# Tag Info

3

The reason is that the index of refraction changes with the temperature. Due to the very hot asphalt, there is a strong temperature gradient in the air. This gradient leads to an changing index of refraction with the distance to the road. This is what you see.

0

The energy of a classical electromagnetic wave is given by the Poynting vector as you point out, and is proportional to E_0^2 or B_0^2 constants defined by the source of the radiation, and antenna, for example. The higher the voltages imposed on the circuits the higher the radiated energy, as Eoin points out. Therefore for calculating the energy input from ...

1

You are looking at two different things assuming they should be the same. In classical electrodynamics the Poynting vector is defined as $\mathbf{S} = \mathbf{E}\times \mathbf{H}$ and enters the variation of energy density as $$\frac{\partial u}{\partial t} = -\textrm{div}\,\mathbf{S} - \mathbf{j}\cdot\mathbf{E};$$ according to the functional form of ...

0

The quantum mechanical proportionality between energy and frequency of a photon originates in Planck's action quantum h, which has the dimension (energytime), i.e. the dimension of angular momentum. All physical interaction occurs in entire units of this elementary action quantum, also interaction of matter with electromagnetic radiation. Now, (energytime) ...

1

No, because we can think of the classical wave as being made up of a large number of photons. If we have a low-frequency wave with the same energy as a high-frequency wave, it simply means that there are a larger number of low-energy photons.

1

The angle of rotation is proportional to the length of the path the light ray spends inside the active material. This needs to happen because each bit of the path only 'knows' what's happening there and it does not interact with the rest of the material's optical activity. This means that the rate of rotation of polarization must be constant, i.e. that the ...

1

I will turn my comment into an answer: One is dealing with classical physics in this question, at the level of the Bohr model for the atom special relativity is unknown. An inertial frame can be defined as a frame where the laws of physics have the same mathematical form and measurements in one inertial frame can be converted to measurements in another ...

4

Most of the mobile towers have no network coverage there in underground chambers, tunnels and cellars. On the other hand, in the lift, the tower fails to catch the link as the lift acts more or less like a "Faraday cage" or even if it does catch the link, the phone conversation gets severely disrupted. Quoting from the wikipedia link by pela in the comment ...

1

Another important issue is impedance matching between the output of the antenna and the receiver input. The length being a full, half or quarter wavelength creates a resonance situation that helps to increase the sensitivity. But then the situation is less favorable if the received frequency deviates from a resonance frequency. For that reason designers of ...

0

Antenna sensitivity for a fixed length depends on the wavelength of receiving radiation. Antennas are usually designed having a length of half or full wavelength or even quarter.

1

The Maxwell equations do not remain invariant in form when changing to a rotating coordinate system and therefore predictions made from them, like the Larmor radiation formula, cannot be held to be true anymore. While that sentence is sufficient to answer your question, if you want to dive deeper, here are some quick resources: ...

0

I believe this apparent contradiction to stem from a misunderstanding that energy can be transferred only by the one mechanism to which the Poynting vector applies: The Poynting vector is defined as ExH, and applies to a "launched" electromagnetic wave in propagation. In your example, the energy is being transferred by a quasi-static E-field in the absence ...

2

The photon is an elementary point "particle". particle between quotation marks because it is not a classical point particle , it is a quantum mechanical entity. Quantum mechanical entities depending on the boundary conditions display sometime classical point-like elementary particle behavior and sometimes have a probability density for their location ...

1

EM waves don't "stop" they just slowly become weaker as $r^{-2}$, so one could conceivably answer "forever." On the other hand, the wave will quickly become so dissipated/spread out that there isn't much to measure, so you have a practical limit where it won't be detectable. However if this is your intent, you haven't given us enough information to answer ...

0

All I can think of is EEG (electroencephalography) signal. It is a diagnostic technique that allows to monitor brain electrical activity by measuring potentials on the scalp. As I know there are some contactless methods of measuring those potentials, but I have no idea if you can measure it from big distance, since it is quite a weak signal. This is one ...

0

Natural germanium contains about 7% Ge-76. To get to Ge-78 would require two neutron captures on the same nucleus within the 11 hour half-life of Ge-77. According to the NNDC there is not data on neutron capture cross section for Ge-77, so if you are in fact seeing two-neutron capture you are in a position to make a very interesting measurement. I would ...

1

Let us say we have a charged particle moving in the positive X direction with velocity v. If we give the particle a nudge in the −X direction causing it to decelerate. Then from the saying 'particles that accelerate, radiate' Let us look at the classical description of this statement. The classical formula for the radiated power from an accelerated ...

