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In a classical way excited particles emit the received energy in discrete quanta, later named photons. As a good description of photon emission one can take the process of stimulated emission. The imagination, that the emission of energy from an excited electron is distributed all other the space is wrong. The best way to predict processes inside the ...

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In the classical picture, an incoming wave excites a (damped) oscillation in an atom. We imagine that the oscillation is of bound electron(s). The oscillating electron(s) then re-radiates electromagnetic waves, but importantly, these waves are emitted in a continuum of directions, following the spatial distribution emitted by an oscillating electric dipole. ...

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The attenuation of a waveguide is minimal when it is excited with the correct mode, but it isn't infinite for field configurations that aren't, so you can also send field configurations trough that don't fit the mode patterns, they will simply be attenuated very strongly for low frequencies. How strong that attenuation is depends on how much of the ...

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A "ray" in geometric optics is a locus of continuous propagation of light. Think of it as mapping where the energy is going in space. In principle there are an arbitrarily large number of them, but we draw a manageable number for visualization purposes. The various [letter]-rays were so named when people didn't know what they were beyond being things that ...

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How do EM waves propagate? Like other waves propagate. IMHO the best way to appreciate this is to shake a rubber mat. When you do this you stretch a portion of the mat, and then the elasticity of the material contracts it back to its original size, but in doing so the rubber is stretched further along. What you then have is a shear wave with speed v = ...

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What Maxwell derived was from radio waves. Radio waves are modulated photon radiation. Electrons in an antenna rod get accelerated at once and emit photons. The density of this photons distributes in space. Detecting this radiation with a receiver one get a sinus wave form. The frequency of this wave has to do with the frequency of the antenna generator and ...

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Individual photons are not considered rays. Because of the wave and particle nature of photons, they are much more complicated than what they are generally thought of: a projectile of light. In fact, they do not have an exact measurable position, but do travel in straight line trajectories. What we consider rays are lines perpendicular to the wave front of ...

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Are the EM fields really moving. My textbooks says it's changes in field that is moving. I don't understand this part. In your post, you mention a wobbling rope. Well, each fragment of that rope does not move along the rope. They just wobble where they are, without moving forward. But the wobbles themselves do move forward. Likewise, the electric and ...

