# Tag Info

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Oscillating currents tend to release radiation only perpendicular to the direction of the current itself. A dipole antenna is maybe the simplest antenna to demonstrate this. There isn't really a good reason why, other than this result pops out from the math. The best explanation I can think of is that currents only tend to create magnetic fields ...

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If black is absolute black, the top bar completely masks bars underneath. If the second layer also uses absolute black, it looks just like the top layer. So you wouldn't be able to discriminate. If you have translucent grey bars, you cannot tell the difference between white-over-black and black-over-white. If you had two layers where one was vertical and ...

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So you're hoping that by measuring the gray level you can tell which bar belongs to which layer? That's potentially possible, but the problem is that a black bar on the top level completely masks any black or white bar on levels below, so there is no way to read all the barcodes. you might be able to reject the lower layers (read just the top) on the basis ...

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I think the main confusion is here is the difference between a scientific measurement and a pedagogical demonstration. You're right that the microwave and chocolate method has to assume the frequency on the back of the microwave is actually the frequency of radiation inside the cavity. A proper measurement of the speed of light would include a measurement of ...

1

Actually the human body emits more than thermal radiation. The Czech military did a study on measuring the extreme low frequency radio band emitted by the nervous system. It can be found on www.measurement.sk by searching Human electromagnetic emission in the ELF band - Measurement ... www.measurement.sk › Lipkova This makes perfect sense when you consider ...

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A key to understanding this is realizing that it's not always true. In fact, at x-ray frequencies, refractive indices are typically less than 1, so that the phase velocity is faster than the vacuum speed of light. The key difference is that x-ray frequencies are well above the natural frequencies of most of the electronic excitations that are involved in the ...

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The radiation of an accelerated or a decelerated charge particle is called "Bremsstrahlung" radiation, which means (roughly) deceleration radiation. The radiation is a continuous spectra. i.e. it's not the emission of one photon. The charged particle which is accelerating emits continuously photons of different energy depending upon the rate of change of ...

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A star with a temperature of 50.000 °K would have to be much larger than our sun to sustain that rate of fusion, and our Earth as we know it would not exist. It would have to be much further out and it would be unlikely to form a similar atmosphere. There would probably be more ozone as a direct result of more UV to break stuff, though all emission at UVB ...

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Yes, the ratio is for example $4/5$ and they are asking for minimum distance $n_1=4k$; $n_2=5k$ (for some $k$), the distance will be minimum when $n$ is minimum, so $k=1$ and $n_1=4$ and $n_2=5$.

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You have a misconception, photons are massless, but there is a momentum term that is usually left out of the energy equation. Since a photon has no mass, so it can travel at the speed of light with no problems. You're thinking of e = mc^2, and since a photon has non zero energy, obviously it must have mass? No! That equation is the simple version for the ...

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If the WiFi antenna is emitting at 2.4 GHz, you could detect a slight improvement of the signal, but unless the door is solid and very thick, I doubt it will make much difference. If it is emitting at 5 GHz the improvement could be bigger, as the wavelength is reduced and the door appears "bigger" to the electromagnetic wave. Finally, if your antenna uses ...

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This is something I never really understood, but Glen Knoll Offers the following in pp. 116 of his book "Radiation Detection and Measurement": "The energy resolution of the detector is conventionally defied as the FWHM divided by the location of the peak centroid H_0 The energy resolution R is thus a dimensionless fraction conventionally expressed as a ...

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In the general case the frequency of a wave and its kinetic energy are not related. As you can derive from Energizer 777's answer one can increase the frequency and this time decrease the amplitude of the wave generator and you approximately need the same amount of energy to support the damping of the wave from dissipation processes. The point is that an ...

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Electromagnetic waves come in all shapes, sizes, and symmetries. If you wiggle an electron back-and-forth, you will produce something-like dipole-radiation which is fairly isotropic. The same type of pattern will be produced if you wiggle your finger back-and-forth in water. The strength of the resulting waves will be roughly isotropic, but somewhat ...

