So, my understanding is that the wavelength of a photon is the distance traveled in the time it takes it's magnetic field to oscillate. And it's inversely proportional to it's energy and it's frequency.

In other places I've read that the wavelength is actually volumetric, but then, how can the radio wavelength be the size of a building, or other wavelenghts (citing Wikipedia) "the size of the Universe"? Is the photon actually affecting things millions of kilometers away?? How is that even possible? I just can't understand this explanation.

Then in the other hand I have this other explanation that the wavelength is actually not a property of each single photon but of a pack of photons. But then I can't understand how can the wavelength be indirectly proportional to a particular photon energy, when it's not a property of a particular photon!

Also, if the wavelength changes with a change in speed (for example entering water), then it means that it has to be actually a proportional to the distance traveled, so I came to accept that explanation.

So, when I have a microwave oven it has a mesh that allows visible light to pass through, but not microwaves. The usual explanation given to everyone is that the mesh holes are smaller than the microwave wavelength, but larger than the visible light wavelength.

But I can't make sense of this. If the wavelength is the distance traveled by a photon, what does it have to do with the size of a hole the photon can fit through? I mean, photons move linearly and are all equal in size, so what's stopping them from actually passing through the empty space?

  • $\begingroup$ I am no expert on this, but what I remember from high school is a wave-particle duality which allows you to describe light as both a set of particles and as a wave. If you have a wavelength that means that you will also have an amplitude of that wave. My guess would be that it is this amplitude which is perhaps assumed (or proven?!) to be similar in magnitude to the wavelength. That would mean that the photons can't pass because their wave will `bump into' the mesh $\endgroup$
    – Michiel
    Feb 27, 2013 at 7:14

3 Answers 3


The classical electromagnetic waves may have many forms. The simplest examples are "monochromatic" waves with a well-defined frequency $f$ – the number of periods per second. The wavelength – the distance between two maxima of the wave – is $\lambda=c/f$.

When you change something about the electromagnetic field at one point, i.e. when you try to modify an electromagnetic wave, you can't instantly modify anything at a distant place. The information and influence can never propagate faster than $c$, the speed of light in the vacuum. This is a completely general fact about Nature that follows from the special theory of relativity and it applies to quantum theory, too. If you start with a wave of a huge wavelength $L$ and modify quickly something about the wave in a small region and in a timescale much shorter than the period of the electromagnetic wave, in a hope that this allows you to "instantly" affect distant points of space, you will fail. Instead of making a long electromagnetic wave collaborate on your project, you will create some electromagnetic waves of a much higher frequency and shorter wavelength. Too long electromagnetic waves just don't allow you to make things "really quickly" – in times shorter than $T$, their period – and too locally – in regions much shorter than the wavelength. Whenever the resolution of time or space is much better, it just proves that some electromagnetic waves of much higher frequency and shorter wavelength are present.

To describe the waves in the microwave oven, it is enough to consider classical physics i.e. avoid the notion of photons. The microwaves – electromagnetic waves whose wavelength is just a little bit shorter than the size of the oven – have a negligible chance to get through the holes because these holes are much smaller than the wavelength. The mechanism is sometimes called electromagnetic shielding. How does it work?

Well, if you want to discuss waves of wavelength $\lambda$ only, their dependence on space must always have the form $A\cdot \cos(2\pi x / \lambda)$: waves with the right distance between the minima. However, the metallic cage surrounding the interior of the oven imposes the potential $\phi=0$ at a very dense network of points. When you try to write down the potential as $A\cdot \cos(2\pi x / \lambda)$, while making sure that it's zero at all points where there is a conductor, you will find out that there is no solution except for $A=0$. The wave of the given wavelength just can't get through at all. Alternatively, you could calculate the reflection from the metallic points of the mesh (not counting the holes). They would interfere with each other and guarantee that the probability of reflection is nearly 100 percent.

In other words, the microwave-frequency photon is "really large", at least as large as the wavelength, and it just "doesn't fit in". It could fit in if it "pretended" it was a shorter-wavelength photon but that would be a different one. The microwave oven prescribes a frequency $f$ and the corresponding wavelength is always $c/f$ and cannot change. You may also look at the situation with a "poor resolution" so that you "neglect" distances shorter than the wavelength. From this viewpoint, the metallic caging of the oven looks solid despite the small holes. That's ultimately why it looks solid to the electromagnetic waves.

All these things may also be phrased in terms of photons – which is needed in quantum theory – even though, as I said, it is not necessary for microwaves because they contain a huge number of coherent photons so that this large group of photons behaves classically.

Individual photons are described by wave functions that mathematically look like an electromagnetic wave. It may be monochromatic but it may be a mixture of different frequencies, too. Individual photons' wave functions propagate through the oven or anything else pretty much just like the classical electromagnetic waves. That's why photons of microwave frequencies can't escape from the microwave, either. But the interpretation is different: the wave function isn't directly measurable like the electric fields. Instead, it encodes (after squaring) the probability density that the photon is here or there.