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The ray theory of light is equivalent to the Eikonal Equation, which in turn is essentially a slowly varying envelope approximation to Maxwell's equations. If we write the electric and magnetic field vectors as $\mathbf{E}\left(\mathbf{r}\right) = \mathbf{e}\left(\mathbf{r}\right) e^{i\,\varphi\left(\mathbf{r}\right)}$, $\mathbf{H}\left(\mathbf{r}\right) = ... 2 Light has a frequency of approx. 1e15Hz. Can light be transmitted in a hollow copper tube? Yes. No need to go relativistic. Can objects move at near the speed of light in a coax cable with inner conductor? No. They can't move in there, at all, not even at walking speed. Does any of this has anything to do with photons? No. Your experiment does have a ... 2 When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ... 3 If you or one of your friends has transition lenses for their glasses, then you can test the UV blocking ability with those. The transition to darker shades in these lenses is initiated by UV-light. So hold your sunglasses over some transition lenses and see if they start to turn darker. You can try this with any photochromatic material really, but ... 1 Without seeing your code, it is hard to know where you went wrong. I did write a few lines of code myself to see what might be going on - and found that the values converge quite nicely as you approach higher temperatures (for the lowest temperatures of 500 K and 1000 K, significant amounts of energy do indeed extend beyond 10,000 nm). updated There is a ... 0 Well a router antenna is simply a dipole. It will have maximum radiation in its broad side direction and it's radiation pattern looks like a donut. Check the link below for the illustration of the dipoles far-field https://www.cst.com/Academia/Examples/Wire-Dipole-Antenna Of course the router itself and any metallic objects nearby will influence the ... 0 Recall that$cos\alpha\cdot sin\beta +cos\beta\cdot sin\alpha = cos(\alpha+\beta)$. If you suppose that$B_1=B_0\cdot sin\phi$and$B_2=B_0\cdot cos\phi$, it turns out that:$B_1cos(kr-\omega t)+B_2cos\phi=B_0\cdot sin\phi \cdot cos(kr-\omega t)+B_0\cdot sin(kr-\omega t)\cdot cos\phi==B_0\cdot cos(kr-\omega t+\phi)$. 0 How can I get from what is given to the direction of$\vec{k}$The E-field has only a y component. And the B-field has only a z-component. Since this is a plane wave, then you know that the direction of propagation must be in either positive or negative x. From the information given, I believe it's actually ambiguous which direction the wave is ... 0 Here's a plot of atmospheric water vapour (fog) which is the green line. The horizontal line is wavelength and the vertical line is attenuation. Liquid water is red and ice is blue. From wikipedia: The absorption of electromagnetic radiation by water depends on the state of the water.The absorption in the gas phase occurs in three regions of the ... 0 Imagine the charges flowing through the wire. In the moving section of the wire there are two components of the velocity one in the direction along the instantaneous direction of the wire and part in the direction the wire is moving. The velocity in the direction the wire is moving generates the emf. But the emf is not work done per unit charge, this simply ... 0 As far as we know, light is mediated by a particle without rest mass. Special relativity says that such a massless particle's speed must always be observed to be$c$, irrespective of the observer's motion. In general relativity, a massless particle's speed as measured locally is also always$c$. No experiment so far has detected a measurable difference ... 0 I think this originated with Hadamard and his Method of Descent. See Lectures on Cauchys Problem in Linear Partial differential Equations--starting on page 7. His results were that waves in two dimensions did not propagate sharply, but had a wake (a tail, ..). Eg. a circular wave propagating in two dimensional space vs. a spherical wave propagating in three ... 0 I'm sure this isn't what you had in mind, but... Reflected optical light (I can't think of a reason that UV wouldn't be just as good, provided there is enough UV radiation around to be reflected, e.g. sunlight) is 'radiation from a human'. A picture is a pretty good way of identifying humans. Provided the picture is sufficiently detailed (for instance, ... 0 There is a nice video on youtube by Royal Canadian Air Force. That explains how directivity works with animations. Antenna Fundamentals 2 Directivity 1 I had the same question once, and scoured the Internet for advice. All I got was conflicting information, much of it from "experts". I ended up getting a signal strength app for my smart phone and one for my laptop and experimenting. In my house, with my router, I found no detectable difference between horizontal and vertical. 1 DC current is organized as following: positive potential applied to one end of the wire, negative potential applied to the other. Electrons move from one end to another with some speed. If you have one electron in vacuum and electric field from A to B, then there will be force acting upon that electron due to$F=eE$. Movement should happen along line ... 0 The heat from Sun, comes in the form of radiation. To be specific, the infrared part of the EM light is responsible for heating. Infrared frequency is closest to the resonant frequency of most molecules. Thus, when these molecules absorb the infrared radiation i.e. the molecules' electron clouds oscillate at the infrared frequency, there's resonance, which ... 