Tag Info

New answers tagged

2

In the chapter The Ultraviolet Radiation Environment of Earth and Mars: Past and Present, the author describes that, despite being closer to the UV source (the sun), Venus's atmosphere is so dense that no UV-C or UV-B radiation reaches the surface of the planet and only very small amounts of UV-A penetrate. As for Mars, the very thin $CO_2$ ...


2

So taking your questions one at a time 1) The concave mirror will form a virtual image if the object is placed closer to the mirror than the focal point of the mirror. The formula for the position of the final image is $s'=\frac{sf}{s-f}$ where s is the object-mirror distance and f is the focal length of the mirror. You can see that if $s<f$ this will be ...


1

Two things that might help you. Finding Sellmeier Co-efficients Firstly, although the refractive index formulas you cite are not in the Sellmeier form, they are in the form: $$n(\lambda) = 1+\delta(\lambda) = 1+\sum_j\frac{K_j\,\lambda^2}{\lambda^2 - L_j}$$ where $\delta(\lambda) = \sum_j\frac{K_j\,\lambda^2}{\lambda^2 - L_j}$ is VERY small. So, squaring ...


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

The answer to your question depends on what you want to study. But first, let's agree on terms: by "optically conjugate" I mean that the deformable mirror/shack-hartmann wavefront sensor is imaging some plane. In case of astronomical AO, the SH WFS is conjugated with the telescope pupil (or, more precisely, with the primary mirror of the telescope). A good ...


1

You can reach the end of a rainbow, if you make one using a hose pipe on a sunny day. It is possible to locate the ends to within a few cm


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, ...


0

There is nothing wrong with working in a completely enclosed box which has an electric safety switch that turns the power off as soon as it is opened. That is exactly how a professional engineer/physicist will solve this safety problem: with deadlocks like this and measures that prevent the beam intersecting your eye under all circumstances of "normal" use. ...


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 ...


0

Here's an idea. Maybe it's kind of cheating, I don't know, you decide. Take a small diameter clear tube, fill with ordinary, unfiltered tap water and line the laser up to fire down the length of the tube. Seal the far end water tight. If you're firing the beam at a downward angle, you can leave the end closest to the laser open. The impurities in the water ...


1

Videos typically have a frame rate of ~25 frames per second. The 'movie' you see is a result of these snapshots. If you assume the propellers were moving at 9k RPM, that is 150 revolutions per second. If the camera recorded at 150 FPS, the propellers would appear stationary as they would be in the same position at each time a new frame was captured. The ...


1

Yes. There are more complete developments based on electromagnetic theory. Exact solutions are known for a few cases. Almost all applications rely on approximations, many of which are very good approximations under the right conditions. But the more complete explanations end up looking like Huygens Principle, and serve to justify it. Huygens' original ...


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.


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 ...


0

It doesn't. You need a parabolic surface, not a spherical one, to have rays parallel to the axis converge at a focal point. I have no idea where you got your failed math from, but it's just plain wrong.


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 ...


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 ...


1

I have been recently performing experiments using UVB (and UVA light) light. The personal protection I have been using have been a welder's mask (to protect the face) and sun glass like material for the eyes and sun protective clothing that did not leave any skin exposed (e.g. hands, arms etc). As mentioned in the comments, surround the experiment in a ...


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 ...


1

This is typically because of optical selection rules which forbid certain types of transitions. The most usual case is where the states are, in order of energy, $S$, $D$ and $P$ states, driven by a reasonably-intense laser. In this case, the coupling to the EM field is usually a dipole coupling, which means that the atomic operator that does the transitions ...


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 ...


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

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

A Pockel cell followed by a quarter waveplate will do this. The Pockel cell acts as an electrically controlled waveplate which will give an elliptical state. The waveplate will then convert this to a rotated linear state. Producing a 10MHz driving signal at the several kilovolts required might be a challenge but not impossible.


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.


0

Faraday Effect. The plane of polarization in a medium is rotated when exposed to a magnetic field. Solenoid, glass rod.


0

Angular magnification is used to describe the magnification of an afocal system, typically a telescope like a Galilean telescope (which a modern microscope's eyepiece almost always is) or Keplerian Telescope. Indeed your question is a good one insofar that it underlines the meaninglessness of "magnification" (or at least calls into question how worthwhile ...


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 ...


1

Although normally considered as photon interactions, any inelastic scattering process will result in the alteration of the frequency of the electromagnetic radiation. An obvious example is Compton scattering, where high energy (X-rays+) light scatters from free electrons. The scattered light has lower energy and longer wavelengths than the light incident ...


1

You can place an object anywhere you like with respect to the lens. In one diagram, the author simply chose the separation to be twice the focal length. In this case, the rules for ray tracing tell you the image will also happen to be the same distance from the lens. This doesn't work with any other separation. Also note that the one place you would have ...


1

Although the method illustrated in the pic below can in principle assure an unprecedented level of collinearity of the beams in a rather uncomplicated manner, in practice it is subject to the availability as well as a reasonable level of performance in usage of fiber-optical components in the $10\, \mu$m regime (I had written a long and complete answer, and ...


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 ...


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 ...


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.


0

I'm not sure the other answers are complete. Quarks could be said to have a photopic color under certain conditions. If there is zero interaction (other than elastic scattering) with photons, then quarks could be considrered to be "white," as they have no preferential absorption. If quarks are capable of absorbing photons, they are either "gray" or have a ...


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

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 ...


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 ...


0

Yes, optical activity can be affected by external electric and magnetic fields. Check out the Faraday effect: https://en.wikipedia.org/wiki/Faraday_effect Explanation of this phenomenon can be attempted with the assumption that the external magnetic field changes the electronic motion inside the molecules in such a way as to modify their effect on the ...


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 ...


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.


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 ...


0

I did some digging and I found a blog that indirectly, but the none the less explicitly, talks about ray tracing chief and marginal rays in thin lens systems: http://taylortechassoc.com/?page_id=617 The equations he presents are the literal mathematical relationships between incoming and out outgoing ray angles (of any type) and it has a spreadsheet linked ...


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 ...


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

One approach is to try to use the original model as a ground truth and then redo the match to the desired form. I tried to do this in MATLAB: xs = 360.0 : 1.0 : 830.0; ys = 1 + 0.05792105./(238.0185-1.0./(xs.*xs)) + 0.00167917./(57.362-1.0./(xs.*xs)); cftool But, quality software that it is, it reliably crashed. So, I wrote my own nonlinear least ...


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.


0

The human eye, in dark-adapted conditions, has been shown to be capable of detecting a single photon (which is to say, the retina reacts, the signal gets to the brain, and [magic happens] to generate a response in the conscious parts of the brain. Since what you're asking about is the release of one photon, it doesn't matter what the source is. So long ...



Top 50 recent answers are included