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6

The red, orange, yellow, and white parts of a candle flame results from glowing soot. The color in this part of the flame is indicative of the temperature. The spectrum in this part of the flame is fairly close to that of a black body. The blue part of the candle flame at the bottom of the flame results from chemiluminescence. Chemiluminescence is not black ...


6

Photons, and in fact all other elementary particles, are not assembled. They exist. We don't know how, or why, they simply do. Photons are the quanta of the electromagnetic field - in quantum field theory, we associate to the classical electromagnetic field particles called photons. For a technical account if how this (roughly) works, read my answer to "The ...


5

Basically, absorption lines exist because absorbed photon are not re-emitted in the same direction, so dark lines can be observed. There are various reason causing this. For example, the extra energy can be dissipated as phonon in solid or strongly interacting system. Excited states can also emit multiple low frequency photon if there are meta-stable ...


5

Someone else with knowledge of the mathematical details may be able to say more, but I think the statement that the deflection of light is twice the Newtonian value is just looking the initial angle that the light approaches the gravitating body from far away, at comparing with the final angle once long after it has passed the body and is far away again--in ...


5

Terrific photo - good that you were able to get it and don't apologize. This diagram might explain it: The sun is "below the horizon" as demonstrated by the green dashed line. A mirage can be formed by rays following the blue line (exaggerated scale, showing a layer where light can be reflected because of a sufficiently large change in density). If the ...


3

It would be better to comment this but I only just joined, so no reputation...you'll also have to forgive me if this is a silly idea, I'm not a physicist :), it's just my 2c. For the sake of argument, assume that (from the perspective of an observer on the "ground") the light deflection really is twice as large as the bagel's rate of fall - or at least ...


3

In the book "Physics of the Plasma Universe" Dr. Anthony Peratt puts candle flames near the bottom of "energy in electronvolts" portion of the 'plasma spectrum'. If you look at the chart below, you'll see candles flames about midway (ok, cosmologically) between the ends: solar bodies and laser radiation terrestrial flames interstellar charged gases ...


3

If "wet" causes the material to become less scattering, and thus darker, because of a smaller change in refractive index between the fibers and the liquid, then the experiment to do would be to change the refractive index of the liquid. If you can see a change in "darkness" as a function of refractive index (making sure to correct for surface reflection ...


3

The optical isolator component is active. It consumes energy and so is no different (thermodynamically speaking) from the heat-pump in a refrigerator. If you are talking about a passive component that draws no power then, a surface that allowed light pass in one direction only would violate the second law of thermodynamics. To see why, imagine two rooms, ...


3

The laser goes through the bagel hole each time until it hits the ground (assuming the mirrors are set up nicely orthogonal to your uniform gravitational field). To see why, the equivalence principle is all we need. You can imagine thrusters powering an elevator without gravitational fields, set up your mirrors and do the experiment, and that's what you ...


3

Your wording suggests a few misconceptions: It seems you are thinking of light as having a corpuscolar nature (nothing wrong with that, you are in good company). Well it turns out that things just do not work that way. Phenomena like diffraction (to name one) tell us that we cannot describe the behaviour of light thinking of it as composed by (classical) ...


3

Whenever one makes light, one makes zillions of photons, so the way to make photons is the way to generate light: burning wood etc, striking stones, make wires incandescent, as in electric bulbs, etc, fusing nuclei as in the sun and stars and maybe more ways. It is much easier to make zillions of photons. How one photon is made takes us to the realm of ...


3

I don't think you can understand how a photon is made without knowing what a photon is, and to understand what a photon is requires understanding quantum field theory - specifically quantum electrodynamics. Quantum field theory is a very odd way of looking at the world, but it works and gives predictions that agree with experiment. In quantum field theory ...


3

A lens is symmetric. By this I mean you can pick it up, turn it round and put it back and the light rays don't change. Or to look at it another way, it doesn't matter whether the light travels from left to right through the lens or right to left. So parallel light travelling from infinity on the left of the lens will converge at the focal point on the ...


