# Tag Info

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My physics teacher gets mad when I talk about the force of sunlight. I still believe👨🏻

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As Arthur Desrosiers mentions, there are 3 main interactions which govern the deposition of energy into the detector. These, however, are not the reason for the effect you are describing. The increase in FWHM is due to the statistical fluctuation of electron-hole pairs created and their contribution to a small current created in the detector, converted to a ...

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The gamma photon energy is a function of the energy levels in the nucleus. There is some uncertainty in these levels. However, the data that you present in the chart is the amount of energy deposited in a detector, where the uncertainty is determined by the uncertainty in the interaction of the photon with the material of the detector and the processing of ...

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In the following paper Professor Pierre-Marie Robitaille has argued that thermal emission is due to vibrations of nuclei within the lattice of a material, and hence also of a blackbody: Robitaille, P.M. On the validity of Kirchhoff’s Law of thermal emission. IEEE Trans. Plasma Sci., 2003, v. 31, no. 6, 1263–1267.

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In a photon electic field vibrate only along a single direction and does not change along its axis. When you see an unpolarised wave you actually see a number of photons. Vibrational axis of each photon is independent of each other. In unpolarized wave we see all these random directions. And talk about changing electric field which is misconception. In ...

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Im think that "perfect vacuum" not able to propagate light. Light is wave it propagate on the principle of domino and wave without transmission medium (domino) cant propagate (in vacuum its have nothing). But perfect vacuum not exist no one can check this and its have only theory we can guest for this and doubt of what is written in books :)

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We say that sound waves need a medium to propagate and we know that light doesn't need such a thing. But is that really how that works? There's no such thing as "nothing" according to Quantum Mechanics, so I'm wondering if light can travel through "perfect vacuum"? Right, but light is not a quantum mechanical concept. It is a well fitted macroscopic ...

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Will the protruding wires emit EM waves? They don't have to. An example where they do not not would be for a spatially uniform magnetic field that increases linearly in time with a wire loop of fixed resistance, and a steady current in the loop and no current in the protruding wires. In this case the current in the wires is steady, and no radiation or ...

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So, there is a current in the open loop and from that you may be very tempted to say that there is necessarily an alternating voltage, which would give you AC currents into the protruding wires. However these last two ideas are not correct. You may choose any of several ways to see this. The simplest is to divide the current up into little segments of the ...

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A coherent state is a monochromatic sinusoidal field. The electric field pulse is inherently not monochromatic, but instead has a spectrum of frequencies which are superimposed on top of one another so as to all add constructively once every pulse repetition time. Therefore, to represent the field pulse in quantum optics you actually need to bring in more ...

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Polarized, partially polarized, and unpolarized light are a phenomenon of classical optics known for a long time. The first successful theory dates back to 1809, and a complete theoretical description was given by Stokes in 1852, though without reference to the electromagnetic field. The correct way to model classical unpolarized light using electromagnetic ...

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Photon energy density does affect the expansion of the universe, similar to the effect of dark matter and dark energy. At the present time, however, the photon energy density has a much smaller effect than either dark energy or dark matter. This is due to the different ways in which the energy densities of the various components scale with the size of the ...

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Answer So my question is, what would electric field of unpolarized EM wave look like when measured (assuming low enough frequencies to be measurable)? To create an unpolarized signal we can start by creating some random noise, allowing it to constructively and destructively interfere with itself. The following is some Mathematica code but written in ...

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I will try to answer this question in two parts: firstly I disagree with the answer in the question you linked. There is "classical unpolarized light" (in some sense) and it is instructive to look at it. Then I will show how this relates to the quantum version. Classical unpolarized light: classically, all the information about the wavefield is contained in ...

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A coherent state is actually a mathematical idealization of a monochromatic laser. Strictly speaking, any continuous wave laser in the laboratory would be a statistical mixture of phase-randomized coherent states. Furthermore, it would also have a finite linewidth, and different frequency components in that linewidth would have no definite phase relationship ...

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EM waves are formed when an electric field couples with a magnetic field. The magnetic & electric fields of an EM wave are perpendicular to each other & to the direction of the wave. The wavelength is just that--the length of the wave through one frequency cycle.

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A properly collimated laser beam is called a Gaussian Beam whose transverse magnetic and electric field amplitude profiles are given by the Gaussian function. The Gaussian beam is a transverse electromagnetic (TEM) mode. The mathematical expression for the electric field amplitude is a solution to the paraxial Helmholtz equation: The width of such laser ...