If you try to catch a photon whose wave function is spread over a large region, you have a certain calculable chance that you will catch it "somewhere here", in a particular nearby region $V$. But when you do so, the probability is zero that the photon is simultaneously at some other, distance place $P$. So whatever you do to the photon here – if you are lucky at all and the photon "shows up" here – will not affect what is happening very far from you. Effectively, if you see the photon here, the "wave function collapses" and you're sure that the photon isn't "over there" so it can't do anything there. The photon was never over there; it only had a chance to be there but your measurement showed that the chance didn't materialize. These points are often misunderstood because people are trying to imagine that the wave function of the photon is a real wave that must leave some traces even if the photon is ultimately seen elsewhere. But it can't and it doesn't leave any traces whatsoever: the nonzero wave function only quantified a "potential" that something may be seen somewhere. When it's seen elsewhere, it's a proof that the potential was never "real".

At any rate, you will see that photons can't be used to send superluminal signals, either.

  • $\begingroup$ The part of the reflection interfering with the rest of the wave was the part that made me understand it (I think). So if this is the case, does it means that the first photons in the wave to reach the "wall" can actually pass through the holes because there wasn't any reflected wave interfering with them? $\endgroup$
    – Zequez
    Feb 27, 2013 at 13:30
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    $\begingroup$ Dear Zequez, if the waves or photons were really monochromatic, i.e. well-defined frequency, they can't have any "beginning". The beginning means that the timing is finite $\Delta t\lt \infty$ which implies, by the uncertainty principle of a sort, that the frequency is a bit uncertain, $\Delta f \gt 1 / \Delta t$. Yes, when the waves are turned on abruptly, higher-frequency waves/photons are effectively emitted at the beginning, including those that can get through the holes. $\endgroup$ Feb 27, 2013 at 15:22

It is a tricky subject - which means you have to keep thinking about it. Short answer is - NO, the "photons" are not equal in size. The size of electromagnetic wave (which consist of many many photons) is equal roughly to the size of device which emits it. So the waves emitted by atoms are roughly atom size the waves emitted by radio stations are the size of radio station transmitter and the microwaves in the microwave oven are the size of the electronic part which emits them. Your main confusion comes from the question - How to understand a wave consisting of particle photons? Look at it as a normal wave - when the wave passes trough breakwater some of the wave gets through but the remaining part is killed, which effectively stops whole wave. Main part in electromagnetism is INTERFERENCE which means that similar to normal waves they can get through barriers but the resulting SUM does not.

SO. To understand how wave is created from "photons" we need to get nice insight into the idea of INTERFERENCE.

PS: Photons are not physically well defined objects, nobody can measure single photon or catch it. It is a very useful CONCEPTUAL idea which helps US to understand how atomic physics and lasers work. Newton, just like you, also thought light consists of photons. But he had troubles using it in optic, same way as you do.

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    $\begingroup$ But we do have single photons: web.physics.ucsb.edu/~quopt/sci_clo.pdf $\endgroup$
    – anna v
    Feb 27, 2013 at 13:25
  • $\begingroup$ Yes, there is much research on this topic. This article is an investigation in process and subject to a great debate. I was trying to talk about the "facts" established through the 20th century. Quote from the article - "A Ti: sapphire laser produces light pulses of 120-fs duration, centered at a wavelength of 780 nm. A tiny part of each pulse is split off at the beam splitter BS and then attenuated below the single-photon level, thus probabilistically preparing the input photon." $\endgroup$
    – Asphir Dom
    Feb 27, 2013 at 15:01
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    $\begingroup$ Don't we detect individual photons in cloud chambers, with photomultipliers, and scintillators? $\endgroup$
    – Qsigma
    Jan 30, 2015 at 22:58
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    $\begingroup$ A rather misleading answer IMO. "The size of electromagnetic wave is equal roughly to the size of device which emits it." An optical wavelength is not equal to the size of a laser. Nor is it close to the size of the emitting particle: a CO2 laser emits 10.6 um light, with the CO2 molecule having the size of 230 pm. And the antennas consist of atoms, too. Also, each single photon can be treated as a wave (Young experiment), so it's misleading to say that "a wave consists of many photons". $\endgroup$
    – texnic
    Jun 21, 2015 at 13:33

In microwave ovens, the front window mesh is made of metal (conductor). Waves cannot pass through holes if the diameter of holes is substantially smaller than the wavelength. Microwaves used in ovens have wavelength of around 12.6 cm and the holes in metal mesh have diameter of around 0.2 cm. A simplified explanation is that EM waves are oscillations in EM field, and the electron gas in conductor of which the mesh is made responds to those oscillations by oscillating accordingly as well; the oscillations of electrons in metal mesh generate changes in EM field that cancel out the original microwaves. EM wave attenuation works in a way that it is inversely proportional to fourth power of the $\frac{wavelength}{hole}$ ratio: in this example, $\frac{1}{(12.6/0.2)^{4}}$ = $\frac{1}{15752961}$ which means that the radiation intensity of microwaves after passing through the metal mesh is around 15700000 smaller than before passing the metal mesh.


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