# Tag Info

## Hot answers tagged visible-light

43

The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons). You can do the calculation: If you're 10 light-years away from ...

20

Although I agree with all three of the above answers let me present a slightly different perspective on the problem. It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light ...

10

The only stars you can reliably see are ones that are spewing enough photons at your eyeballs to appear stable. Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough. So your question is basically embroiled in observer bias; it ...

7

Allow me to channel something akin to the anthropic principle here. You can only see the stars that have a lot of photons reaching your eye. If a star were so far away that photons were reaching your eyes only occasionally then the star would be too dim for you to see in in the first place. Even if you could see the photons, the star would appear to ...

6

It is extremely hard to say what would happen because the only way to reliably test those regimes is to do it experimentally. Single-atom thin layers have only been realized, so far, in graphene, which is a single layer of hexagonal carbon crystal, and which is strong enough to exist by itself without any support. The Wikipedia section on its optical ...

5

A star radiates in all directions. You would still see the star regardless of the number of steps you take to any side, just not the same photons. A laser radiates in only one direction (or in a very small cone). If you took a large enough step to the side (larger than the angular size of the emitted beam) so as to exit this cone, then you would no longer ...

5

To simulate unpolarized light, you need to do two separate simulations using Maxwell's equations. In the first simulation, assume the incoming light has some polarization (any polarization will do). In the second simulation, assume that the light has the opposite polarization (y is opposite to x, right-circular-polarized is opposite to ...

4

Electromagnetic waves obey the principle of superposition. In other words, the light waves pass right through each other. There are situations in which this is not the case. It's the same reason you can hear everyone.

3

To answer your question without getting boggled down in quantum mechanics, we can assume the light is of sufficient intensity(*) or that we are talking about run-of-the-mill radio waves (again, with sufficient intensity). Well, regardless of the nature of the light, at any point in space there will be a definite $\vec E$ and $\vec{B}$ directions, so ...

3

To model a diffuse surface, imagine a house that is on fire inside (!) so that everything inside is emitting light equally in all directions. You can also imagine a very hot oven or kiln in which the interior walls are aglow. Now, if you look through the door of the house, the flux of light entering your eye is obviously proportional to the area of the ...

3

From your point of view it would be instant. Sun is there, then poof! The sun is gone and so is its influence. We only know this event happened ~8.5 minutes before we saw it because we're so clever :-). However there's no way to detect it early and warn ourselves because no information could reach us any faster than the sun's extinguished light and ...

2

Gravitational waves travels at the speed of light, thus you would feel it at the same exact moment you saw the sun disappear. By general relativity, spacetime acts like a trampoline being bent by a central mass. When the mass is removed the trampoline does not go back to the unbent state instantaneously.

2

It is indeed a topic that is discussed in many books but only a few give a rigorous mathematical description of the phenomena. For stringency in non-linear optics topics I always trust HARTMANN ROMER: Theoretical Optics, An Introduction. 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim. A book which is also mathematically rigorous is BOYD, ROBERT W: ...

2

Under normal circumstances, what you are seeing is the steady state condition where the rate of absorption is equal to the rate at which energy is conducted away or radiated away again, so the material doesn't heat up. As I understand, unless a material fluoresces, the de-excitation happens in the infrared. Also, you should realize that just because a ...

1

Strictly according to Double Slit experiment calculations, yes infinite fringes are possible if (1) the slits are infinitesimal and placed infinitely apart, but actually, the fringes will be very close to each other so your eyes won't be able to differentiate and secondly each of the fringe will have almost $0$ intensity.So for obtaining infinite fringes you ...

1

The main idea behind polaroid sunglasses is that reflexion from water, snow and other glary reflectors is mainly polarized in one direction. To understand this, witness the behaviour foretold by the Fresnel Equations (the graph below taken from the Wikipedia "Fresnel Equations" page): so that you can see for a wide range of scattering angles from these ...

1

It's all because of the wavelength of light. In most bands the radio telescope is about the size of the wavelength it's observing - so it can only see a single point at once anyway. It would be like having an optical telescope that was a tiny microscopic pinhole - there wouldn't be much point in having a megapixel camera behind it. Radio telescopes that ...

1

I suppose it depends on application. For example, broadband dielectric mirrors sold by ThorLabs do not specify their temperature-dependence for the obvious reason of redshift magnitude you specified. Even narrowband dielectrics and laser line mirrors don't usually specify this. However, other devices such as crystal optics for wavemixing can strongly depend ...

1

I found a good paper that can help you. However, due to copyright issues I cannot put the spectra here. Try to get this article: "The Distribution of Energy in the Visible Spectrum of Daylight". A. H. TAYLOR and G. P. KERR. J. Opt. Soc. Am. 31 no. 1, pp. 3-8 (1941) . Also available here (pdf).

1

In the specific case of slowing light with a Bose-Einstein condensate there will be a limit because the slowing of the light is due to an interaction of the light with the BEC to form a polariton. If you put too much energy in you'll destroy the BEC and it will stop slowing the light. Offhand I don't know what the limit is, but it will be a very small amount ...

1

There seem to be two fundamental misunderstandings here. 1. The nature of electromagnetic radiation Take a look at the depiction of the electromagnetic spectrum from Wikipedia. There are lines drawn there, but they are as "real" as lines of longitude drawn on maps of Earth. There is no fundamental difference between different types of light. Instead, we ...

1

But as I've learnt at, we see an object- any object- because of the White light reflected by that object. Not only white, any type of light. In other case we would only see white. But White light is just one of the 7 types of radiation What are the rest? gamma ray from and object into my eyes, even if my eyes were able to perceive ...

1

In essence, what is the difference between scattering (elastic or nonelastic) and fluorescence from an electronic transition point of view? Elastic scattering of photons happens on the collective spill over electric field of the material and the quantum mechanical formulation is comparable with the classical formulation as far as the directionality ...

1

Reflection can be calculated in a completely classical way requiring only that the material have a bulk polarisibility. It's true that ultimately the polarisibility is a result of the electron configuration, and this arises from quantum mechanics, but you would not call reflection a quantum process any more than you'd call viscosity of a fluid a quantum ...

1

I'll assume that "enhance the vibrancy of colour" means something like turning up the saturation in a photo editing application. I think it's highly unlikely that this could be accomplished using analogue optics, because it's a highly nonlinear operation. This is because we need to block or let through light in three different frequency ranges (red, green ...

1

In one sense there is no such thing as either a antumbra or penumbra. Neglecting minor diffraction effects, a point source of light is either visible from a given point or it is not visible from that point. In order to deal with an extended light source, like the Sun, we need to answer two questions to determine the light level at a point. First, what ...

1

You seem to have answered your own question. The only requirement seems to be that the light source behave as you have assumed in your derivation. When there is a lot of scattering, as in a cloudy day, there are many secondary sources that will wash out this arrangement, and obviously, somewhere in between there is a point in which the approximation breaks ...

1

Monochromator rail setup If you have access to a white light source, a pair of lens (to columnate the light source and focus it into the monochromator), a monochromator and a photodetector, then the experiment is fairy easy. Measure the spectrum of the white light source, $I_0$ Place the filter in between the source and the monochromator, $I_f$ The ...

1

How is light created at the atomic level? Atoms are composed of nuclei, agglomerates of protons and neutrons, which have electrons in orbitals around them; as many electrons as protons so that normally atoms are neutral. What keeps atoms stable is that the orbitals exist at specific energy levels given by the solution of the potential problem in ...

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