# How Electromagnetic Waves Pass Through Holes

1. Dec 12, 2011

### jonlg_uk

Hi everyone I have a question that has been bothering me for a while now. I have basically built a Faraday cage using wire mesh. I know there is a relationship between the mesh hole size and the wave suppressing ability of the cage.

As I understand it an electromagnetic wave can be thought of as a series of peaks of energy. Each peak being separated from the next peak by a given wavelength...What has this got to do with the size of hole that a electromagnetic wave can travel through? Why would smaller waves, such as the 3G signal on my phone pass through the hole and larger waves not?

Can somebody explain it in layman's terms so I can form a graphical representation of what is actually happening?

I remember back in my university days, the professor was telling me how they managed to detect red LASER light when shining it through a hole of 1nm. He said it was the equivalent of squeezing an elephant through a keyhole. This got me thinking and has led my subsequent confusion.

I thank you in advance

Jon

2. Dec 12, 2011

### Blibbler

Imagine if you will a fence running alongside a road.

The fence is made of metal strips a few centimetres wide, stuck vertically into the ground, with the strips a few centimetres apart.

Now imagine standing directly in front of the fence - it is half metal, half air. A vertically polarised EM wave could pass through the gaps; a horizontally polarised EM wave couldn't, in all probability, and waves of polarizations in between could pass with varying degrees (sorry) of success. Now get closer to the fence and look along its length to its farthest point. The closer you get to it the smaller the gaps in the fence will appear to be and, sure enough, the closer the fence will become to a mirror to any EM waves you shine on it.

So the answer to your question involves EM wave polarization, conductor to gap ratio and general geometry of the cage and antennas and the ratio of wavelength to geometry.

I hope this helps a bit.

3. Dec 12, 2011

### sophiecentaur

@Blibber:
From your description of the fence (with vertical conductors) you appear to have got the polarisation thing the wrong way round. Vertically polarised waves will excite currents in the vertical conductors and these will re-radiate the wave back the way it came. This may be counter intuitive but the same thing applies to thin wire antennae - they behave as if they had a very large effective cross section to parallel polarised waves. Horizontally polarised waves, however, will not excite currents in the conductors (assuming their horizontal dimension is small compared with the wavelength) so a large proportion of the wave's energy will pass through.
The actual width of the slats, the angle of arrival and their thickness will have an effect on exactly how much HP power will pass through. Fine square mesh works better than slats, of course.

Unfortunately, when you require good screening, the small currents that will be induced around on the other side of the screen, through the holes, may radiate an inconveniently high level. It may seem small in percentage but the effect of interference on equipment tends to be more logarithmic than linear so you may still have signals leaking through at perhaps -60dB less than what you got outside and that can be an embarrassment. This is why good screened enclosures are so expensive. Doors, hatches and cable entry points need to be given special attention if you want really good screening.

4. Dec 12, 2011

### Blibbler

@sophiecentaur

Many thanks for the correction - I'll look into it.

5. Dec 12, 2011

### Drakkith

Staff Emeritus
I think what the OP is asking is more related to the wavelength and not the polarization. Why does the wavelength matter when it comes to absorbing EM radiation or letting it pass?

6. Dec 12, 2011

### sophiecentaur

He's discussing a grid and the spacing and size of the conductors, wrt wavelength, affects the way the currents will flow on the structure which will affect how much energy gets radiated on the other side.

7. Dec 12, 2011

### Drakkith

Staff Emeritus
Yes, but what does the wavelength have to do with it? Why does a shorter wavelength pass through holes smaller than a longer wavelength can?

8. Dec 13, 2011

### jonlg_uk

@ Blibbler and sophiecentaur- Thanks the reply, although it was really interesting, it never quite answered my question.
As Drakkith said, I am specifically looking for the reason why shorter wavelengths may pass through the holes and longer wavelengths do not...

9. Dec 13, 2011

### jonlg_uk

@ sophiecentaur

When I place my mobile phone inside my Faraday Cage, the signal is attenuated but it doesn't disappear completely. In cellular and most commercial applications signals are vertically polarized.

According to your explanation the reason for this is because the wavelength between each successive energy peak is greater than the horizontal distance between the vertical conductors. As a result current is not exciting in the vertical conductors, and no wave is re-radiate back the way it came, due to the fact that their horizontal distance separating the vertical conductors is much shorter than the wavelength. As a result a portion of the waves energy will pass through the cage.

Is this right? If so, what effect are the horizontal conductors having on the vertically polarized mobile phone waveform signal? Surely they should be blocking it like it like a polarized filter does?

