Question

I keep seeing explanations that say a microwave can't get through the metal screen because the holes are so much smaller than the wavelength. But EM waves are transverse, so if it's true that they can' fit through the holes, then they are being blocked by their amplitude, which also can't be true.

Anyone know the real reason an EM wave can't pass through a conductor opening smaller than the wavelength?

Answer 1

An electromagnetic wave can pass through a conductor opening that's smaller than the wavelength, it's just a question of how much gets through. If the wave is incident on a solid (perfect) conductor, then you know that regardless of the wavelength it can't get through because the electric field must be zero at the boundary.

Now consider the mechanism responsible for this cancellation. What's really happening is that the electrons in the metal feel the electric field and they move to cancel it out. This movement is extremely fast, which means that you can basically say that the electric field at the boundary is zero -- however, you must bear in mind that the electric field is actively being cancelled off by furiously moving electrons.

Let's ask the question "What happens if a moving electron bumps into a hole in the conductor?" It is reasonable to assume that it would hit the hole and take a certain amount of time to move around it, right? The time required should probably also be proportional to the width of the hole, which we'll call L (if you imagine running around a lake to get to the other side, your time is proportional to the circumference, and therefore the width, of the lake).

If a bunch of electrons encounter these holes, then their oscillation frequency will be interrupted. It will go from
\[ \omega_{max} = \frac{1}{T} \text{ to } \frac{1}{T+\delta} \approx \omega_{max}(1 - \frac{\delta}{T})\]
where delta is proportional to the width of the hole. So we might instead write
\[ \omega_{max} (1 - \alpha \frac{L}{T}) \]
there we find our answer. The electric field can adequately be cancelled off by running electrons if that factor
\[\alpha \cdot \frac{L}{T} \]
is quite small compared to one. I don't know what alpha is (You could try to work it out in detail, it shouldn't be too hard) but the thing to notice is this: T is one over the oscillation frequency, which is c over the wavelength:
\[ \alpha \cdot c\cdot \frac{L}{\lambda} \]
So if the width of the holes, L, is small compared to the wavelength lambda, this factor will be small, the oscillation frequency won't be interrupted appreciably, and the field will be killed at the boundary. However, if L is comparable to the wavelength lambda, this factor may get close to 1. That means the oscillation frequency will be severely interrupted, and the electrons will not be able to move fast enough to cancel the incident electric field, which will then simply pass through.

Whew. Longest answer in a long time...

Answer 2

@Jemurray3 i do salute you for such good reasoning...........

Answer 3

microwaves have smaller wavelength ............so as wavelength is small so the no . of crests formed per unit distance would be greater...........

Answer 4

No, that's the other way around. Microwaves have long wavelengths. And you are mistaking the amplitude of the electric field for some spatial thing -- it is not. Electromagnetic waves have no "size" in the transverse direction.

Answer 5

well iam sorry for saying anything about the size of microwave's wavelengh because its relative.........................

Answer 6

well with regards to the second query i did not get it rightly what do u tend to oppose if electric field propagates in space then would not have any amplitude..........

Answer 7

It's amplitude represents the intensity of the electric field. It's not like the amplitude of a water wave, which is the water physically moving up and down.

Answer 8

Thanks, Jemurray3! My distant recollection of EM was along the same lines, but not nearly as articulated as your explanation. I think you've got the best explanation on the web right now. I really do hope that this myth about "the wavelength is too big to get through" will be replaced by something a bit more sophisticated than "It's like a bus flying through space side-ways and it can't get through the hole." I'll amplify the time you spent answering my question by getting a better answer into the hands of elementary physics teachers, and in a way that helps them learn more about EM.

Answer 9

Well, I appreciate the compliment. However, it is a fairly hand-wavy argument. Hopefully the explanation given by most others is not that the wave somehow gets "stuck" trying to get through the mesh, but rather that the wavelength of a wave is the only measure of its spatial extent, and it seems fairly obvious that in the limit as hole size goes to zero, the behavior should mimic that of a conducting wall. Thus, as the dimension of the holes shrinks relative to the dimension of the wavelength, the mesh becomes impenetrable.

Important parameters which determine physical behavior are usually dimensionless -- ratios of quantities with the same dimension. In classical Electromagnetism, the size of the holes is the only spatial parameter in the mesh, and the wavelength is the only spatial parameter in the wave -- it can be reasoned then that the quantity of interest is
\[ \frac{L}{\lambda} \]
But again, that's quite hand-wavy.


Of course, electromagnetic waves are really just photons whizzing around, and metals are predominantly empty space with loosely bound electrons floating around stationary nuclei....quantum mechanics is quite mysterious indeed. :)

Answer 10

Hey \[\Huge{\color{red}{Welcome\ to\ OpenStudy}\quad\color{green}{\ddot{\smile}}}\]

Answer 11

i do consider both the particle and wave nature of the electromagnetic wave ...........................and when i get to my conclusion the actual length of the path of the wave is greater than the wavelength and the hole which is smaller to the wavelength occupies a small region along the path where the wave can pass along so the ratio of this small length of space to that of the whole path travelled by the wave is very small and when subtracted by 1 we get the probability of the wave not passing through the whole............

Answer 12

and with regards to the consideration of mere electric and magnetic field ................if just analyse the electric field the field would not be transmitted along the conductor and it needs a cavity so again............we have the same thing......it can be compared to bouncing particle ............................too....

Answer 13

I have absolutely no idea what you're talking about, I'm sorry.