1

The question is: how do you "give the particle a nudge"? The way you interact with electrically charged particles is to let them interact with photons: either real photons, as in the Compton effect, or the virtual photons of the electric and magnetic field. When your particle interacts with your photon, the particle's momentum is changed and the photon's ...

0

Short answer, No. Longer: gamma rays are of MeV energies. The magnetization of the material is a solid state phenomenon of order of electron volts. The gamma will go right through the material, or interact with a nucleus. It will not see the long wavelength structure of the magnetic field. In addition, gammas will be emitted randomly and in random ...

3

To add to the other answers: without detailed analysis of the light passing through fog, one cannot infer that, just because fog dims the Sun to a level that makes staring at it comfortable, therefore it is safe. The "brightness" of the Sun, and the discomfort that staring at it induces, is only very weakly related to the damage it can do. Indeed, a healthy ...

7

I think that from a medical perspective your advice was correct, but your physical explanation of why was not. The wavelength dependence of the extinction due to fog depends on the distribution of particle sizes. If the particles are bigger than the wavelength of light, then the (Mie) scattering and extinction become independent of wavelength, and this I ...

17

The damage to the eye from looking at the Sun is thought to be due to high intensity light creating free radicals that attack the cells in the retina. Contrary to popular belief it isn't a simple burning process. To a first process fog attenuates all (visible) wavelengths equally, which is why it's white. If it preferentially absorbed some wavelengths it ...

11

No - assuming they don't hit anything they don't decay. The distance dependant "decay" is the drop in the number of photons per volume as the volume gets bigger - it's not a decay of individual photons. It's the same as a crowd dispersing as it leaves a subway exit - nobody is disappearing. Photons can lose energy as they collide with gas or dust in space ...

1

First, in the specific case of hearing, it has to be considered that the perceived loudness varies according to the frequency. So if the frequency scale of a sound is stretched, some parts of the sound may sound louder than others before stretching. For instance, if you want to focus on the bass line of a song, transpose it an octave above and it will be ...

0

Your question mixes two different notions of "color." There's a "physical" definition where it's used loosely to mean a pure frequency of light, as would come from a laser. There's also the view from human perception, which is the "intuitive" notion of color used in everyday language. They are not the same. From the physical point of view, if the ...

18

The unit of illumination is the lux, lumens per square meter. What is the minimum lux required for reading? How many lux does the Sun provide at distance D? What is the minimum lux required for reading? You can plug all sorts of numbers into this depending on how good your eyes are, how big the print is, and how close you hold it to your face. I'm going ...

7

Based on the information here which claims: Daylight is between $10^4$ to $2.5 \times 10^4\,\mathrm{lux}$, and 1 candle at 1 foot is $10$ lux (I'll use this as the readability limit) using the $1/r^2$ scaling yields that "daylight" will fall to $10\,\mathrm{lux}$ at somewhere between $30\,\mathrm{au}$ and $75\,\mathrm{au}$; Pluto is at around ...

1

That information is not contained within the bb radiation - all that can be gleaned is an emitting area and a temperature. In practice the radiation can have arisen from any process where it is feasible for a photon at that frequency to be produced. Of course to actually be a blackbody emitter there must also be a 100% chance that a photon at that ...

30

This is very rough and based on eyeballing without careful measurements: I've got a four-watt nightlight. I can read by it (not comfortably) at a distance of about a meter. The sphere of radius 1 meter has a surface area of about 12 square meters, so it appears that 1/3 of a watt per square meter will (barely) suffice for reading. The earth gets about ...

22

One point, the difficulty of seeing colors in dim light is due to properties of the human vision system. Most cameras will not have the same effect and will be able to show vivid colors in even dim light (as long as the light is sufficient for imaging). But as a good guess, with accommodation, you can read (to some extent) under a full moon. The sun ...

1

That information is not contained within the bb radiation - all that can be gleaned is an emitting area and a temperature. In practice the radiation can have arisen from any process where it is feasible for a photon at that frequency to be produced. Of course to actually be a blackbody emitter there must also be a 100% chance that a photon at that ...

3

You are almost certainly seeing fluorescence here. Even 532nm is still a fairly short wavelength as far as fluorophores are concerned and many fluorophores commonly used to color plastics absorb powerfully at 532nm. For example, rhodamine B has an absorption peak near 532nm, and it fluoresces strongly in red yellow. Indeed, if you see the Wikipedia Page for ...

1

http://www.nndc.bnl.gov/nudat2/indx_adopted.jsp Appears to be the answer I was looking for, if anyone else is interested.

