# Tag Info

1

Everything scattering light has a characteristic spectrum. The spectrum is defined by the way wherein the particular thing interacts with light, and this is set by the (1) chemical makeup and (2) texture (at the wavelength-of-visible light scale). You cannot change the spectrum without changing one of these two things, either by e.g. (1) altering the ...

1

Many light bulbs already do this. See for example this article. I was in the lighting technology business at one point. At that time it was done in some tungsten-halogen incandescents. I don't know current state of things.

0

As CuriousOne says, look carefully for a test suite within your software installation. MEEP is widespread, notwithstanding the LISP interface (gotta love MIT's confidence in its own creations), so if you seek carefully, you are bound to find MEEP analysed examples on the web. As for your proposed simple test: it is a good idea, and there are many, well ...

8

Interesting question! Cherenkov radiation would definitely be inefficient for illumination. You only get Cherenkov radiation from charged particles moving faster than the local speed of light in a medium. If you have a transparent medium with index of refraction $n=2$ and you're sending fast electrons through it, you'll only get Cherenkov radiation while ...

4

Quite aside from the issue of ionizing radiation, Cerenkov generating particles also lose energy by other processes and that ends up as heat. Moreover, all the kinetic energy of the particles once they drop below the Cerenkov threshold is lost in non-optical channels (i.e. more heat). So no, they could never be anywhere near as efficient as diode ...

1

Better late than never? Yes, what you found in the Matlab package is correct. The luminosity function is the exactly the same curve as the green part of the three XYZ Tristimulus Curves used in modeling human perception of color. Note the XYZ color space is not the same as the RGB color space of a display; XYZ represents the totality of what humans can ...

0

You are totally right about photons and their reflection and absorption on bright or dark surfaces if you illuminate these surfaces with visible light. The photons in the range of visible light will be reflected or absorbed (and re-emitted with a longer wavelength) and that is the reason we see these surfaces as bright or as dark. And the dark surface has a ...

0

When I was in high school I wondered the same thing. According to the Inflation models space time expanded exponentially between $10^{-36}$ and $10^{-32}$ seconds after the Big Bang. According to the theory a piece of space the size of a nucleus expanded to the size of a galaxy (rough order of magnitude, possibly even larger) in that very short time. ...

0

For a parabolic mirror, the answer is "all rays go through the focus". For a spherical mirror, you can see the answer most easily by looking at the limit as the spherical mirror is almost a hemisphere: Clearly the rays that start further off axis are focused closer to the mirror.

1

Unfortunately, as interesting an idea as this is, and as creative as you must be for thinking of it, it's not an actual possibility as far as I'm concerned. A one-way mirror works much in the same way that a metallic screen door works. It allows you to see from the inside of your house, outward. However, this is due to the fact that there is far more ...

1

In addition to @Floris response: You have missed a lot of wavelengths in your list of wavelengths that would experience interference. Take your example of a $6,000,000 \text{ nanometer}$ pane of glass, and consider that 15,000 waves of $400 \text{ nanometer}$ wavelength light exactly fills this space. So, indeed, this light will experience some sort of ...

3

Very simply, when a plate is quite thick, the fringe patterns will be very close together - because a tiny change in angle will result in an additional wavelength's worth of path difference. Different colors will have a different repeat distance (because of different wavelengths); and light will typically arrive at the eye from more than one direction ...

0

The answer I gave before was wrong, as was kindly pointed out by CuriousOne. The fading of the intensity isn't because of the $1/r^2$ fall-off of light, since the diffraction formulas are only for small angles anyway. First of all, it's clear from the second figure in the question that the only relevant thing is the diffraction and not the interference; ...

0

Is this not a question about human physiology rather than physics? We are discussing the variations in observed response of the human eye to light with different qualities: overall intensity, and distributions over the visible spectrum. This question seems to ascribe all of the observed differences (or lack thereof) to the quality of the light; "the Moon ...

1

But I've never seen that happen. You haven't looked then. The rising or setting Moon is rather reddish, just as is the rising or setting Sun. However, there is a difference between the Moon and the Sun. You can look directly at the Moon, even a full Moon, regardless of where it is in the sky. On the other hand, you can only glance at the Sun when it is ...

2

This is just an opinion, but the moon on the horizon is simply less visible than the sun is. I suspect that color changes it makes are more subtle and less easily noticed. However full moons are often noticeably orange. Here is a page with a wonderful time lapse view. http://www.pikespeakphoto.com/moon-rising.html

1

Reflection,refraction and transmission of light are macroscopic manifestation of a phenomenon called scattering.In this incoming photons are absorbed and either the quantum energy level of an atom is raised (as in case of resonance absorption) or the outer electron cloud is set into motion(this is responsible for light around us).Almost instantaneously ...

1

The spot of light isn't below the line of sight of the laser gun, and the outside observer shouldn't expect that to be the case. The laser gun is attached to the elevator wall, so according to the outside observer, the momentum of photons as they come out of the laser gun must have a non-zero upward component, or else conservation of momentum would be ...

0

If the elevator is going up rather fast, the light from the back of the elevator takes longer to reach the observer than the light from the front. This means that I see the back of the elevator at an "earlier time" - put differently, it looks to be a bit lower. The angle by which it appears to be lower is given by the velocity of the elevator divided by the ...

0

Neither the antumbra nor pentumbra have a fixed brightness. It varies across the region based on the amount of the solar disk that is visible. Each has regions where it is brighter than places in the other. It does not make sense to call one "brighter" than the other.

0

I'm going do do the annoying teacher thing where I answer a question with a question. Which is darker, a partial solar eclipse, in the penumbra of the moon, or a planetary transit, in the antumbra of an interior planet?

