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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 ...

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 ...

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 ...

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 ...

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 ...

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.

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 ...

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 ...

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 ...

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.

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 ...

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.

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 ...

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 ...

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

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 ...

2

The problem for photons is that you cannot observe them time to time. If you do measure their state, photons will collapse to some Eigen state and become classical. In fact, a photon can have a lot of eigenstates in theory. That is special compared to EM for photons.

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: ...

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 ...

1

There are two types of energy involved, and the blurring of this distinction is cause of a huge number of misunderstandings. Light comes in discrete packets called photons. The energy of each photon is proportional to the frequency of the light. On top of that, a light beam can have any number of photons in it, and this gives it its overall power. The power ...

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 ...

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

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 ...

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 ...

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

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

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 ...

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, ...

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