10. Dec 13, 2011

### sophiecentaur

That diagram is just plain wrong. EM waves aren't like waves on a wiggling piece of string so the 'picket fence' analogy doesn't apply. Treat robotroom with care if that's the sort of thing they are saying. In fact, a 'polaroid' filter consists of molecules that have been stretched inside a plastic matrix and these behave like tiny wire dipoles and absorb light with a polarisation parallel to these dipoles. Thus, they block light with polarisation which is not like that diagram would suggest, at all.

The reason for the reflection of VP waves (VP means that the Electric field is vertical, btw) by vertical strips of conductor is because currents are induced in the long lengths of wire and re-radiate energy (the current is mostly induced on the side of arrival of the wave, too). It behaves just the same as a solid sheet. This is because a single (infinitely) thin wire 'intercepts' much more than its cross section might suggest; it has an effective cross section many times the wire thickness. However, if there is a big space (wrt the wavelength) between the vertical conductors then this will leave 'gaps' between the regions of influence of the wires and the screening gets less.

If there are no horizontal wires connecting the vertical conductors, HP waves will not induce any current so they will not affect the waves. The small horizontal extent of the wires would not allow any significant horizontal currents to be induced.

A mesh is more complicated but consider a mesh with very small holes compared with the RF wavelength. This will behave more or less like a sheet of metal because currents can flow along it in virtually any direction - finding their way across the surface even when they flow diagonally.

If the wavelength is very very short and the wires of a large mesh are very fine then most of the energy will go straight through as it will not interact with the conductors to produce currents. You can see through a chicken wire fence, which tells you that em of visible wavelengths gets through the holes undisturbed. For cm microwaves and a mesh size of a few cm, each hole becomes like an individual antenna. Wire dipoles and slots in metal sheets are equivalent and both behave as radio antennae. This is not very well known but an example is the antenna in your mobile phone that will almost certainly be a slot type (they don't use sticky up whip antennae any more do they?). If you take the back off some phones, you can actually see a system of L shaped slots, which is the antenna; have a look. Anyway, there are currents flowing around the inside of all these slots and energy will also be radiated towards the back of the mesh too.
Of course that's not the end of the story. Low frequency waves have a habit of inducing currents around the outside of a cage and producing nodes and antinodes. This can mean that a lot of current can be present at some points on its surface. If there happens to be a high resistance gap (seam or door) at this point, a relatively high voltage can appear, which can produce an E field inside the cage.

11. Dec 13, 2011

### Gobil

I think what the OP wants to know is the relationship between the LONGITUDINAL component of the EM wave, the WAVELENGTH, and the TRANSVERSE component of the EM wave, THE ELECTRIC FIELD.

In vacuum, light travels in the direction of propagation, at velocity c.
It has an electric (and magnetic) field point in the transverse directions.
The velocity these fields propagate in the transverse direction is.....also c!
The electric field changes sign over half a wavelength, meaning the E field also propagates half a wavelength.

Over a full wavelength, the transverse E field distance traveled is ALSO one wavelength.

So in vacuum, the extent of the transverse E field in space, is equal to the wavelength of the light. That´s why, if you want to affect this electric field (Faraday cage), you have to have a cage with gaps the order of the wavelength. Red light gets through a 1 nm gap because it is not absorbed by the gap completely, the fields are affected on both sides, but the beams are diffracted, or scattered. not necessarily absorbed.

Hoped this helps.

12. Dec 13, 2011

### Born2bwire

Building off of sophiecentaur's explanation, you can think of the reflection of the electromagnetic wave as being the result of the waves excited by the currents induced on the surface of an object by the incident wave. The secondary waves produced by these currents will cancel the incident wave on the transmission side and behave as a reflected wave on the incident side of the surface.

So if we have a solid infinite sheet of a perfect conductor and a vertically polarized plane wave strikes this sheet, then the induced currents on the surface of the sheet run in the vertical direction too. Each of these currents can be thought of as a source for the reflected wave. If we look at two adjacent current strips, we notice that the phase difference between the currents is going to be infinitesimal. The result being that the waves produced by these two strips are not going to strongly interfere with each other. However, if we take two current strips that are separated by half a wavelength, we see that the two strips are 180 degrees out of phase. The waves sourced by these two strips will strongly interfere with each other.

Now we replace the solid sheet with the vertically running wires that sophiecentaur has explained. Since the induced currents originally ran in the vertical direction, the use of a set of wires instead of a solid sheet of metal does not affect the physics of the problem. In addition, if we remove a small number of these wires in a regular fashion we also find that it does not strongly affect the behavior because the wires adjacent to the holes will act as sources that are very much like the removed wires. It is only when we remove enough wires that the adjacent wires differ by a significant amount of phase that the strength of the reflection becomes diminished.

The rule of thumb for the allowed gap is usually around a quarter of a wavelength.