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Just as Jon Custer wrote in his comment, even a perfectly collimated laser beam with a planar wavefront will diverge. The way it happens is determined by the Huygens principle, and depends on the beam profile: When the light intensity is abruptly cut by a sharp flat obstacle, the light will indeed diffract in almost all angles. A razor blade cutting a laser ...

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Ultraviolet would be the best the short wavelengths carry a lot of energy.

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I will answer this question in two parts: first comparing plane waves, spherical waves and cylindrical waves (which are really the same thing as I will explain). how this relates to ray optics (which is completely different). Plane waves, spherical waves and cylindrical waves are 3 different examples of doing a decomposition of a wavefield. The method is ...

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You are describing a spherical wavefront i.e. light radiating outwards from a point source with spherical symmetry. But suppose you have two such point sources near enough to each other that their wavefronts overlap. Now your expansion model has to have space expanding simultaneously in opposite directions. Consider also that a wavefront can be any shape. ...

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Surely, when you consider remote point charge $e$ circling on the ring it emits radiation. That means it creates fields $\mathbf{E}$ and $\mathbf{H}$ such that they decade as $1/R$. That field is discribed by Lienard--Wiechert potentials $$\mathbf{E} = \frac{e}{c^2R} \frac{[\mathbf{n}\times[(\mathbf{n}-\mathbf{v}/c) \times ... 1 We know that light is an electromagnetic wave and it does interact with charges. It contains magnetic field and electric field oscillating perpendicularly but when we apply an electric or magnetic field in any direction to the wave the applied electric field or magnetic field vector doesn't alter the magnetic or electric field in the electro magnetic ... 3 An applied electric or magnetic field don't alter the field of an electromagnetic field because, as you said, the superposition principle holds. This principle is a principle of linearity, and comes from the linearity of electromagnetic equations : there is no interaction between photons at low energies. You can see it from a field theory point of view, as ... 2 When a receiving antenna picks up a signal, a current flows. This current acts as a secondary "transmission", and this will partially cancel out the electromagnetic field that is incident - this is how you get power from the EM field into the antenna. If you look "slightly downstream" from the antenna, you will see a reduction in the EM field (assuming for a ... 3 A radio wave may "bend" due to diffraction or scattering off objects, but it will not "bend" just because a receiving antenna exists. 1 Putting a dark material onto snow does increase the melting rate, and indeed soot from pollution is having exactly this effect. The mechanism is exactly as you suggest. Soot absorbs sunlight and heats up, and the heat is transferred to the snow by conduction and convection. Using clean oil wouldn't work very well because oil does not absorb sunlight ... 1 a planewave impedance E/H =377 ohm, and it is said that for plane waves none of the components electric and magnetic dominates they are indistinguishable or same. The electric and the magnetic fields do behave in the same way in an electromagnetic wave, and a combination of both is needed for the wave to propagate and both are equally important for ... 0 A travelling wave in the direction of the picture(or whatever direction) can have two polarizations, each perpendicular to its direction of propagation. Now, in the black body radiation derivation, we usually use a box as a black body and inside the box standing waves are formed. But standing waves are nothing more(mathematically as well as physically) ... 3 Photon energies in that frequency range are far below the input noise of the receiver, so one can't do photon counting. The smallest detectable signal will consist of a large number of photons and can be treated like a classical electromagnetic wave. The technically most often used figure of merit that is similar to the quantum efficiency would be the ... 0 The fact that Marconi's machine worked! It relies on the electric field component to affect the electrons in a long wire whuch today we call the antenna. On an optical scale we have modern devices: nonlinear materials in optic fibers, wakefield particle accelerators, and metamaterials bending light as-designed. The E-field and the B-field are very real, ... 1 How can the position of that charge be determined without EM waves Accelerated charges and Cerenkof radiation can be useful in experiments, but if one really wants the location of charges one needs detectors, i.e. consecutive small interactions. These are mainly electromagnetic scatterings but so soft that they are included in the measurement error. To ... 0 In your first sentence you mentioned charges. Electrons, protons as well as their antiparticles have permanent an electric field. This field seems not to be only an induced one during any measurement. The permanent state of the electric field of charges is a postulate because without any interaction a measurement is not possible (see the comments to your ... 0 Take a look at the Hall effect at the Hall Effect Wiki. It does not necessarily pertain to electromagnetic radiation, but the same general principle applies for all electromagnetism. See below for a general description for this relationship. We use Hall effect ammeters to non-invasively measure the current through a wire. James Clerk Maxwell studied ... 2 The reflectivity of the atmosphere, and of the surface itself, is strongly wavelength-sensitive. So while some percentage of any given wavelength is reflected -- and some percentage is absorbed rather than transmitted, the variation over wavelength is what leads to the somewhat misleading statement you refer to. Here's an example of atmospheric absorption, ... -2 It does. Consider this: you can see the Earth from space. Therefore, not just infrared light gets reflected but also light on the visible spectrum. Here's a graph (by NASA) of various planet's radio emissions. The ways that Earth can release radio waves is a bit limited. Because of that, it is safe to assume that at least some come from the Sun. 0 There are two things: The light and the cable. Light moves at speed 18600 miles/second to the north, cable moves at speed 98000 miles/second to the south. The light has nothing to do with the speed of cable. If the observer stands at the end of north, he will read the speed equals to speed of light only. It is to be understood that the light is a ... 0 In dielectrics, no energy can be absorbed for you can just elastically deform the diectric object (e.g. the sphericity on an electronic orbital around the nucleus), just like a spring. So the deformation is transformed back into an EM wave (with polarisation and non-isotropic intensity depending of the axis of compression). It occurs with a delay depending ... 0 The answers to this problem can be more easily obtained by having the light source "attached" to the optic cable. This way the frame of reference for all cases, is the optic cable, and its motion will have no effect on the light pulse propagation speed in the cable. Under this condition, it is easy to see that the pulse will take 1 second to travel the ... 1 If you wrap your electronics in aluminium kitchen foil then the appropriate equation for the electric field transmission factor, that takes into account reflection from the foil and attenuation in the foil is$$\frac{E_t}{E_i} \simeq 4 \frac{\eta_{\rm Al}}{\eta_0} \exp(-t/\delta) = 0.47 \omega^{-1/2} \exp(-22 \omega^{1/2} t), where $t$ is the foil ...

