New answers tagged

3

As mentioned in comments, a photon is "delocalised", so it feels the whole system. You may imagine a photon as a long-long wave (to have a defined frequency) and as such it interacts with the whole material. More strictly, one can say that the source of photon is the whole set of charges, so the photon is a collective mode of excitation of a given system. ...


1

You are not the first to question this; see for example the work by Leiner et al. from 2013 and 2014. Here the Ray Tracing technique is interfaced to Finite Difference Time Domain (FDTD) Maxwell simulations with ''the Poynting vector representation of either rays or wave propagation directions''. As noted in these works you should be aware of the absence ...


1

From a recent paper, Direct detection of a single photon by humans. JN Tinsley et al. Nature Comms 7, 12172 (2016) (open access). Abstract: [...] Here we report that humans can detect a single-photon incident on the cornea with a probability significantly above chance. This was achieved by implementing a combination of a psychophysics procedure ...


0

I don't know why this hasn't been answered yet, but the simple answer is: no, the wave function doesn't collapse when it goes through the polarizer. The "photon" remains in a superposition of two states.


1

you are basically looking for fluorescent materials. These materials are coated inside the fluorescent lamps which absorb the ultraviolet light and convert it into visible line radiations. Rare earth compounds such as Gadolinium oxy sulphate are (were ?) very commonly used phosphors. However new research is going on you may find this article interesting. ...


2

There is likely no practical and straightforward way to do what you want. An acousto-optic modulator can shift an optical frequency by a fixed amount, but the amount of the shift is small, not big enough to cause a difference of color perception. It is also not particularly efficient; typically a few percent of the energy in the input beam is converted to ...


1

There are several ways to make a line focus. You can try to use two perpendicular cylindrical lenses of different focal lengths. Or you can try to use one spherical and one cylindrical lens. Or you can put the spherical lens at an angle the astigmatism of the lens will generate the line focus. If you use only one cylindrical lens then you can not ...


1

Hints: The Jones matrix for e.g. right circular polarized light is: $$\frac{1}{2} \begin{pmatrix} 1 & i \\ -i & -1 \end{pmatrix}$$ To see this try applying this matrix to your linearly polarized Jones vector. It will give a right circularly polarized Jones vector. The Jones matrix for a quarter wave plate is:$$\begin{pmatrix} 1 & 0 \\ 0 & e^...


2

The curvature of the drop is influenced by the liquid-solid, solid-air and liquid-air interface forces. It can be determined by the Young-Laplace equation. For more details see this article


-1

Transparent object allow light to pass throught them and we can see throught these objects clearly translucent object allow light to pass through them partially.


0

Light as a Particle The photons in the beam of light are continuously being absorbed and re-emitted by the glass atoms (though this is also true in the other mediums light slows in). The level by which the light is slowed is dependent on how often this happens. As @garyp commented below this question, the delay also depends on how long the photon stays in ...


4

The answers given here make me wonder, because I sense in here perhaps a misunderstanding. Or maybe I'm wrong, which might be more likely. :-) The answers here refer to distances light travels. But as far as I understood, light is never slower than 299 792 458 m/s. I guess it may "look" like from a point of reference that light has slowed down, when a event ...


-2

How would you define speed? In case it has a direction: The maximum slowdown you can get is 2C, this is achieved by using a mirror. In case it is defined by the time taken to get from A to B In case you would define speed by looking at the width of an object, where the light enters at point A, and leaves at point B, the following can be derived: The ...


18

The short answer is that there is no known theoretical strictly positive lower bound to the speed of light. Any positive number, no matter how small, is possible, although limits are set for each candidate material, as I explain at the end of my answer. One has to be pedantic to understand the lack of limitation to a more generalized "speed of light". From ...


16

How the light slows down in a matter, it depends on the recipe of its refractive index. For common, ordinary materials it is in the range of 1-3. Bose-Einstein condensates have an extreme refractional index, even millions or billions. In a BEC with a refractive index of $10^9$, the speed of light is only $30~\mathrm{cm/s}$. Here is a relative old article ...


3

As in ptomato's answer optical path length is generally expressed with units of length. However, a related quantity is optical path difference for a system or rays, which measures the degree of optical aberration (not to be confused with stellar aberration). Optical path difference is the RMS deviation of the optical path length of rays through a system's ...


2

Optical path length generally has units of length, as its name implies. I haven't seen the dimensionless quantity $OPL/\lambda_0$ referred to as optical path length.


