uses for a reflector that is too thin for the wavelenght being reflected? - Page 1
Forum section: electronics / rf-microwave

author: coppercone2
author_rank: Super Contributor
timestamp: March 26, 2025, 03:51:36 am
content: So you can make some metal coatings very thin, much thinner then the wavelength.To me this seems like a partial mirror.Does it have any uses for RF?It reminds me of a partial mirror for lasers, but those seem to be built differently.ITO glass seems like it fits the description..But are these structures used for anything other then shielding? I.e. in the case of a glass display EMC shield, you would ideally want it to be 100% reflective like solid metal. But are there any devices that use it on purpose, where its not just a limitation?

author: MarkT
author_rank: Frequent Contributor
timestamp: March 26, 2025, 09:03:18 pm
content: The skin depth is the relevant comparison, not the free wavelength. Or do you mean that?

author: coppercone2
author_rank: Super Contributor
timestamp: March 26, 2025, 11:58:52 pm
content: I am not sure, I guess skin depth?I thought something like membranes inside of a wave guide. People describe the effect as 'leakage' in terms of shields.I forgot about masers. I wonder if they use something like that.. and it looks like they don't.

author: TimFox
author_rank: Super Contributor
timestamp: March 27, 2025, 03:06:00 am
content: Within the reflector layer, the wave is "evanescent": a real exponential decay function rather than a propagating sinusoid, with the amplitude at the output exciting a traveling wave in the next layer.

author: coppercone2
author_rank: Super Contributor
timestamp: March 27, 2025, 03:25:32 am
content: I see the little drawing. does that drawing tell you the output power of the wave that goes through the barrier based on how thick it is? So that its just that the exponential means that the coupler behavior is not linear with thickness?That graph is confusing as hell because it makes you think of phase. What happens to the phase. its like compressed in the membrane for the part that gets through? For the ones that get through? How about the part that bounces off? Does it bounce off all on the surface or doese it disperse?You know what im saying because it looks like its being distorted.. something with velocity factor"?some stupid crime robot attacked my picture. No one wants to pay or use your server, arvixthat one is for a particle though, but there is similar ones for waves, but it all looks similar, but I can't find another pic that shows the wave form after it exits the material, so you need to deal with the particle one.********But that's what im understanding so far. It hits a barrier, depending on how thick the barrier is, there is a attenuation that is proportional to the thickness of the barrier as per a exponential equation. Then it leaves the barrier as a wave with lower amplitude.If you draw two circuits, of the same length, one of which has a thin membrane, and you draw the wave forms on top of each other, is there any frequency distortion (as small as it may be) over the area which the membrane occupies? (I think the answer has to be yes?) Is the attenuated wave when it leaves the membrane perfectly in phase with the wave in the circuit without the membrane in the same position? I know its thin. But assume you can measure 0.000000000000000001 degree phase shift. (like its smaller but is it shifted?)

author: radiolistener
author_rank: Super Contributor
timestamp: March 27, 2025, 11:15:34 pm
content: A thin metal plate provides limited shielding, allowing a greater portion of the wave to pass through with minimal attenuation. The portion of the wave absorbed by the metal plate is partially dissipated as heat, while the remainder is re-emitted. The re-emitted wave overlaps with the direct wave behind the metal plate in antiphase, reducing its amplitude and creating the shielding effect.However, due to this mechanism, it is impossible to create a perfect shield for a pulse. The initial phase of the pulse passes through the metal plate as if through a transparent material, with shielding beginning only when the plate starts re-emitting the wave in antiphase. As a result, any shield appears transparent to microwave pulses, making complete shielding unachievable. Conversely, a thinner metal plate is more transparent to high-frequency pulse, as it provides less impedance to the initial wavefront.In the near-field zone (within one wavelength), directivity is negligible, allowing the wave to be captured from any side of the metal plate. Effective shielding primarily occurs in the far-field zone.