2

The explanation you give is correct. A white body reflects all wavelengths. We call it white when all colors (all wavelengths) are reflected from an object and hit our eye. Black is the opposite. I would say that white is all colors, as you do. But maybe he sees it from the perspective that since all is reflected and nothing is absorbed, there is "no ...


2

32,000 - 100,000 lux is the typical range of illumination that the Sun provides. You don't have to look at the sun, you look at the world it illuminates. Lux is a "per unit area" quantity - not a "per solid angle" quantity. The variation in values mostly depends on the position of the sun in the sky - when it is low, there is significant scatter of sunlight ...


2

As an analogy, consider the photon that strike your face and reach my eyes, we say that that photon carries information about your face which then helps me to identify you, You are confusing the individual photons with the electromagnetic wave that is light, which is composed out of a zillion photons. but don't these photons collide midway with air ...


2

This is the plot of sunlight, red at ground level. Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power. As all light comes from the sun during daylight this should suffice. One can get the number of photons by ...


2

To my knowledge, there is no discretization of the light wavelength (they form a continuous spectrum). On the other hand, there exists no infinitely narrow absorption "potential". I mean that all transitions of electrons that may correspond to a photon absorption have a finite width. Consequently, the photons have a non-zero probability to get absorbed. ...


2

It is not correct saying that no force is applied. A photon carries momentum see PE here so on reflection there is momentum transfer. This is the idea behind laser propulsion discussed here. Concerning the speed it is even more complicated. The fact that light gets reflected usually requires an abrupt change in the index of refraction. To get reflected, ...


2

Without GR, there is no reason for Mercury to not be tidally locked with the Sun, always showing the same face to the Sun (much as the Moon always shows the same face to us). There is also no reason for clocks at higher elevations to desynch with clocks at sea level. If you take these as strong evidence for general relativity, then all that the authors' ...


2

The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...


2

Darkness is nothing, it is not a physical entity. Darkness is what you have when there are no/very less photons, and thus you can't see anything. When you light a lamp, the photons from the lamp bounce all over the room, and hence the darkness "disappears", so to speak. Absence of light doesn't exist when there is presence of light.


2

I'm not sure if this is worth an answer but anyway- darkness is not a physical entity in itself. It is simply where light isn't. As it is not an entity, it doed not have to go anywhere when you turn on a light, it just seizes to exist as light enters the place where beforehand it wasn't.


2

… photons undergo twice the deflection from gravitational fields as do physical objects. I don't think this is a correct assessment of the situation. Rather, there is a Newtonian way to predict the deflection of light due to gravity: assume that the light is made of corpuscles with effective mass $m=E/c^2$ which enter the gravitational well with speed ...


2

The technology is described in this Wikipedia article. The bottle is acting as a light pipe. It doesn't create any light itself, it just transfers sunlight outside the building into the building. Here in the UK light pipes aren't much used, probably because electric light is cheap and natural light tends to be rather variable. And of course light pipes are ...


1

The Poynting vector $\vec{N}$ is the power per unit area of your beam. If the beam is perfectly absorbed, then the force is given by $$ F = \frac{1}{c} \int \vec{N} \cdot d\vec{A}$$ So, providing you have the beam incident normally upon something, the force on it will just be the power of the laser divided by the speed of light. Of course, if the light is ...


1

The radiative transfer equation is a simplified model for describing light transfer. Of course it is possible to derive the radiative transfer equation by the Boltzmann equation for a photon density function $f(x,t)$: $$ \partial_t f(x,t) + v_x \partial_x f(x,t) = (\partial_t f(x,t))_{coll}. $$ Here, the term $(\partial_t f(x,t))_{coll}$ is the gain and ...


1

Light does indeed bear momentum. If you have a quantity $E$ joules of light in a plane wave beam, it has linear momentum $\frac{E}{c}$ in the direction of propagation. So if it bounces off a mirror, the impulse transferred to the mirror is $2\frac{E}{c}$, in the direction of propagation, at least at first. Most certainly, the light bounces back. If the ...


1

No. The only effect that the red color would have would be to attenuate all the colors that are not red. With white light in, you'll still get a spectrum out, but the blues and greens will be dimmer than they would be with a clear prism. (The positions of the colors might be shifted by a very small amount because absorption comes with a small change of ...



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