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39

The Nobel website scientific background is good. Basically, when you try to make gallium nitride, you usually end up with a material that is (1) chock-full of defects, and (2) n-doped (even when you were trying to p-dope it). So blue LEDs required The invention of MOCVD technology for growing crystals (early 1970s); Finding the right recipe to grow good ...


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 ...


8

When working with any laser above class 1, you should seek appropriate qualification for dealing with your specific laser system (which includes both its wavelength and its power), and you should inquire with your institution as to any formal safety requirements. Specifically, you should not take laser safety advice from what are essentially random ...


7

The book The Blue Laser Diode: The Complete Story deals with the issues of p-type doping of GaN. The difficulty of growing high quality GaN crystalline films lies in the problem of finding a suitable substrate material. (...) The link above points to the chapter you may be interested in.


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 ...


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 ...


4

The "critical" part was in finding and producing a structure with a large enough bandgap to produce blue photons. The first LEDs produced relatively longwave infrared (IR) photons, which have far less energy than the green or blue photons now available from LEDs. In general, the larger the desired bandgap, the harder it is to manufacture a suitable ...


4

The wikipedia article is correct. The relation between the actual response function and the conductivity is not immediately obvious, however. Let us consider the case of the longitudinal conductivity for example. The susceptibility, $\chi^L(\textbf{q},\omega)$, which is the true response function, is related to the conductivity using the following equations: ...


3

Great setup! I don't envy your task: IR is very tricky and frustrating to align. Do not expect to solve this problem overnight: this is a many month project, unless you can find someone who has done exactly this before, in which case it would have been a many month project for them. Part of the translation is easy: for fine ajustment, you use an angled ...


3

Carbon nanotubes, or similarly manmade materials, might be able to accomplish this in the future. The current record is a tensile strength of 63$\times$10$^{9}$ Pa (Pa = pascals = N/m$^{2}$ or force per area). So if we assume a circular cross section, then the area is defined by 4$\pi$r$^{2}$. We can take the tensile strength and the area formula to find ...


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.


3

You have to have a second polarizing device that you can use to analyze the polarization. A common choice is to use a reflection from water or a glossy floor that happens to lie near Brewster's angle, which makes the light strongly polarized in the plane of the reflecting surface. Look through your polarizer at the reflections, and rotate your polarizer ...


3

As Olin said, for a parabolic mirror lines that start out parallel to the principal axis all converge to a point. This can easily be confirmed with a bit of simple math. For a mirror, angle of incidence = angle of reflection. I show that this results in a parabolic shape in my answer to an earlier question (see the last part of the answer). Now a spherical ...


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 ...


2

The positions of the diffraction peaks do not depend on the size of the atoms. The peak positions are determined by the spacings of the crystal lattice and it does not matter what the atoms are or how big they are. That's why the atom size does not appear in the Bragg formula. However the intensities of the lines, both absolute and relative, depend very ...


2

You can create wavelets anywhere - the propagation of a wave is always represented by the Huygens construction. You need to keep in mind the phase: it is usually convenient to draw a new wavelet starting at a boundary, and with a known phase, because then it's easy to draw a series of concentric circles with appropriate spacing (wavelength). If you draw ...


2

Why can't we reach the ends of rainbow? The WIkipedia article you linked to contains the explanation The rainbow is not located at a specific distance, but comes from an optical illusion caused by any water droplets viewed from a certain angle relative to a light source. Thus, a rainbow is not an object and cannot be physically approached. Indeed, ...


2

Assuming your screen is "far away", then the pattern you see on the screen is the Fourier transform of the aperture function. So if you take the (inverse) Fourier transform of the image on the screen, you get back to the aperture function. Small complication: the intensity image no longer contains the phase information which you need to fully reconstruct ...


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 ...


2

As you point out in your ray diagram, the point source no longer looks like a point source after the diffraction grating. You could get around this by collimating the input beam with another lens (which you did), with an aperture, or by moving the point source far away from the grating (effectively using the grating as an aperture). So, I think the answer ...


2

Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of: We can always detect uniform motion with respect to a medium by a positive result to a Michelson-Morley experiment that is confined to a region of spacetime ...


2

Here are some topics to read about: Frequency doubling, also called second-harmonic generation as Johannes mentions. Here, you put one wave into a medium, and some fraction of it is converted to a wave with a different frequency. By carefully engineering the medium you can get quite a high conversion percentage. Other nonlinear optical processes, not just ...


2

There are "circular-polarization-maintaining fibers" (you may wish to google them). One can get such fibers by introducing circular birefringence, say, by twisting the fiber.


2

All you need to know about the chief ray is that it will show up in the right place at the focal plane. Any path that gets there is valid. You will see that this is indeed the case for your example - the ray starts up at the top of the image (tip of the arrow) and passes through the tip of the arrow in the "internal image" plane. Any pair of straight lines ...


2

Additionally, blue was the last of the primary colours so its invention made the production of white LEDs possible. Ordinary lamps could then be replaced with extremely energy efficient LED alternatives.


2

I wanted to post this as a comment, but it grew too long. The final question that was a bit hidden, but that several other users seemed to be also interested in, would be about why the blue LED was maybe so much harder to construct than the red one. Reading through Steve B's link to the nobel scientific background provides me with enough information that I ...


1

Roughly speaking, the color of an object is the wavelength which hits your eyes when [typically non-coherent] light interacts with the object. This interaction includes refraction and reflection and the spectrum of incident light. But for objects that are single particles, there aren't really any refraction/reflection properties, there is just stimulated ...


1

For usual way of looking, trying to see an object with size smaller than wavelength you shine at it will give you nothing: you'll basically not notice the object at all: it can't be resolved. But, there's a trick, where instead of using normal "shine and look from a distance" way you watch the object very closely, taking advantage of the properties of ...


1

As opposed to type II phase matching that produces orthogonally polarized photons in parametric down conversion (PDC), the type I PDC process produces identically polarized photons in the output signal and idler modes (labels $s$ and $i$ below). Normally the output state from type I PDC is not entangled: to get the required phase matching in the nonlinear ...


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 ...



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