15

What you are describing is anomalous dispersion. This happens when a material becomes strongly absorbing, typically near an absorption line, and the refractive index becomes complex. In these circumstances the phase and group velocities are different and indeed the group velocity can be greater than $c$. This isn't a problem since the group velocity is no ...


10

This may be a little technical, but I always thought it was cool: one of my professors once pointed out that transparency only happens because the material is (approximately) a linear dialectic over the frequency range that you care about. Turns out water is a linear dielectric over precisely the range of frequencies our eyes can detect. coincidence?


8

Yes, it's not only possible but relatively easy. For a gravitational lens the deflection angle at a distance $r$ from the black hole is approximately proportional to $1/r$. So your lens is going to look something like a hyperboloid of revolution: Note that the light rays are bent increasingly strongly as you approach the axis. This type of lens has been ...


8

There is a lot of nonsense around about this. It is NOT a very thick stiff or cold liquid, nor is it due to how ordered the structure is. In simple terms it is all about the electrons in the substance. When photon of light enters a substance, it will interact with an electron changing its energy state. KEY POINT — Electrons can only exist in fixed (...


6

It appears that the questioner wants to know if a laser can burn the jelly balls in water even though they are invisible in the water due to having the same refractive index. The answer is "yes". Having the same refractive index simply means that light will not be reflected at an interface between the jelly and the water. If the jelly can be burned by a ...


5

I'm not an expert, but it seems to me that they used a magnetic field to Step 1: Turn the field off so that the light energy 'stores itself' into a lattice of cold $^{87}\text{Rb}$ atoms. Step 2: Turn the field on so that the stored energy becomes light again. So it's not light itself that's stopping or slowing down, but they're just 'temporarily storing'...


5

The diameter of the lens is its physical diameter. It is sometimes advantageous to reduce the area through which the light travels and that is then the aperture. In the case of resolving power making the aperture smaller reduces the resolving power which means it is more difficult if not impossible to distinguish between objects whose angular separation ...


5

No, because in a dielectric medium light isn't light, and that's why its speed isn't $c$. The speed of light in a dielectric medium remains unchanged. You'll find lots of questions discussing exactly what goes on with light in a medium, but basically the EM field of the light interacts with electrons in the medium to form an entangled system that has an ...


5

The Greek letters are not related to wavelength but to the directions in crystals. In anisotropic crystals the speed of light (and so the index of refraction) depends on the polarization of light and the direction of propagation relative to the crystalline axes. As your link shows, Barium sulfate has an orthorombic structure so probably is biaxial. You can ...


5

Short answer is no. In that wavelength range, the effect of the magnetic polarizability is almost always much less than the electric polarizability. So even though the index of refraction is defined as $n=\sqrt{\epsilon\mu}$, it is often just written as $n=\sqrt{\epsilon}$ since $\mu\approx 1$ in most materials. And as you pointed out, ferromagnetic ...


4

Let us take the example of the Hubble primary mirror. It has a diameter of 2.4 m and a mass of 828 kg. It is actually made in a sandwich structure - glass-honeycomb-glass - making it about 30 cm thick (for stiffness) but light. The mirror is coated with an aluminum coating of thickness t = 65 nm, with a 25 nm MgF2 protective coating on top. Coefficient of ...


4

(Linear) magnification is equal to $\dfrac{\text{image size}}{\text{object size}}$. Problems arise with applying this definition when objects and/or images are a very long way away - usually termed "at infinity". It is easier to describe a telescope and then move on to the microscope? How can you reconcile this definition of magnification if you are told ...


4

You could look at Optical coatings and dichroic filters. Optical coatings are many thin layers of material of two of more different refractive indices, about one light wavelength in thickness. They structure of the coating can be designed for high reflectivity at some wavelengths and low reflectivity at others, but the switch-over cannot be too abrupt. ...


4

The refractive index is defined by the ratio $$n=\frac{v}{c},$$ between the speed of the light in the medium and the speed of the light in the vacuum. From the wave equation we read the velocity $v$ in terms of the permittivity and permeability of the medium, $$v=\frac{1}{\sqrt{\epsilon\mu}},$$ and, in particular, for the vacuum we have $$c=\frac{1}{\sqrt{\...


