# Tag Info

1

The characteristic modes can be excited by photons whose energy matches these modes. However, heating mechanisms are dependent on several other parameters, not the least of which is the surface shape. Take a look at the literature on black-body cavities and on ultra-dark roughened surfaces for some info on how to maximize radiative absorption.

0

Well, obviously it would be the speed of light times two. However, this can be misleading. Equations are all fine and dandy, but if you do not understand them, they are not of complete help. Imagine that you have the following... 1) A 300,000 km long spaceship which is at rest in space. 2) A clock is located at each opposite end of the spaceship, and these ...

1

You vastly overestimate the meaning of frame. A frame is a (local) choice of coordinates on the spacetime manifold $\mathcal{M}$. All physical laws can be directly formulated on the manifold itself, without referring to frames at all. That is at the heart of relativity, and that is what Lorentz invariance means. Let's go through your numbered points one by ...

2

From your question, I guess the double mirror configuration is just an example you thought of. I suppose your question actually is about if a photon can be trapped. Then basically yes. A device able to confine electromagnetic wave or light or photon is called cavity. You should understand a photon does not necessarily means a propagating plane wave. It can ...

0

The best complex dielectric mirrors, see http://en.wikipedia.org/wiki/Perfect_mirror may reflect up to 99.999 percent of the incident energy. The loss is about 1/100,000, so after 100,000 reflections, the total intensity decreases $e=2.718$ times or so. If the distance between the mirrors is 3 meters, the light travels 3 meters times 100,000 = 300,000 ...

0

Yes & No, You can however create an perfect mirror, which does not absorb any of the photons energy however its simply not possible or even feasible at this time to create such a device without energy being conserved in the photon, but it will however it will loose its energy due to Gravitational red shift after a long-time or Red shift due to moving in ...

1

A full classical answer to your question is actually rather subtle and tricky, but its summary is pretty much as in Davidmh and DarioP's answer, both good answers: the electric field vector's direction is changing wildly. The classical description of the nature of depolarised light is bound up with decoherence and partially coherence, a topic which Born and ...

9

An object is white when it reflects more or less any color – any frequency – and it reflects it in a random direction: the light gets scattered. For this reason, the photons arriving from a specific point of the white object (e.g. a point on the paper) are photons that arrived there from random directions in the space and they have random colors. Because ...

2

This is similar to why a glass of water and a glass of milk look different. We can use geometric optics (used in 3D animation ray tracing) to explain this. When a light ray (of any color) hits a mirror, it reflects as another light ray. But for a white nonreflective surface, it is scattered in all directions. From the perception side, a eye gathers ...

2

No, it doesn't mean that. One must distinguish two things: "laws of physics that apply to an object" and "laws of physics formulated from an object's viewpoint". These are two different things. Laws of physics apply to all objects. And the behavior of the objects may be described relatively to many coordinate systems or "frames of reference". The special ...

3

I guess you have this image in mind: This hods just for a single photon, or elementary packet of light. An unpolarized beam of light contains a bunch of photons with many different polarizations. The total electric field will randomly jump all around, still the interactions with matter typically involve a single photon at a time. This means that if you ...

2

They will look all over the place. If you take a particular photon, you will see it in a particular polarisation state, but it will have nothing to do with the photon next to it, or the one before. Actually, getting a perfectly unpolarised source of light is quite difficult. One of the best cheap options are the sodium discharge lamps, used extensively in ...

7

You don't even need highly specialised equipment to see the colour separation of the sun at low sun angles, a decent zoom lens on a camera will see it, and it's the origin of the "green flash" effect as the sun drops below the horizon. This site offers a good image: http://www.atoptics.co.uk/atoptics/gf15.htm

5

There are better answers than this, but I just want to contribute. As Joshua said, The quantification of how much light bends when transferring from one medium to another is called the "index of refraction," and air's index of refraction is very very close to that of a vacuum, so the bending of the light is very small, and the spreading apart of the ...

1

If we incident a monochromatic light (assume red light) on a glass layer starting from thinner to gradually thicker and thicker layers of glass, partial reflection increases to $16\%$ and returns to zero-a cycle that repeats itself again and again. If the the layer of glass is just the right thickness, there is no reflection at all! And it is to be noted ...

1

The fact that you see the sun as red means that the shorter wavelengths (green, blue, purple, etc.) were significantly attenuated as the sun rays traversed the atmosphere, due to having undergone scattering. It seems to me that while some rainbow effect indeed theoretically takes place, the extent to which it happens is relatively small. Adding this to the ...

15

First, it must be said that the picture you provided in your question is extreme. The concept of light bending is true, but the amount that the light bends is nowhere near as large as the picture shows it. The quantification of how much light bends when transferring from one medium to another is called the "index of refraction," and air's index of ...