2

Listening to the recording that the article was derived from, it appears that the article is: Donald Lynden-Bell. Exact Optics : A unification of optical telescope design. Mon. Not. R. Astron. Soc. (2002)


0

In the limiting case, consider that the object and your lens are finite in size and infinitely far apart. Then each appears as a point when viewed from the other. Two rays passing from the object to your lens would then follow the same path and would thus be parallel.


3

The answer is because our ken (field of view) subtends an extremely small angle at the source. Even though the source may emit over a wide angular range, we can only receive a small angular range of that light if we have a limited aperture instrument and our distance from the source is large compared with the aperture. Suppose we look at Alpha Centauri ...


2

The radius of curvature depends in detail on the beam profile and the aberration present in the beam. Indeed, in an aberrated beam, there is no simple definition of the radius of curvature. One of the problems with the $M$ parameter is that there is no simple relationship between it and the beam details; it is a very blunt characterization. If you want to ...


2

The PhD. Thesis: Hannah Dunstan Noble, "Mueller Matrix Roots" gives a gentle introduction to the concepts and José J. Gil, "Characteristic Properties of Mueller Matrices", JOSA A, 17, pp328-334 derives necessary and sufficient conditions for a matrix to be a physical Mueller matrix. The situation is not quite as simple as you assume, although the ...


0

It's because you want to extrapolate the derivative of the reflected power, which is the error signal. This is implemented by adding a frequency $\Omega$ such that you can Taylor expand $F(\omega \pm \Omega)$ and get $\frac{\textrm{d}F}{\textrm{d}\omega}$. So you want the term that contains $F(\omega \pm \Omega)$. The $2\Omega$ term comes from the ...


0

You will still have light on the other side. The eigenmodes of the fiber are complete set so any light will be happily decomposed. Light is still incoherent. In principle you could excite single mode.


1

Protection of the coating layer that reflects and is easier to clean. And the other reason is glass is cheaper to make with small surface defects.


3

In a mirror the thin layer of metal at the back provides the reflection. If this metal is exposed to air it will 'tarnish' - ie become dull due to formation of oxides on the surface. It can also be scratched quite easily. Both these reduce the quality of reflection. Glass provides a layer of protection for the metal. Glass is transparent and also hard,...


0

It seems like you have established that your photons go through both paths! Not only do you not know which path a photon went through, you know it went through both because of the changes in interference pattern you see. You might want to do this in a controlled way!


0

@Flojo1 - I know more about RF atmospheric refraction than optical, so this won't give you anywhere near a complete answer. But I can say a few things that may be of some use. First, looming is no more than enough vertical refraction in the earth atmosphere that an object on the surface looks like it is above the surface. Sinking is the opposite. There's ...


1

Coherent states are classical in a precise way which hasn't been stated explicitly yet, although Rod suggests at it. Suppose you want to time-evolve the interaction between a coherent EM state and matter. This amounts to solving the Schroedinger equation for: $$ i\hbar\frac{d}{dt} |\psi \rangle= H |\psi \rangle$$ with $$\hat{H}= \hat{H}_{0A}+\sum_k\omega_k ...


3

The blue end of the spectrum is higher frequency, meaning shorter wavelength. A cage that can block blue is going to block (technically attenuate) the given wavelength and any that are longer (lower frequency). If you start blocking at blue, you'll be blocking the entire visible spectrum and on down into the IR, microwave, and radio. To filter out UV, the ...


1

The dimensions and meaning of the B coefficients are not the same as the A coefficient. The probability of spontaneous emission does not depend on the radiation environment of the atom, whereas absorption and stimulated emission do. Given that, one has a choice of how one encodes that in terms of the B coefficients, which are only a property of the atom ...


0

Well, assuming they are bright enough to affect the other light, then, it will depend upon if they have exactly identical timing or not. Theoretically, with identical timing and mechanism, they should blink. Even a tiny fraction of a second difference will be important. If one light switches on a fraction earlier, the other one will stay switched off. The ...


0

in order to obtain the wave nature of light you need to make your slits as close to each other as you possibly can as well as make the slits as thin as possible. The two slits/holes need to be VERY close to each other I applaud you for taking interest in experimental physics! Often times experimental physics is a matter of changing all the variables you ...


4

Assuming that each sensor sees only the light from the other night light, and assuming that each night lights is bright enough to reliably trigger the other's sensor, then you have discovered a configuration that computer engineers call a "flip-flop". https://en.wikipedia.org/wiki/Flip-flop_%28electronics%29 It has another name, "bistable multivibrator." "...


