# Physical reason why the wavelength defines the minimum sized opening an electromagnetic wave can pass through? [duplicate]

I'm having a hard time understanding why wavelength restricts a wave from passing through a hole smaller than that wavelength. For example on a microwave, the front grating prevents the microwaves from escaping but allows visible light to pass through.

The picture I have in my mind is a string in space representing a single photon, this string has a wave propagating down it with wavelength, lets say 1 meter, but the amplitude of the wave is 1 mm. So even with a wavelength of 1 meter it seems to me that the string can easily fit through any hole of size 1 mm or greater. I'm sure this analogy is wrong, but I don't understand why. Could somebody please explain to me the physical reason why the wavelength defines the minimum sized opening an electromagnetic wave can pass through and why my analogy doesn't work?

## marked as duplicate by Rob Jeffries, Kyle Kanos, user36790, Sebastian Riese, MartinNov 17 '15 at 14:08

• Maybe this can help until someone writes a more thorough answer. The amplitude of an electromagnetic wave is more or less irrelevant, because it's not a length. Therefore, you can't compare it to the size of a hole. The only quantity with dimensions of length is the wavelength. – Javier Nov 15 '15 at 19:09
• You may find the answers in here useful: physics.stackexchange.com/q/154648 – Phonon Nov 15 '15 at 20:04
• @jodag please read it carefully again. I never even speak of amplitudes!! The wavelength is the spatial periodicity of the wave, whereas the amplitude (just a compression or expansion of the wave along the coordinate axes) is rather arbitrary, i.e. it can take whatever value, the wave will still behave the same. Take e.g. two harmonics $y=\sin(3x+\frac{\pi}{3})$ and $y'=A\sin(3x+\frac{\pi}{3}),$ they both have the initial phase of $\pi/3,$ period of $2\pi/3$ and shifted on the x-axis by $-\pi/9$ regardless of their amplitudes. – Phonon Nov 15 '15 at 21:20
• Possible duplicate of Diffraction by small holes – Rob Jeffries Nov 15 '15 at 21:51
• – Rob Jeffries Nov 15 '15 at 21:52

You need to distinguish between conductors and non-conductors. If your material is non-conducting then the EM radiation can pass through any size of hole regardless of whether the hole is larger or smaller than the wavelength. The power transmitted is just the incident power per unit are multiplied by the hole area, exactly as you would expect. All very boring really.

But you mention the screen in a microwave, and in this case the screen is conducting, which completely changes the behaviour. The incident EM wave induces oscillations in the metal of the screen, and these oscillations reradiate EM that interferes with the incident wave. It's this process that blocks the incident wave. The process is purely classical and requires no appeal to the existance of photons.

Conducting screens like the screen in a microwave are generically known as Faraday screens or more commonly Faraday cages. Calculating the relationship between the hole size and the transmitted intensity is a somewhat tortuous business, but as it happens there is an excellent description of the calculation here on thi site in David's answer to What is the relationship between Faraday cage mesh size and attenuation of cell phone reception signals?.

The problem may be that you you are thinking about a string. It may also be that you are mixing classical and quantum views of physics. Never the less, let's try some visual images.

Let's move to 2 dimensions. A classical electric field at a point tells you which about the force on a test charge at that point. You can represent that as an arrow. If you put an arrow on every point of a plane, you might get something resembling a field of wheat. An electric field fills space. So you really have to imagine stack of wheat fields. But 2D is useful to get the idea.

Now imagine a gusty wind blowing through the field, disturbing the arrows this way and that. This gives the idea of an electromagnetic wave. Textbooks talk mostly about waves that oscillate regularly back and forth because they are the simplest to analyze and because waves often are like that. But they don't have to be regular.

For an electromagnetic wave, the "wind" is really changing electric and magnetic fields generated by moving charges. Most moving charges are electrons, and most electrons are found in atoms. You need quantum mechanics to describe how electrons in atoms behave, particularly if you want to think about a single photon. A simple image of a vibrating particle doesn't really work.

Likewise, the idea of an electromagnetic wave changes in quantum mechanics. A single atom can emit a photon. Sometime later, a distant atom might be affected by the photon. The atom next to it won't feel anything. In quantum mechanics, waves describe probabilities. They tell you where atoms are likely to be affected and where they are likely to not to be.

The classical electric field works when you have lots of photons. Regions where the probability is high receive more of photons. Test charges there feel a larger force. The intensity of the electromagnetic wave is high. The intensity of light is high. Regions where the probability is low are darker.

Going back to the gusty wind, one might ask how quickly the wind can shift, so that wheat blows left then right then left again. A related question is how far apart are stalks bent left and bent right. These questions correspond to the frequency and wavelength of the electromagnetic wave.

One of the rules of quantum mechanics is that high frequencies require high energies to generate. I can't give any better reason for it than that is how the universe works.

Microwaves are generated by low energy processes. They have low frequencies and long wavelengths. Visible light are higher energy, higher frequency, and shorter wavelength.

Going back to a classical picture, the grating on a microwave oven is made of strips of electrical conductor with insulator in between. Electrons can flow freely in a conductor, but are held in place in an insulator.

When an electromagnetic wave hits a conductor, the forces on the free electrons cause them to vibrate. The wave is absorbed when this happens. The vibrating electrons emit a new electromagnetic wave.

In a smooth mirror, the emitted wave is just like the incident wave, except that it goes in a different direction.

If the mirror has holes in it, you have to add up parts of the wave from all parts of the mirror and holes to find out much is transmitted and reflected. When adding waves together, keep in mind that they reinforce when in phase and cancel when out of phase.

For gratings with holes smaller than the wavelength, it works out that very little gets through the grating. Everything is cancelled.

It seems logical that a photon has to have a cross-section. But since the QED is applied to everything and not only to inner nucleus interactions, the photon as an indivisible particle is "unfashionable". Now a photon is a excitation in a global electromagnetic field. This field is endless and so even from methodological side it is not possible to search for cross-section.

That each photon has a centre is without doubt, only in this case photons are able to knock out electrons from metals. I never have seen an answer to your question, or a vague only like "slit width has to be x times of the wavelength".

Furthermore it is displaced that intensity distributions appear behind every edge, not only behind slits. So the slits width plays a role for letting through light and the slits width is important for the quality of the intensity distribution. Take two edges, you get in any case two high quality intensity distributions behind every edge. Move the edges closer together and you get a slit. During the movement the intensity pattern overlay each other and at some time one get a nice pattern and at some time a smeared pattern. This will help you to understand, how the phenomeneon works in detail. But to say something about the cross-section of photons would be possible only if the photon has finit electric and magnetic field components.