So the reason is centered around the behavior of wave physics and how well the source currents induced on the wire grid can approximate the ideal current distribution needed to perfectly reflect the wave.

There are other reasons why radiation can be transmitted. The primary reason is that a metal is not a perfect conductor like we assumed above. All metals have a finite conductivity which will impede the metal's ability to generate the currents we need to perfectly reflect the wave. One consequence is that the wave is able to penetrate into the metal some distance (related to the wavelength of the wave and the conductivity of the metal). If the wire is very thin compared to the wavelength of the incident wave then some of the wave will be able to transmit through the wire directly. There are also the physical defects inherent in any Faraday cage. Even seemingly minor points like adding a hole in the cage for a feed line can create areas where induced currents can couple into the interior and radiate the fields inside. This is similar to wait sophiecentaur was describing in the end of his post.

13. Dec 13, 2011

### Drakkith

Staff Emeritus
So does the electric and magnetic field just not travel any further somehow? After the photon travels one wavelength the fields don't continue to propagate out?

14. Dec 14, 2011

### Gobil

the fields change direction over half a wavelength, so it can only travel a certain distance during the time it is going in that direction, which is one half wavelength. the next question I had was this; are they photons? they answer is, not really, they are VIRTUAL photons, which represent the forces between particles, they travel at c, and are used to describe forces between subatomic particles, they don´t have a frequency. If you think of two electrons close to each other, there is a repelling force... virtual photons are being exchanged. If you have an isolated electron, and another electron miraculously appear next to it, there will appear a force between them. the force cannot travel faster than c, and we call it virtual phtotons, so this is my interpretation. I am open to correction.

15. Dec 14, 2011

### Born2bwire

The observed electromagnetic wave is the result of the observation of a large set of photons. If I have a detector that is observing a continuous incident wave then that represents a continuous absorption of photons (the rate of the photons incident upon the detector being proportional to the magnitude of the electromagnetic fields). It is not very instructive to talk about the electromagnetic field of a single photon because the fields are the observable of the photon states and only when we have a large number of photons do we get a statistically consistent electromagnetic field. That isn't to say that we cannot have inconsistent fields in a sense. The granularity in the photons in a low power/high frequency wave gives rise to shot noise. So a well spatially localized electromagnetic pulse can be roughly correlated to a well spatially localized set of photon states.

Virtual photons do have a frequency, this is an important aspect in using virtual photons to explain the Casimir and van der Waals intermolecular forces. In terms of electromagnetic force, the virtual photons are only used as a mechanism in static fields. In the DC case, there are no real photons but one can use the model of virtual photons to explain the Coulombic force.

16. Dec 15, 2011

### jonlg_uk

Thanks for your replies, this is what I am struggling with, what do electromagnetic wave look like when travelling from a vertical antenna?

An electromagnetic wave is a traverse wave. It's wavelength is the distance between two successive peaks separated in time. The peaks arrive one after each other. How I picture the wave doesn't tie into your description.
In order to understand your description I have to picture the faraday cage being perpendicular to the wavefront rather than head on, facing the antenna.

17. Dec 15, 2011

### sophiecentaur

It doesn't "look like" anything. Any wiggly lines you may want to draw on a piece of paper are only 'graphs' to show the strength and direction of the electric and magnetic fields as they vary in space and time.
The wave that is launched from an electric dipole has its electric field parallel with the wires and, if you took an instant measurement of the fields at any one time (A photograph of the graph) they would, as you say, vary from a maximum 'up', through a zero, down to a maximum 'down' and back through zero again as you move out from the dipole. The distance between successive positive maxima is called the wavelength. In time, the wave progresses outwards and if you measured the field at any point, as it varies in time, you would again get a variation up and down and back again. The rate at which this happens is called the frequency. If you multiply the wavelength by the frequency, that will tell you the speed at which the wave progresses. NO NET FORWARD MOVEMENT of fields or matter takes place.

btw, it is bad enough trying to discuss most aspects of electromagnetism, sensibly, using photons and it's even worse if you try to describe what goes on if you want to use photons and waves at the same time. You want waves, then talk in terms of waves. Sort that out first and then let QM come into the game - but only when you're quite sure you have classical wave ideas well sussed.

18. Dec 15, 2011

### Drakkith

Staff Emeritus
So would it be safe to say the EM radiation is a "disturbance" of an EM field that propagates outward but only transfers energy and interacts through specific quanta called photons? Or is that just way off?

19. Dec 15, 2011

### sophiecentaur

I think that is absolutely right and, furthermore, there is very little else you can say, incontrovertibly, about photons. We'll be back on that other thread if we're not careful!

20. Dec 15, 2011

### Drakkith

Staff Emeritus
Interesting. That's alot less confusing to me now. Not sure if thats a good thing or a bad thing!