author: coppercone2
author_rank: Super Contributor
timestamp: March 28, 2025, 01:58:21 am
content: When you say antiphase you mean if the wave hits the plate at 90 and the plate is thin enough the wave will come out of the plate at exactly -90? Even if the plate is varied in thickness? So the phase is not proportional to the thickness (within a certain range?).. that it is essentially a 180 degree phase shifter between say 1% to 99% of the thickness in the exponential region?And it still does it if the wave hits at 0 degrees or 180 degrees? It sounds like a teleporter

author: radiolistener
author_rank: Super Contributor
timestamp: March 28, 2025, 02:41:24 am
content: No, that’s not what I meant. In reality, a metal plate does not directly affect electromagnetic waves - they pass through the metal regardless of thickness. However, as the electromagnetic wave passes through the metal, it causes oscillations of charged particles (free electrons in the metal). These oscillations result in the emission of a new wave with some delay. This re-emitted wave propagates in all directions: both toward the source of the original wave and in the direction the wave was traveling and in all other directions. Due to the slight delay in the re-emission of the wave, the new wave traveling in the same direction as the original wave ends up being in antiphase with it and synchronized with it in terms of phase.As a result, through wave superposition, the total amplitude of the waves decreases behind the metal plate, which is the shielding effect. While for the wave emitted by the metal plate which moves in the opposite direction toward the source, there is no amplitude reduction in that direction. Behind the plate, however, the waves superimpose, leading to a decrease in the total amplitude in that region. This causes the shielding effect.Additionally, the waves re-emitted by the metal plate propagate in all directions, including toward the source. Thus, the superposition of the waves from the metal plate and the original wave results in interference, which leads to the reflection of the wave. The angle of the reflected wave is due to the superposition of the source wave and the waves from the metal plate, creating the appearance of the wave being reflected at a specific angle. Essentially, the reflection is a result of the interference between the two sets of waves, not a physical reflection as one might expect from a smooth surface.This is why the near-field region has different properties than the far-field region. In the near field, the waves have not yet fully formed due to interference between the waves from the source and the plate. Therefore, in the near field, it is not accurate to speak of reflection or shielding, as these effects only become noticeable in the far field, where the interference has already created the final pattern of the summed waves.As you can see, the initial pulse edge passes through the metal sheet without any shielding. Shielding begins to take effect with a delay. Therefore, for the rise edge of a high-frequency pulse, the metal sheet is essentially transparent, as this initial edge passes through the screen as if it weren’t there. Shielding starts to work only after a delay.The delay in the re-emission of waves by charged particles introduces a complex component into the impedance of the medium containing such particles. Unlike in a vacuum, where there are no free charges, the wave impedance of the vacuum is 377Ω with no reactive component. The presence of charged particles in the medium makes the impedance of the medium complex, as these particles interact with the electromagnetic field, causing a delay in their response and adding a reactive component to the overall impedance of medium.The delay in the response of charged particles gives rise to the reactive components in capacitors and inductors. In capacitors, this manifests as the displacement current due to the movement of charge, while in inductors, it appears as the opposition to changes in current caused by the induced magnetic field. These reactive components are a direct result of the interaction between the electromagnetic field and the charged particles within the medium.

author: coppercone2
author_rank: Super Contributor
timestamp: March 28, 2025, 03:02:04 am
content: I like the extra details but I am confused now. It sounds like it IS a 180 degree phase shifter. To me that is what anti phase is. At least a end result is this???Or do you mean the mechanism of attenuation is anti-phase annihilation? That all wave goes through but the majority of it is annihilated on the other side of the plate?So is it that the attenuation is caused by anti phase annihilation and the phase shift of the attenuated (mostly canceled) wave is just the result of the delay from speed of light in the medium that it had to pass through?

author: radiolistener
author_rank: Super Contributor
timestamp: March 28, 2025, 03:09:06 am
content: Quote from: coppercone2 on March 28, 2025, 03:02:04 amWhy 180 degrees? The phase shift can be different depending on the material properties. A 180-degree phase shift of the re-emitted wave would cancel out the original wave, but in most cases, it results in a phase shift that depends on the medium’s characteristics. This phase shift is what causes the wave propagation speed in the medium with charged particles to be slower than in a vacuum.