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To add a little detail about radiative thermal equilibrium: As atoms at nonzero temperature collide with each other, they do emit electromagnetic radiation, and if they were in an empty universe, they would approach zero temperature. However, since the universe isn't empty, they also absorb electromagnetic radiation coming from all the other atoms around. ...

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You're right that the classical idea of radiation emission from an accelerated charge cannot be applied to electrons in orbit around nuclei, and thus they do not emit radiation (unless they're in an exited state and decay to a lower state). The same thing does not apply to the nuclei. As you suspect, they will, over time, lose energy and vibrate less and ...

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The sentence can be proved by Fermat's principle and it is not relevant to wave nature of light. Waves have some properties like velocity (and could be different in different media), phase and the important property called interaction. When we say light is a wave, we must check the above properties in it. And the first experiment that showed that, light is ...

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This is mostly because we're usually more interested in the spatial part of a plane wave than in the temporal part, so that plane waves are most convenient when written as $$e^{i(\mathbf k\cdot\mathbf r-\omega t)}. \tag 1$$ The normalization follows from this choice. In general terms, it's hard to call which factor has more weight. There are plenty of ...

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I think we can think of this as a wave although a bit different from water waves or sound waves.. I hope you agree that wave has energy .afterall by propagating waves energy is transferred..when we think of waves in a medium we can plot the displacement of different points in space and that's how wave propagates. Now in this case (em radiation) the amplitude ...

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Intuitively, it is merely energy. It is transmitted by photons. This can take many forms depending on the size of the wave the photon travels upon.

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I have read about an electron making a transition between two energy levels and electromagnetic radiation will be emitted. The problem is how and why e.m radiation is emitted. An intuitive understanding can come from considering the Bohr atom. Bohr imposed the fixed energy levels for his hydrogen atom to explain why radiation from hydrogen gave ...

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Electrons produce radiation anytime their agitated. They don't need to be in orbit with an atom to do this. Synchrotron radiation is produced anytime a charged particle is accelerated.

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When we are calculating the energy levels of an atom (or anything else) we are generally solving the Schrodinger equation to calculate the energy eigenstates. These eigenstates have the property that they are time independant i.e. they do not change with time. So if you consider the ground state and an excited state we end up with the surprising result that ...

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In general terms as per my understanding. an EM wave is generated as follows imagine a piece of wire of points A to B placed like this A-----B when you pass an alternating current (voltage +v 0 -v) Emission of Electric wave. E1: lets say for negative voltage's peak value the electron concentrates on one side of wire, lets say it at the end of wire i.e ...

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I too had the same problem, so I started constructing EM wave just as Maxwell did.To get a EM wave he combined two simple laws. When an electron moves with velocity $v$, some magnetic field will be associated around it. As an electron moves, the electric field associated with it also moves; this results in a change in the electric field in space around the ...

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So yes, if you compute the Poynting vector (energy flux density), $\vec E \times \vec H$, for an exponentially decaying evanescent wave, you indeed find zero time-averaged energy flux perpendicular to the reflecting plane. Ask you say, this leads to a conundrum --- how do evanescent waves transfer energy across barriers? For sure, we know they can transfer ...

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The only reason I could imagine a frequency shift is fluorescence (ok, this is cheating :-) ) ultra slight Doppler effect due to the thermal motion of scatterer/reflector atoms. (at macroscopic scale for many photons and atoms, it's more like an ultra slight frequency blur).

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Now generally, it is true that light incident upon an object is slightly shifted in frequency, but when we come to point of your question there are other factors weighing in. In these cases it is better to look at the phenomena in Maxwell's wave model of light. When light falls upon a body, it energizes the lattice to vibrate (this can be thought of as ...

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I'll rephrase the question as I interpret it: Electromagnetic waves are drawn like this: source: https://commons.wikimedia.org/wiki/File:Onde_electromagnetique.svg Suppose this wave comes up to a vertical slit (a slit in the z-direction). What if the red arrows are longer than the slit? Then the wave won't fit through. But if the arrows ...

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remember that the amplitude is not an amplitude in space, it is an amplitude in the sense of the intensity of the electromagnetic field. The spatial amplitude is given by the wavelenght

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