4

There are transparent things at any wavelength, and everything is intransparent at some wavelength. Cutting of 'clear' acrylic or glass happens at wavelengths that are strongly absorbed in those materials. Could your eye see those (and only those) the mazerial would appear black. To cut the jelly in water, you need a wavelength that water is mostly ...


4

Glass has a band gap of about 5 eV, which corresponds to photon energies at UV frequencies. Photons with lower energy do not have enough energy to excite the valence electrons to unoccupied states. The values depend a bit. Quartz glass has a larger gap than lime glass and borosilicate glass.


3

In the Drude model, the electrons are bound to the material by a Hooke's law restoring force, and this coupling with the material provides the damping. In the usual plasma model, this is not the case. Rather, the electrons are free to move. They respond to the electric field in the plasma by producing currents, and the changing currents produce time-...


3

The vacuum is polarizable. The polarization can be with respect to electric charge or color charge. In the presence of an electric field, virtual electron-positron pairs briefly exist (created from virtual photons of sufficient energy). The virtual pairs act as dipoles and orient with respect to the field. For example, near a proton, the virtual electron ...


3

It's a black coating to stick on surfaces of devices of an optical setuo to reduce spurious reflections. So it's very very black ("99,99%") on one side, and possibly sticky on the other side. For an example, see Edmund Optics: Acktar Light Absorbent Foil


3

In short, no, not all fluorophores are dipoles in the permanent dipole sense. As a counterexample, anthracene has zero permanent dipole moment, but fluoresces blue under UV illumination. The reason why many fluorescent molecules are called "dipoles" is because electric dipole transitions between quantum states occur due to nonzero values of the transition ...


3

Having spent rather a lot of years designing and building adaptive optic systems, I've seen this sort of problem in a number of guises. Certainly you can apply some fitting function to the image of each of your spots separately and determine the centroid of each spot alone. If your detector has sufficient resolution, these two measurements will yield the ...


3

Say you have a convex lens with a focal length of 50mm , standard on many SLR cameras, the diameter of the glass may be 30mm and there may be an aperture of 25mm , again diameter. In this setting the F-stop setting is f-2 because 50/25 = 2 . This is why F-stop gets larger as aperture gets smaller.


3

III-V photovoltaics will indeed photoluminesce quite nicely (particularly if not loaded). As you note, they are direct band gap materials so if there is any overlap of electrons and holes you can get recombination with the accompanying photons. If connected to an external circuit, the electron/hole pairs generated will get swept out of the junction fairly ...


3

What is the "travel time" of a beam of light in a medium with negative refractive index? Do you save time by traveling a greater distance? Time = distance/speed, and speed = c/n. When $n<0$, the part of the trajectory in the meta-material will contribute "negative time" to the over all travel time of the beam. Remember, Fermat's "shortest" means least ...


3

The reason that metals are generally opaque (and, in fact reflect light) is that they have so many conduction electrons that their plasma frequency, which is proportional to the square root of the electron density, is very high and in the ultraviolet. Visible light is below this frequency, which is why you can't see through typical metals. But what about a ...


3

Non-absorbing waveguides, such as optical fibers, are reciprocal optical elements, which means that as easily as light can get in, it can also get out. (This is also true for greenhouses, which work by converting the light to heat and preventing the heat from getting out by eliminating convection with the outside air) So there is no fiber that behaves ...


3

The index of refraction can indeed be less than $1$. This means that the phase velocity of light can exceed $c$ (the vacuum speed of light), and in fact it does exceed $c$ over a large range of frequencies in the ionosphere. The article [1] is a relatively concise and easy-to-read review. It says: As the refractive index [at radio frequencies in the ...


3

For lower frequencies, we can find examples. One practical example may be at very low/zero frequency. Magnetic permeable materials distort static (or almost-static) magnetic fields. Technically, we can call this refraction. Oscillating magnetic fields (radio waves) are routinely used to measure thepermeability of materials; what is measured is the effect ...


2

What you should be comparing is the time it takes for direct propagation (which I would guess is the "energy transmitted without total internal reflection") versus the time it takes for guided propagation at the critical angle, which is the longest delay/broadening you will get out of the fibre at the other end. Modes at angles higher than $\theta_c$ will ...


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