author: coppercone2
author_rank: Super Contributor
timestamp: March 28, 2025, 03:10:48 am
content: antiphase means 180 degrees commonly, and it leads to canceling/wave annihilation if 'mixed'i.e. i heard a inverting amp called a antiphase amplike if I imagine a wall being hit by a wave in the start of a oscillation the other side will be in the 2nd half of a oscillation if its antiphase

author: radiolistener
author_rank: Super Contributor
timestamp: March 28, 2025, 03:19:56 am
content: The superposition of a wave with itself, where one of the waves is shifted by a certain angle, results in a change in the amplitude of the resulting wave. If the shift is 180 degrees (in antiphase), the waves completely cancel each other out. For smaller phase shifts, there is partial attenuation of the resulting wave’s amplitude, as the waves do not fully align, but their interaction still leads to a decrease in the total amplitude. This phenomenon is called interference, and it depends on the magnitude of the phase shift between the waves.Quote from: coppercone2 on March 28, 2025, 03:10:48 amThe key point here is that for the wave's rise edge, the wall does not have an immediate effect - it will pass through without noticing it, like it is transparent. Re-emission will begin after a certain delay, and it is at this point that the shielding effect starts to occur. To put it simply, if we create a high-frequency pulse using a laser and place a metal plate in the path of the laser, with using very high frequency capturing, we would observe that the initial front of the pulse passes through the metal plate as though the plate were transparent. Only after a some time delay would the shielding effect of the metal plate begin to attenuate the laser beam.

author: coppercone2
author_rank: Super Contributor
timestamp: March 28, 2025, 03:21:47 am
content: its also called wave annihilation (but that is more popular for things like matter/antimatter) or canceling alot tooTo me it sounds like you are saying super position stuff is why its attenuated? That instead of going into heat like a resistor it all goes through bt its canceled depending on thickness?I think I need a drawing with a graph scale (and perhaps a really magnified portion of the graph redrawn near the thin barrier) to understand because its confusing in text formI think I am interested in a steady state solution (like a CW wave thats been running for a while), it sounds like there might be something more confusing going on with just pulses (transient analysis)?

author: radiolistener
author_rank: Super Contributor
timestamp: March 28, 2025, 03:40:23 am
content: Quote from: coppercone2 on March 28, 2025, 03:21:47 amYes.What I mean is that any metallic screen (and any other material) is initially transparent to electromagnetic waves. It don't shield it all the time. It only becomes opaque after a very short delay, once the wavefront has passed through it and re-emission begins.

author: coppercone2
author_rank: Super Contributor
timestamp: March 28, 2025, 04:05:08 am
content: How does it 'catch up' to the front of the wave though? If its all happening at the speed of light it sounds like some part of the wave gets away? Is that what you mean by total shielding is not possible? Does that not imply super luminal...??

author: radiolistener
author_rank: Super Contributor
timestamp: March 28, 2025, 09:56:54 am
content: Yes, some part of the wave still gets away.The reason the delayed response of charges in the metal does not result in visible light passing through a thick metal sheet is due to the extremely short duration and minimal intensity of the transmitted wavefront. When the initial wavefront reaches the metal, it induces charge oscillations that re-emit electromagnetic waves with a slight delay. This re-emission quickly interferes destructively with the incoming wave, significantly reducing the transmitted wave’s amplitude almost immediately after the front has passed.For a very thin metal sheet, this brief transmitted wavefront might be measurable under certain conditions. However, for a thick metal sheet, any residual wave that initially passes through is both extremely short in duration and drastically attenuated, making it undetectable to the human eye or standard instruments. The energy that "gets away" is negligible, and the total shielding effect becomes nearly complete after the initial transient period. This does not imply superluminal behavior - it is simply a consequence of wave interference and the finite response time of charged particles within the metal.To observe the passage of the pulse front before shielding starts, the pulse frequency must be sufficiently high (with a short wavelength) so that the front can traverse the metal thickness within this short time frame. In other words, thicker metal requires a higher-frequency (shorter-wavelength) pulse to detect this effect.

author: coppercone2
author_rank: Super Contributor
timestamp: March 29, 2025, 12:15:38 am
content: What is this effect called?That is what I meant by super luminal, if the thing works as you say it does, then there should be part of the wave that gets through because anything with delay cannot 'intercept' it (have the correct position in space at the right time) unless it is super luminal.In some sense it does make sense because its electromagnetic and metal is a conductor so in some principle its 'hollow' or transparent or whatever... but its still very weirdI assume that some interesting experiment can be made with unusually tough foils (refractory metals that won't melt easily?) and high power lasers? Tantalum or tungsten foil?I think I saw a formula for tungsten electroplating somewhereThere must be some experiment involving photo multipliers or other sensitive detectors that characterized this phenomena? Perhaps just do a light plating of a PMT and shoot it with a pulse laser?