0

Interesting question. What you have arrived on is a classic conundrum. Let me make my argument clear through a few scenarios. First, is the light from the nightlight bright enough to affect itself? I mean, if the light is turned on, won't the environment become bright again and hence and trigger a change? I am assuming it does not work that way. Second, ...


2

There is actually 2 questions here. The van Cittert-Zernike theorem is essentially used to calculate the free space propagation of the electrical field/light. It can thus be used to reconstruct the original field at the astronomical object from our measurement on earth. However as mentioned correctly it only applies under certain incoherence conditions. So ...


2

I agree with Peter R. In the case of coherent astronomical sources, The radiation arrives only from one direction. All the interferometers will measure the same visibility. You get nothing more than a point. In 1989 K. R. Anantharamaiah, Tim Cornwell and Ramesh Narayan (NASA) wrote a paper on Imaging coherent and incoherent astronomical objects. Take a ...


0

Here is why the angle of incidence shift in band rejection filter spectra is minimized by high refractive index materials? Miroslav wrote in this blog [http://quantum.opticsolomouc.org/archives/464] that "The majority of interference filters are designed to be used at normal angle of incidence (AOI). The primary effect of an increase in the incident angle ...


0

I think the original question is, why the white laser is electrical pump instead of optical pump? Because electrical pump is much easier for practical use. You just inject current then get light, especially you have multiple lasers with different colors integrated together. Why they made white laser? Because the light can propagate much longer and have ...


2

If by "transmission metals" you mean "transition metals", then there are transition metal doped fiber lasers, indeed I believe erbium doped fibers may have been the original gain medium for a fiber laser. As for a Ti:Sapphire fiber laser, I'll stick with the most salient physics reason hindering the building of such a device: there will of course be ...


2

As your question is really broad, I'm only going to cover a couple of fiber optical items, give the resources I used, and give a few more general resources you can use to look into other items. Concrete Fiber Optical Item Descriptions: A multiplexer is used for multiplexing (though it can be used in electronic switching). multiplexing (sometimes ...


0

Pure water is transparent because it is a liquid. Objects that are not transparent either scatter light, due to difference in refractive index between air and the substance or they absorb all the photons at the wavelength you are observing. Skin for example is opaque largely due to scattering of visible light. In fact liquid water is blue if you look ...


4

Assume you are looking with just one eye (in order to isolate perspective from stereo vision effects), and choose any Euclidean coordinate system that has your eye at the origin. The light emitted from any point on a given straight line will fall onto your retina at essentially the same position, i.e. straight lines through your eye are collapsed onto a ...


1

although the hint by @WetSavannaAnimal aka Rod Vance is sufficient to solve this problem but I would like to elaborate his hint. you know diameter of the sun, and distance from earth to sun. Now take a point, the rays emerging from two extremes of the sun and passing from this point will have maximum angle between them and that will set an upper limit on ...


0

To understand this fully, you really need to be thinking of rays as normals to wavefronts. In this representation, a ray's tail gets assigned by a phase factor to represent the total phase of the plane wave. This phase factor advances by $2\,\pi\,\times$ an amount calculated by multiplying the distance advanced along the wavefront, in wavelengths, by the ...


3

Many ways to detect single particle radiation, charged or not. Scintillation counters can use photomultiplier tubes to detect single particles, single protons, neutrons, positrons, uncharged gamma rays etc. It depends on them having enough energy to ionize the material used for the scintillation. Plenty other ways, for instance a modified version of MRI ...


3

In general, the interaction of matter with light is dominated by electrons because they are lighter and this makes their coupling with light stronger than that of nuclei. However, this doesn't mean that nuclei cannot interact with light: they are still electrically charged particles, and they do interact with electromagnetic radiation to some extent. On the ...


2

The Heisenberg Uncertainty Principle has two distinct aspects: One is the identification of matter as a wave and, in particular, the relationship between a particle's momentum $p$ and its wavelength $\lambda$ through de Broglie's relationship $p=h/\lambda$. This is the crucial bit of physical input. The second one is purely mathematical, and it's the ...


0

From refractiveindex.info you can get n and k for many different materials, as a function of wavelength, and the site even includes a calculator at the bottom with which you can calculate the reflectivity R. All of the provided data are backed up with citations.


2

Whatever your personal definition of "optimal", I will go ahead and assume that among other things you're aiming for is seeing as much image detail as possible - and get as immersive, or "large", an image as possible - without starting to perceive the pixels distinctly (or more precisely to perceive the mesh of thin dark lines separating the pixels, which ...



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