5

The 511-keV photons in the $^{22}Na$ spectrum are annihilation photons. They definitly have Compton interactions as seen in the diagram. The 1250-keV peak is a gamma in the daughter of the sodium positron decay. It also has a Compton edge and a backscatter region.

3

Photons are photons. If photons from one source experiences a certain kind of physics, then photons from other sources do too. So, short answer: yes. And they can produce Compton edges as well.

0

It can be done. But the frequency has to be very high which makes it useless for practical purposes.

1

Why don't electromagnetic waves need a medium to propagate? That's a "Why question". It's dangerous to ask "Why" in physics, because the answer is simply "they do". "How" is more interesting, and in this case it is very complicated. Just know that for a very long time, people really thought that electromagnetic waves needed a medium to propagate, ...

0

That a rigid body can move without medium - in vacuum - is without any doubt. The same is possible for gas or single atoms or electrons ... Beside matter exist energy. Max Planck explored the black body radiation and found, that energy consists of energy packages. Planck not believe, that the energy packages in his formula are real. Einstein stated, that ...

0

An oscillating magnetic field always induces an oscillating electric field. So, the "magnetic field oscillating at radio frequency" is just shorthand for a radiofrequency electromagnetic field: both electric and magnetic fields are present. I believe the manual chooses to emphasize the magnetic field because it is the magnetic field that couples directly ...

0

From Wikipedia: The hertz (symbol Hz) is the unit of frequency in the International System of Units (SI) and is defined as one cycle per second. It is named for Heinrich Rudolf Hertz, the first person to provide conclusive proof of the existence of electromagnetic waves. In English, "hertz" is also used as the plural form. As an SI unit, Hz can be ...

2

Absolutely pure water has an absorption coefficient of about 0.01 $m^{-1}$ in the visible part of the spectrum, however in general terms you might beat this with very long wavelengths. https://en.m.wikipedia.org/wiki/Electromagnetic_absorption_by_water Even "fresh" water has some conductivity $\sigma$. In which case, the "skin depth", which is the e-folding ...

2

There's no critical angle for light traveling from air (or another transparent medium) into a conducting medium; even if you shine a light at a perfect 90° angle to such an interface, it will still be almost entirely reflected. Rather, the reason for reflection is that if you try to set up electromagnetic waves in a conducting medium, the electric field of ...

0

At the fundamental level a beam of light is composed out of a huge number of photons with energy=h*nu . The photon is an elementary particle and obeys quantum mechanical equations. A classical beam, a solution of Maxwell's equations, emerges from the synergy of the wavefunctions of all those photons, here is a simple example: Left and right handed ...

0

A pentagon double prism in a globe shape format made from flat glass and silicone filled with water is a great way to disperse light into its different wavelengths. I have been told it is a great house fly deterrent as it confuses their many eye focal points, confuses them and makes them move on. would this be valid hypothesis or just an old wives tale?

2

In this description of the experiment the radiofrequency is supplied externally from the B field that splits the states. The application of the magnetic field then provides a magnetic potential energy which splits the spin states by an amount proportional to the magnetic field (Zeeman effect), and then radio frequency radiation of the appropriate ...

4

"Night vision", in the sense of a device to help humans see in the dark, can refer to different things. The most commons are: 1. Residual light enhancement These rely generally on the principle of image intensification in which incoming light is converted into electric charge via the photoelectric effect and then amplified with a microchannel plate ...

0

Mount the lamp on the outside of a really fast spaceship. If you want a longer wavelength, fire up the engines of the spaceship and get it to travel away from you, so that the light is red-shifted. If you want a shorter wavelength, do the same but with the spaceship travelling towards you, so that the light is blue-shifted.

0

Why doesn't night vision have color? The normal human eye has two basic kinds of photoreceptors, rods and cones. Amongst the cones, there are three subtypes, each of which is sensitive to a different range of frequencies of visible light. These three different types of cones are what lets you see colors, but only if the lighting is sufficiently intense. ...

3

The colors we see in a day-to-day basis are the result of a source emitting light in a broad spectrum with varying intensities for different wavelengths. If you shrink that spectrum to specific wavelength or to a narrow band of wavelength, then your experience will be of seeing a specific color in different tones. Night vision cameras achieve their results ...

-2

Below is my opinion, Actually, human body not only emit infrared radiation, it also all kind of different frequency waves or spectrum, but infrared radiation emitted from human body is higher intensity compare to other wavelengths. Our universe is containing few kind of current known fields such as electromagnetic field, gravity field, quantum field etc, ...

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