Wavelength and Aperture SizeWhat determines distance between grid lines in faraday cage?But when we are talking about something like a faraday cage to block out microwaves for example, we say the spacing of the cage 'bars' should be shorter than the wavelength, I have trouble understanding this, as the wavelength is along the normal to the cage surface, and the wavelength shoul not have a bearing on whether it 'gets in' or not. Why is it that the spacings (which are paralell to the polarisation vector) determine what kind of waves get through? It might be useful to remember that the microwave radiation exists as a series of fields propagating through space. The electric field and magnetic field are at right angles to each other, and both of these fields are at right angles to the direction of propagation. While there is a wavelength along the direction of propagation, the wavelength along the direction of these two fields is what you are concerned with in your question. As long as the spacing of the bars is greater than one wavelength, and the diameter of the bars is very small, the fields pass through largely unimpeded. At a spacing of exactly one-half wavelength, theoretically the fields will be shorted out and at spacing of less than one-half wavelength they will be reflected. The mathematical treatment is somewhat complex, involving characteristic impedances, reflection coefficients, attenuation and phase shift constants, it gets a bit involved! I do understand your question but it does take a bit of visual conceptualization to get the feel for this. I hope my explanation makes some sense: I usually refer to the E and H fields, but no matter, these fields are right angles to the direction of propagation and at right angles to each other. Of course, they do move along with the wave in the direction of propagation, but they move as a wave front. If you were directly in the path of the oncoming signal, and if you could see it coming, the E and H fields would appear as the wall of a steradian moving towards you. They extend out to the sides at a distance of one-half wavelength on each side, or one wavelength across. Depending on the type of polarization, if orthogonal the E field may be horizontal and the H field vertical or vice versa, or if circular polarization they may be rotating CW or CCW. They extend out to the sides, but they propagate forward, towards you. They cannot extend any further than one-half wavelength on either side, (in a perfectly directional wave) because the wave is moving forward at the velocity of light and these fields are changing in accordance with the amplitude of the wave. So by the time the forward wave reaches max amplitude, the field reaches max strength (one-half wavelength) then it must collapse in again. The amplitude of the wave does not determine the distance the field extends to the side, only the frequency and speed of light determine the wavelength. They don’t propagate to the side; they are dragged along in the direction of propagation of the wave front. The amplitude determines how dense the field is, or how many lines of force it contains. It may be difficult to visualize at first but if you give it a try you will see it. I think I have it now, if we have a higher frequency, the E and B fields don't have enough time to propagate any further out than half a wavelength of the light from the centre of the circle (if you're looking at it head on). So passing through an aperture larger than the wavelength, does not effect the E and B field, only the photons that are very close (1/2 lambda) to the edge of the aperture are diffracted. "The amplitude determines how dense the field is, or how many lines of force it contains." So the amplitude of the EM wave determines the density of the field. If we have a high power laser for example, does this mean that the E and B fields are the same spatially (1 lambda in total) but the collective amplitude of the EM wave is huge. People tend to talk about large E fields in laser physics, but a large field has the same spatial dimensions (1 lambda)? Is this correct?
but where do the eddy currents get the energy to produce these waves? they must absorb it from the incoming wave, yet the incoming wave never actually touches the metal wire. the answer seems to be that 'its a transverse wave and transverse waves con do that'.
Copyright ©2000 - 2012, Jelsoft Enterprises Ltd. © 2012 Physics Forums http://www.physicsforums.com Cell phones are usually 2ghz and up, some dinosaur phones are 900 mhz, your gmrs or frs radios are 462mhz, which penetrates concrete, and some metal (if not grounded) pretty well and your handheld will have an output of .5 to 5w max. If you are using vhf its a bit harder to escape even a car body with a rubber duck antenna on let alone any signal 54 mhz or lower getting thru your car (6 meter wavelength). Try the small galvanized trash can, a small oily rag sized one should do the trick for most hand held or mobile radio applications. If the radio still recieves well with the lid on, try a ground rod to wet dirt making sure to use a flat tape, and not a round wire that may be resonant length. Im fairly sure you will get the results your looking for. The water pipes in your house, although connected to ground ARE RESONANT. I think the ground rod, to ground strap to trash can should be as short as possible, with no resonant lengths past the can. http://www.thesurvivalistblog.net/faraday-cage-emp-protection/ |