author: radiolistener
author_rank: Super Contributor
timestamp: March 29, 2025, 12:36:17 am
content: Quote from: coppercone2 on March 29, 2025, 12:15:38 amI don't remember exact name, I just remember the effect description from physics lecture. The effect is essentially the transient transmission of the initial wavefront before the metal starts to re-emit and reflect the incoming wave.The apparent transparency of the metal to extremely short pulses is indeed a fascinating and somewhat counterintuitive effect, but it is well understood within the framework of classical electrodynamics. In essence, any finite response time creates a brief window where the metal does not yet behave as a reflective or opaque surface.However, this does not imply superluminal behavior. The wavefront itself always propagates at the speed of light. The reason part of the wave seems to “get away” is simply that the metal cannot instantaneously respond to the incoming field - it takes time for the electrons to oscillate and re-emit the wave.

author: coppercone2
author_rank: Super Contributor
timestamp: March 29, 2025, 01:03:49 am
content: I am saying that if there WAS NO PULSE TRANSMITTED THROUGH and the effect you claim is TRUE, it would imply superluminal behavior, and if the effect you are saying is true and a pulse gets through then there is NO super luminal behaviorthe idea being that if it is transparant but the pulse can't get partially through something has to be faster then lightActually that does bring something to mind, has anyone tried this with super conductor thin films? It sounds more related to inductance and capacitance then resistance though. Non inductive metal could be another type of super material lol, but something thinks it would fundamentally break conduction?Apparantly they are on type III super conductors, whatever that is., something with magnets.. and theoretically up to IV. romulan war bird hull material lol

author: CatalinaWOW
author_rank: Super Contributor
timestamp: March 29, 2025, 02:50:07 am
content: I would suggest getting a good EM text like Plonsey and Collins, or Ramo, Whinnery and Van Duzer and go through this. Full understanding requires some math, at the least differential equations and preferably vector calculus.There is much in this thread that is misleading or an approximation.One simple example. Most of what has been said describes normal incidence to a plane conducting surface. The answers are similar conceptually, but details are quite different.Another example. A microwave short pulse can be shielded as can longer and lower frequency pulses. But the skin depth may indeed not be the best metric for the effectiveness of the shielding. Remember that just as a capacitor theoretically never discharges, and rules of thumb like one time constant or five time constants don't mean that total decay has occurred, the definition of shielding depends on how good you need it to be. 80dB of attenuation may not be enough.I don't feel competent to do a tutorial here, it has been fifty years since I was deep into this. I would be likely to make misleading errors much like those above. Just want to warn you to get some more sources of information.

author: coppercone2
author_rank: Super Contributor
timestamp: March 29, 2025, 02:51:46 am
content: the main question in this thread is if there are uses for thin surfaces in microwave, free standing but it helps to know what it does exactlyits kinda like a apature but kinda not

author: CatalinaWOW
author_rank: Super Contributor
timestamp: March 29, 2025, 02:59:15 am
content: As to your original question, uses are probably best described by analogy to optical uses.One fun one would be to repeat the Michelson-Morley experiment at millimeter or microwave wavelengths. Couldn't really go much lower than the upper end of microwave due to the size of the partially reflecting mirror required (along with the other mirrors involved).Maybe some alternate architectures for doppler speed measurement. Probably not something that would replace existing methods, but again an interesting demo.The problem with adapting optical usages is that most of them are ways to get around the fact that we can't collect phase information from optical detectors. Since we can in the RF world that is usually the easier way to go.

author: coppercone2
author_rank: Super Contributor
timestamp: March 29, 2025, 03:06:30 am
content: the thing that stands out at me is that you can get up with some pretty thick materials, meaning free-standing materials, potentially very big because it seems structurally robustin space you can set them up without dielectric supports too, theoretically. That one is interesting because you can have very thin and long things that don't need supports.