# Tag Info

5

No, in a vacuum light sources will appear dimmer as you move further from them because of the inverse square law. In a medium, a light source suffers from both the inverse square law and absorption/scattering. Below is a diagram illustrating the inverse square law: As you move further away from a light source, your pupil (assume it remains the same size, ...

0

Well just what do you mean by "Brightness." There is no such unit of measurement in the field of Photometry. A more appropriate and meaningful term is "Luminance" measured in Lumens per steradian, per square meter, or alternatively, in Candela per square meter; since the candela is one lumen per steradian. And Luminance is a property of a light source; so ...

5

It's because the amount of area "covered" increases as the square of the distance. Imagine a sphere, centered on the source, at a radius equal to your eyeball's location. If the source generates X watts (or whatever unit you like) total, the brightness, i.e. the percent of light which hits your eyeball, is X divided by the ratio of your eyball's area to ...

1

Your coloured object is absorbing light, i.e. light is changing into mechanical energy, while the atoms in the Bunsen burner are emitting light, i.e. mechanical energy is changing into light. If you have, for example, sodium atoms in a flame those atoms are continuously colliding with air molecules. The velocities of the air molecules are a function of ...

2

In Bunsen burner atoms get heated up which means they absorb energy, and go to some excited states (or even get ionized). And then they de-excite in any way they can which means radiative decay is dominant, as there is almost no other way to dissipate energy. There is no crystal lattice, and compared to luminescence rate collisions are infrequent. So yes - ...

2

The phosphors lining the glass tube of a fluorescent light do a pretty good job of smearing the atomic mercury line emission spectrum into something closer to black body radiation. There could be a phosphor mix that accomplishes what you want...

2

When you scatter light off of a material there is a photon-phonon interaction which will shift the photon frequency depending on the phonon energy (Raman scattering, Brillouin scattering). The effect is quite small, however. How much broadening do you need? Rayleigh scattering through a warm, high density gas will probably go quite far in messing up the ...

3

This might not be feasible for your setup, but you could try rapidly rotating your light source, which would Doppler broaden your spectral lines. You're correct in that the speed would need to be a substantial with respect to the speed of light. As an example, if your frequency is 500nm let's say. If you'd like it to spead out on the order of a single ...

0

Try this: Make a paper cut-out of the letter F (or any easy-to-cut-out letter which is different from its horizontally and vertically flipped images). Shade or mark one side of the cut-out to distinguish it from the other. Get a second piece of paper and write the same letter on it several times. Hold the cut-out in front of you while standing in front of ...

0

Vsauce answered this question exactly in one episode. I suggest checking it out, it explains the different ways the eyes see the mirror, and gets right to the point.

1

This very nearly a duplicate of Does infrared rays pass through polarized glass? if you ignore the bit about polarisation. There is a useful collection of articles about the optical properties of glass on the Schott web site, and in particular there is one titled TIE 35: Transmittance of Optical Glass. The article is freely downloadable, though you need to ...

0

That sounds wrong. Also, later it somehow contradicts itself by saying A greenhouse keeps warm inside because of the properties of the glass. The short wavelength radiation from the Sun can get in but because the plants and soil are cooler they emit longer wavelength radiation which cannot escape through the glass. Heat radiation is mostly ...

3

The reason is not quite as intuitively put as for ropes, but it is essentially to make the fields consistent with the electromagnetic boundary conditions, which in turn can be traced to (1) Kirchoff's voltage law and (2) no conduction currents can flow in a dielectric. Consider a tiny, thin rectangular loop running parallel to the interface with one half ...

-1

The hypothetical matter you are referring to is called neutronium. It is basically a huge nucleus of pure neutrons. The problem is that current physics predicts that it would instantly decay via radioactive decay and explode much like an atomic bomb. thats how it would look like. Edit: My answer concerns the only qualitatively different case from the ...

5

Actually, it can be theoretically derived from D'Alembert equation (that is satisfied by each component of ${\bf E}$ and ${\bf B}$ in absence of sources in view of free Maxwell equations). The idea is to compute the field (any component of ${\bf E}$ or ${\bf B}$) in $p$, when it is generated by a spherical point source localized in $q$ emitting a spherical ...