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It all depends on your construction of the Faraday cage... for a sufficiently well constructed cage (multilayer, continuous, RF gaskets on all seams) the answer is "yes". It's much easier to add another 3 dB of isolation than to double the power of your EMP generating device. Imagine you have a cage that provides just 3 dB of shielding. If you put that cage ...

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Let's go for a somewhat more unconventional answer. Consider this: if you have a radio wave of certain frequency and wavelength, you can "cancel out" its effect by emitting another wave of same frequency, wavelength and amplitude, just offset by half of its period. Something like this: (forgive the crude drawing) The two waves would neutralize each ...

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Speaking more generally, waves of any kind are just a physical expression of the flow of energy through time and space. And energy can be 'stored' by either converting it to another form (for example electrical to chemical as in a battery) or by trapping it within a confined space using resonance, such as in the wave guide Floris suggested.

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Every radio wave is "stored" in space between the time it is sent and the time it is received. You could prolong that time by send the wave into a long, low-loss waveguide. In reality waveguides do have some associated losses so it would only be a short term solution, but any RF delay line is in essence a short term RF storage device.

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The problem is that you are confusing light intensity with energy of a single photon. The photoelectric effect requires a certain energy per photon to work. But low light intensity just means fewer photons come - you can actually see the grain if the conditions are too dark: every pixel can get ~10 photons or less... and yet still, each photon that comes has ...

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Let's replace the fiber optic cable with one laser source and one photodetector a distance $L=186,000$ mi apart in vacuum and at rest relative to each other. The laser source is pointed straight at the photodetector. Alice observes the laser source and the detector moving at constant velocity $v = 93,000$ mi/s $= c/2$ with respect to her inertial frame, in ...

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Under the vast majority of practical circumstances, there would be no difference between the blocks of metal and layers of metal. The reason is that only the surface of the metal (more precisely, a thin layer with a thickness on the order of a few microns for milimeterwaves/microwaves e.g Wifi frequencies) interacts with the EM wave. If the metal is roughly ...

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There are a few related things here. Let the light move right and the cable move left at speed $v$. First, suppose the light is not in the cable. If the light is moving right and the cable is moving left, can the light clear the length of the (light-second long) cable in less than one second? Yes! The relative velocity of two objects can be greater than ...

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There are two things: The light and the cable. Light moves at speed 18600 miles/second to the north, cable moves at speed 98000 miles/second to the south. The two things have speeds they are allowed to have. The distance between the light and the cable changes 284000 miles/second. Now let me remind that there were two things with those speeds that the two ...

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I'm going to assume that you somehow built a vaccum-fiber optic cable. Although you CONVENTIONALLY don't add your speed to the speed of light, when your speed approaches the speed of light, your speed does come into play. If your cable is flying at 99% of the speed of light, and the light beam is right next to it,relative to the cable, the light is traveling ...

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