14 You probably are under the misconception that heat travels only via molecular interactions. (i.e, heat transfer by conduction, which needs a medium of sorts). Heat also transfers by radiation, which the sun is an enormous source of. Electromagnetic radiation does not need a 'medium' to travel through. All types of electromagnetic radiation carry energy, ... 4 The heat 'comes' as electromagnetic radiation, that is light. Light from the sun is electromagnetic radiation, that is a wave having energy and momentum or a very big amount of quantum particles, photons, that have energy and momentum. The interaction of this electromagnetic interaction is what heats the earth. Hope this helps. 1 A Gaussian beam has a width that changes with distance because of diffraction, which is an effect that takes place in any wave phenomenon. It has a pretty similar description to the Heisenberg uncertainty principle in QM if you're familiar with that. Namely, as the position in the$x$and$y$directions (with the optical axis pointing in the$zdirection) ... 1 Both \begin{align} \tilde{\bf E}&=\hat{x} E_0 \exp(-i(\pm\frac{\sqrt{2}}{\mathrm{m}} (\hat{x}\frac{\sqrt{2}}{2}+\hat{y}\frac{\sqrt{2}}{2})\cdot(x,y,z)-\omega t))\\ &=\hat{x}E_0 \exp(-i(\vec{k}_\pm\cdot \vec{r}-\omega t)) \end{align} are part of a valid electrodynamic field travelling in opposite directions. (1/m is just the unit 1/meters) with ... 1 Remember that\vec{E}=E_{0}e^{-i\vec{k}\cdot\vec{r}}\hat{x}$, so that, just by identifying the variables$\vec{k}\cdot\vec{r}=k_{x}x + k_{y}y + k_{z}z$; from that you should be able to get the wave vector$\vec{k}$. 2 CMB hasn't a frequency but a typical black body frequencies x radiances distribution. With the Planck law, the curve distribution gives the temperature of the radiation. I found this image in a previous question Relationship between temperature and wavelength? As you can see, each temperature has a typical curve. Yes, red/blueshift affect the ... 4 No. Light travels at the speed of ... light, when measured locally in inertial reference frames. And the relationship between wavelength and frequency is$\lambda = c/f$. As the universe expands, the wavelength of the cosmic microwave background photons is "stretched" and thus their frequency must decrease by the same factor of$(1 + z)$, where$z$is the ... -6 Has the frequency of CMBR changed at all since the beginning of the universe? The usual answer is yes. It's thought to have redshifted by a factor of a thousand. But there is an issue: conservation of energy. Where did the energy go? This is an intriguing thread to pull, because we don't know of anything that's in breach of conservation of energy. There ... 0 Theoretically, the CMBR is what we see of the early visible universe. If we could be there to observe it, we might see much higher frequency radiation. However, due to the expansion of the universe, the wavelengths got stretched out and the frequency redshifted. Or perhaps you can say in our reference frame, we happen to measure these photons to be ... 0 I want to get an answer in quickly because your question is on the cusp of being closed. So I'll answer briefly then revisit this answer when I have time. The solution to the Maxwell's equations in a vacuum is a plane wave, and any combination of plane waves is also a solution. So you can take an arbitrary waveform like the one you propose and Fourier ... 0 In general, if no further information is given, you can assume that the focal length is approximately correct for the visible spectrum. Note that, in your example, since the focal length is specified as 25 mm, as long as the FL is within 0.5 mm of 25 mm the lens is within spec. Even with window glass, you should be able to get this performance from 400 nm to ... 2 No, the photons do not travel in a helix, they travel in a straight line but with a phase delay that is dependent on position. Looking across the beam's wavefront there is a phase delay that is dependant on the polar angle$\theta$around the beam axis. If we take a simple helical mode's complex amplitude as$\zeta(r,\theta,z) = u(r,z) e^{-ikz} e^{il ...

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Classically EM radiation is just a wave (not a particle) and the double slit experiment is only the result of the well known interference proprieties of waves.In quantum mechanics we've got the wave-particle duality, and so the light could be seen as composed by particles, photons. So the first thing to point out is that mixing the two approaches (talking of ...

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"How exactly the RADAR works? Is it possible for RADAR to work with 1940s clocks?" is, of course, two questions, and the second is easier to answer: in the sense that we talk about it today, where a signal is analyzed and a digital readout provides timing information, 1940's radar did not do that at all, and therefore did not "work with clocks" at all. The ...

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Although glass is an amorphous material, it behaves surprisingly similar to crystalline materials in some respects. In this case, you can imagine glass to be a semiconductor with a large bandgap, at least large enough to be beyond the visible wavelengths. Therefore, all visible light passes through, which makes glass transparent. Obviously, there will be ...

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