9

In theory, perhaps. It is possible, using multilayer dielectric coatings, to produce a surface which is reflective in very narrow bands (in this case, the Sun's dark lines)and transmissive (or absorptive) elsewhere. In practice, the spectral "blurring" caused by atmospheric transmission/absorption/re-emission effects would make this effect pretty much ...

1

Since the data is captured in Red,Green,Blue and you know the correction filter's transmission in each of these bands you can simply scale the RGB output to give you the colour shift you want. All you need to know is the Red,Green,Blue bandpass of the Bayer filter on your camera's chip. You probably need to do this with your camera's raw mode. Other modes ...

2

Any warm body radiates electromagnetic radiation with a spectrum that depends on the temperature. Above 500 degree centrigrade there is enough radiation in the visible part of the spectrum to be visible but at lower temperatures most of the radiation is at infrared frequencies or lower. Our eyes are not sensitive to infrared radiation so we do not see it.

4

The human eye is only capable of perceiving a very limited range of electromagnetic radiation, with wavelengths ~400-800 nanometer. Objects at low temperatures (room temperature) do not emit an appreciable amount of radiation in this range. The fact that we CAN see objects when it's light is due to reflection. For more info, take a look at this wikipedia ...

3

Cold bodies radiate mostly in the infrared zone (invisible to the human eye), but as the temperature increases the body will emmit higher frequencies with more intensity. So room temperature obects will not be seen due to black body radiation. As you can see, hot bodies are visible because they emmit visible light mostly.

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 ...

0

If we assume that mirrors will leak some energy, then is it possible to put objects such as a photo multiplier tube (in combination with a mirror) and adjust it in such a way that only the amount of energy lost by reflection of the mirror is recovered and sent back to the other mirror. This cannot be done even as a thought experiment. Photomultipliers ...

0

Since this came up again I will add my two cents, which consists in delving down to the quantum mechanical level. Current day physics accepts that the fundamental framework of nature is quantum mechanical. Classical mechanics, classical electromagnetism are emergent theories from the quantum mechanical foundations, in an analogous way that thermodynamics is ...

0

Ben Crowell's answer is good; let me just add a useful way to think about it, in a sort of quantum-mechanical way while retaining the theme of the question: the fully developed superposition of waves (including the reflected, in your thin-film example) says where the energy actually can be transferred. The places where the waves add up to 0, there the ...

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

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 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).

0

I have since found this pdf from CVI Melles Griot giving a temperature coefficient of 0.016 nm/°C at 400 nm, increasing to 0.027 nm/°C at 820 nm. This will vary between coating types but it is enough to get started.

0

It's for you to judge if the following is relevant. "Another example of a drastic change in surface properties by one-atom-thick layers can be a silicon monocrystal surface. When the surface of a silicon plate is etched in hydrofluoric acid (in order to dissolve the thin native oxide layer), the atoms of silicon on the surface are terminated with hydrogen ...

0

This was just asked on another forum last week, a student from Penn State made a sheet of graphine to see for himself. He reported that he could not feel the sheet, but he could see it.

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 ...

11

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 ...

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 ...

23

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 ...

45

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 ...

8

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 ...

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 ...

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 ...

0

I wouldn't say it is a myth. Like you say, it's complicated. When you are in the sun, the predominant source of heating is from incident radiation, whereas in the shade cooling would take place primarily through convection, conduction and evaporation. When you are in the shade, you are just not that hot compared to your surroundings for radiative cooling to ...

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

Do you know that you are glowing right now? Especially your brain that glow alpha and beta wave, known as aura When electron was shaken by any action it would always generate electromagnetic wave. However, normally electron was not shaken in enough frequency to generate visible light so you never see it. But sometimes it can generate visible light and when ...

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: ...

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 ...

0

You see image not in white because when it come to your eye. It not white When you see red object. It because red light come into your eye more than other light See your computer monitor, see deep into one by one pixel. And imagine that each pixel will shot one light into your eye. When you see white pixel, it because it shot 1 white light into your eye. ...

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

I like thought experiments ! No need to replace those expensive melted bits !! And that is relevant to this experiment. We will wind up the RF output of the transmitter until the aerial itself is glowing a nice cherry red . Still intact (just !) as a functional aerial but emitting light (you can see its red). So assuming we have still got a good SWR and all ...

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