it shines on metal induces a similar effect but at much lower efficiency and power it moves electrons around but much of the energy is reflected. More notable is when the photoelectric effect is induced by certain metals and often higher frequency light.

Part of this difference in response is measured in a terms call permeability and skin depth. In electromagnetism, permeability is the measure of the ability of a material to support the formation of a magnetic field within itself. In other words, it is the degree of magnetization that a material obtains in response to an applied magnetic field. On a metal the electric current flows mainly at the "skin" of the conductor, between the outer surface and a level called the skin depth. As frequency increases the skin depth gets thinner. For aluminum at microwave frequencies (say 10 GHz, microwaves are around 2.6Ghz) it is only a 0.8 micons deep. At 10 THz, there is really no material left that light could effectively couple too because the skin depth is almost zero. Gold for example at 10THz has a skin depth of about 0.025 microns making the energy almost impossible to penetrate.

Unfortunately I'm out of time to try to elaborate more.


As far as I understand, the sparks are due to microwave discharge in air - aluminum just facilitates such discharge. As far as I remember, for a given radiation frequency, there is an optimum air density where the field required for discharge is minimum (at this optimal density, the (electron-neutral, if I am not mistaken) collision frequency is of the order of the radiation frequency). For laser radiation, this optimal density is much higher than the atmospheric density.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.