0

Depends on the size of the sun. If it's capable of illuminating the entire surface of the earth at once, then the transition would be sudden. Take a torch, and a coin, and sit in a dark room. Shine the torch obliquely onto one face of the coin. You can then choose whether the coin spins so alternating faces are exposed to the torchlight, or the torch ...

1

Well, assuming a lot of things aren't true, then no. For example, if the Earth was shaped like a pizza, and the sun revolved around it, then it would have the same transition.

0

The one reason behind this is Interference! The bouncing light rays interfere constructively or destructively which affect the propagation through the fiber. If you want to go one step further, at the scale of optical fibers and using lasers, the coherence length of the light allows for these interference to happen between the waves before and after ...

1

I'm not entirely sure that I understand your question, but what leaps out at me is it's wrong to suggest that the energy of the light rays are decreased by an amount proportional to the difference in gravitational potential between the place where the light source is located and the potential of the place where an observer is detecting the light. ...

2

This is a partial answer for the first part of your question. You are not completely right. In the Schwarzschild metrics, for instance, the ratio of the energies (or frequencies) for the photons, is given by (in $c=1$ units) (here no Doppler effect is included, we only check gravitational effect): $$\frac{E(r_1)}{E(r_2)} = \sqrt{\frac{1 - 2 \Phi(r_2)}{1 - ... 1 This is a fun question! Concerning the cross section for absorption of resonant light. I can speak to the case of Rubidium atoms and the 780 and 795 nm 5P doublet. The cross section is greater than the atomic size, and also greater than the mean of the wavelength and atom size. In fact it's much nearer to the wavelength of the light! The experiment is ... 2 Even classically the ability of an antenna to radiate or absorb is not related to its size but to it being "resonant" with the "ether" being its loading impedance. The size is related to the bandwidth over which this resonance can be achieved. For an antenna whose characteristic size is much less than a wavelength the relative bandwidth over which it can ... 1 It is also important to note here the difference between a single molecule or atom and a solid comprised thereof. A single atom is indeed not terribly likely to absorb a photon - a large bulk of atoms or molecules in a solid, however, is quite a different thing. The branch of physics that deals with this is called Solid State Physics. A relevant topic ... 2 I'm assuming you're talking about absorbtion lines. In that case it's a mistake to compare the size of the molecule to the wavelength of the photon. These absorption lines appear as a result of the electron energy levels of the molecule in question. These span a range of energies that, for most gases, will usually include visible light. Let's take a look at ... 5 In quantum mechanics you calculate the charge density by taking the square of the wave function. If you do this for a hydrogen atom in a superposition of the ground state the first excited state (1s and 2p) you get an oscillating charge density. If you analyze this oscillating charge using Maxwell's equations, you get all the properties of the hydrogen atom: ... 2 The absorption/radiation mechanism for antenna is purely classical. It depends on the current oscillation on the structure. For a molecule or atom, it is purely quantum. What only matters is energy level difference. There's no direct relation with the size of the particle. However, the two mechanisms are not unrelated. They get connected for the case of ... 2 Light (all the electromagnetic radiations) is something like raindrops-each little lump of light is called photon-and if the light is all one color, all the "rain-drops" are the same size._1 The size makes photons of visible light and other waves different. Credits: _1 Richard Feynman-QED, The strange Theory of light and Matter. 11 Yes, X-ray, UV, and even radio-waves ares made of photons. The differences is the Energy (or equivalently the wavelength). See the picture of the Electromagnetic spectra . The different nomination comes from the time of the discovery. Youre eyes can see the visible part. the radiowaves can be observes with antenna etc... The only differences is the way we ... 2 Does that mean the speed of an individual photon is c even with respect to another photon? I mean, shouldn't the relative velocity be zero? When we write "the speed with respect to X" we mean precisely "the speed as observed from the inertial frame of reference in which X is at rest" Thus, if it is true that the speed of an individual photon ... 0 The idea behind special relativity is that the laws of physics do not change from one reference frame to another (this is key). So, if you're moving at velocity v, a photon traveling in that reference frame will travel at c (not c - v, which is the classical Galilean transformation). I believe this is what your book means when it says "with respect to ... 0 The whole point of special relativity is that imposing this postulate (on the constancy of the speed of light) forces a reunderstanding on the very geometry of space and time. This re-understanding can be summarized by the Lorentz transformations. These describe the coordinate transformations required when transferring measurement from one observer in ... 0 John's answer is clear for the ensemble of photons that make up the electromagnetic wave If you are really asking how individual photons end up making the classical electromagnetic wave, whether reflected or not, you have to dig in further in to quantum electrodynamics. Lubos Motl has in his blog an entry of how classical waves emerge from a large ensemble ... 