-1

In the penumbral area, part of the disc of the moon obscures a section of the sun. In the antumbral area, the entire disc of the moon obscures a, logically, larger section of the sun. Therefore the antumbra cannot be brighter than the penumbra during the same event.

4

Your eye has a lens in it. Without a lens, the light is all spread out and overlapping, just like you say. The light from any given pixel goes out in all directions, but a lens can make it re-converge back to a point. Hold up a sheet of white paper. Is there an image on it? No, of course not. It has light on it---light coming from each object in the ...

2

Let me go a little further than @mark-h's answer: The behavior of light at an interface is described by the electro-magnetic field solution to the Helmhotlz equation. It gives the reflected and transmitted electric and magnetic components as a function of the refractive indices of the incident and exiting media. From those solutions we can derive the ...

0

The small angle approximation is very reasonable here. The angle of deflection predicted by the approximation is ${4GM \over rc^2}=8.49\times10^{-6}$ radians. Writing $\tan{\theta}$ instead of $\theta$ is just clutter. The difference between the two is $2.039\times10^{-16}$. In order to resolve this difference in angle in visible light, you would need a ...

6

To follow the information in Chris White's answer - essentially, you would want a medium that allows you to see the spectra. There are several online resources that could help you in this experiment, in particular, the CD spectrometer, which can be constructed simply and on that website, it shows several examples of how everyday light sources can be ...

9

Lasers by definition only emit a single wavelength of light. You use one if you want that wavelength or if you want your photons to be in phase. You don't care about the photon phases, and you want to sample all wavelengths, so a laser is very much the wrong tool. If you just want collimation of the light, mirrors, lenses, or even just well-separated ...

1

Binocular vision has already been discussed, but it left out an important aspect. A single eye is sensitive to distance. The shape of the lens changes to focus on near/far objects. The reason this is needed is that our pupil has finite size and cannot be modeled as a pinhole. The same physics is going on here as in a lens of a camera focusing on an ...

3

To have depth perception two eyes are needed. Our two eyes are some distance apart which causes the photons from an object to arrive at slightly different angles. The brain then reconstructs the depth field from these differences. Similarly, we can figure out how far nearby stars are by using images made by a telescope at two different times of the year, ...

89

You have created a rather poor pinhole camera (camera obscura). You can see an "image" of the sky, a green space (trees) and even a reddish brown blur that is your driveway. This is not diffraction or refraction - it's geometrical (classical) optics. Because the hole is pretty big, you see a very blurry image. But basically, the light from the sky falling ...

-9

Different wavelength light diffracts at different angles.

2

Yes. The boundary conditions for Maxwell's equations gets you this. This reasoning is much more simple than it sounds. High refractive index means phasefronts of a plane wave are nearer together than for a low refractive index. When the waves in both mediums line up at the interface, the spacing between the intersection of the phasefronts and the interface ...

5

In general, reflection and refraction happen when light passes from one medium to another. You can see this if you see your own reflection in a window. Now, as a light ray approaches the critical angle, not only does the refracted ray get closer to the surface, but the amount of light transmitted gets less and less. At the critical angle, the refracted ray ...

1

If the light ray is normal to the surface, the maximum amount of light is transmitted. As the light ray bends, as in your part (b), a percentage of the light will be transmitted (refracted) and the remaining will be reflected (at the incidence angle). Very near the critical angle $\theta_c - d\theta$, likewise, some of the light will be transmitted ...

4

It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ...

0

1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body). 2) It will be reflected back and forth, but only a finite amount of time. This is ...

1

Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone ...

0

This question has another interesting aspect which has more to do with neuroscience than physics: why do we perceive metals with a neutral colour (such as silver) as grey, even why they are shiny and therefore simply reflect the colours of their surroundings? One answer is that such metals always have some roughness and therefore scatter light from a range ...

4

The following passage is extracted from Stephen Hawking's book "A Brief History of Time": In fact, various contemporaries of Newton had raised the problem, and the Olbers article was not even the first to contain plausible arguments against it. It was, however, the first to be widely noted. The difficulty is that in an infinite static universe ...

7

Who is interested can find detailed information at wiki, or here The problem is known (as you added in your edit) as Olbers' paradox, and was posed already in the mid 1500's, by Johannes Kepler in 1610 and even later by Edmond Halley in the eighteen century, and curiously, even the novelist an poet Edgar Allen Poe anticipated possible explanations as to why ...

13

I'm going to respond to (v1) of the question, which asks why the night sky is dark (black and unlit) compared to the day sky even though there are many light sources at night. The updated question references Olber's paradox, which has been answered many times before. Like most things we see in everyday life, there are a number of reasons contributing to ...

18

The sky does not skip over the green range of frequencies. The sky is green. Remove the scattered light from the Sun and the Moon and even the starlight, if you so wish, and you'll be left with something called airglow (check out the link, it's awesome, great pics, and nice explanation). Because the link does such a good job explaining airglow, I'll skip ...

19

Note well: What we perceive as color is bit of a tricky subject. This is a different question, one that has been asked and answered multiple times at this site. Per the typical human eye response, sunlight at the top of the atmosphere is about as "white" as "white" can be. Some of that incoming sunlight is reflected back into space, some is absorbed by the ...

8

The hand waving explanation in your question is called Rayleigh scattering Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle therefore becomes a small radiating dipole whose radiation ...

5

The sun is technically green because the peak of its black body spectrum is near green wavelengths. When light scatters parallel to the plane of incidence (i.e., during the day time), it is blue-shifted. When light scatters perpendicular to the plane of incidence (i.e., sunset or sunrise), it is red-shifted. The light that is not scattered but makes it ...

3

Today we know that Collins is wrong. He appears to be unaware of Newton's finding, and of course, advances made after he wrote his book.

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