1 If you know when to expect it, light can be detected at the single photon level. For very long distances, the expected photon flux will be less than one per detector per unit time of your choice. Go look up the Lunar Laser Ranging Experiment, in which light was detected at this level. Also, the Lunar Laser Communication Demonstration, which allows high ... 6 Infinity as long as you have a detector strong enough to detect it. Light keeps on travelling in a straight line forever as long as it doesn't bounce off some object. The problem with the lights you are talking about is that their intensity is really low and you can't resolve them because of the other stronger sources of lights you have around yourself. ... 4 Your observation is linked to the "Optical window in biological tissue". Like you already suspected, the absorption of blue light in tissue is higher than the absorption for red light. Best read the related wikipedia article, where all relevant effects are nicely illustrated. http://en.wikipedia.org/wiki/Optical_window_in_biological_tissue 1 Darkness is not something. it's the absence of something (photons, Electromagnetics waves) So if you want to emit the absence of photon. It's kind of hard. Therefor I would say that it's not possible to have such a device. Now, you can have material that absorb photon. 4 White light consists of many wavelengths. Hence, the interference pattern using white light appears different than that for monochromatic light. At the center point, all the waves travel the same length and hence no path difference is produced at the center point. Thus at the center point we get the maxima of all wavelengths and we obtain the maximum for ... 3 r=1.5r_s for the Schwarzschild solution corresponds to the unstable maximum of the effective potential for a photon, therefore you won't be able to see much in practice, since practically every photon on this orbit will either fall in the black hole or escape to infinity. 8 The distance where light has a circular orbit is actually 1.5r_s not the event horizon. This distance is known as the photon sphere. In principle a shell observer hovering at this distance could indeed see their own back. The proper distance is indeed just 2\pi r, however the object would look bigger than expected because the curvature of spacetime has ... 1 Legolas probably only needs one eye if he has enough time and can make sufficiently accurate spectral measurements. First, note that Legolas was watching on a sunny day; we'll assume that between incident intensity and albedo that object were reflecting on the order of  100 \mathrm{W}/\mathrm{m}^2 light, which is about 10^{22} photons per second. At 24 ... 0 Suppose you have an object which is a perfect absorber/emitter of electromagnetic radiation, a.k.a. a "black body". Suppose we try to compute the electromagnetic power radiated by this object using statistical mechanics and classical electromagnetic theory. We would find that the object emits electromagnetic radiation at all wave lengths, and the wave length ... 7 Take the following idealized situation: the person of interest is standing perfectly still, and is of a fixed homogeneous color the background (grass) is of a fixed homogeneous color (significantly different from the person). Legolas knows the proprotions of people, and the colors of the person of interest and the background Legolas knows the PSF of his ... 4 One thing that you failed to take into account. The curve of the planet (Middle Earth is similar in size and curvature to Earth). You can only see 3 miles to the horizon of the ocean at 6 feet tall. To see 24 km, you would need to be almost 100m above the objects being viewed. So unless Legolas was atop a very (very) tall hill or mountain, he would not have ... 4 Deconvolution can work but it only works well in case of point sources as e.g. pointed out here. The principle is simple; the blurring due to the finite aperture is a known mathematical mapping that maps a hypothetically infinite resolution image to one with finite resolution. Given the blurred image, you can then attempt to invert this mapping. The blurred ... 1 Yes, this is possible. The device that makes this possible is called a polarizing beam splitter, which will transmit or reflect light according to its polarization. Thus, it will split diagonal or circular light into its horizontal and vertical components, and when used in reverse it will undo the process (it has to). Note, however, that you will in general ... 71 Fun question! As you pointed out,$$\theta \approx 1.22\frac{\lambda}{D}$$For a human-like eye, which has a maximum pupil diameter of about 9\rm mm and choosing the shortest wavelength in the visible spectrum of about 390\rm nm, the angular resolution works out to about 5.3\times10^{-5} (radians, of course). At a distance of 24\rm km, this ... 15 Let's first substitute the numbers to see what is the required diameter of the pupil according to the simple formula:$$ \theta = 1.22 \frac{0.4\,\mu{\rm m}}{D} = \frac{2\,{\rm m}}{24\,{\rm km}}  I've substituted the minimal (violet...) wavelength because that color allowed me a better resolution i.e. smaller $\theta$. The height of the knights is two ...

2

In the spirit of your question, having two eyes and assuming you can use them as an array (which requires measuring the phase of the light-something eyes don't do) allows you to use the distance between them for $D$ in the resolution equation. I don't know the spacing of an elf's eyes, so will use $6 cm$ for convenience. With violet light of \$\lambda = ...

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