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| bio | website | sjbyrnes.com |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 11 months |
| seen | 12 hours ago | |
| stats | profile views | 248 |
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Sep 18 |
answered | Formula for polarized “light” transmission through close filters? |
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Sep 17 |
comment |
Interference of EM Waves with Orbital Angular Momentum yea, there's no such thing as light with L=3 in free space. The electromagnetic wave equation requires that the electric field components all have the same period as the wavelength. They can't have structure at one-third the wavelength. Also, in case readers don't know, +1 and -1 angular momentum mean the same thing as "right circular polarization" and "left circular polarization" respectively (or maybe vice-versa). |
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Sep 17 |
comment |
Focal Point vs where you see the images Mentioning "retina" makes this question confusing. Your eye has a lens in it. So sometimes you need to think about two lenses: One lens made of glass plus one lens which is your eye. The two-lens problem is different from the one-lens problem. I don't think that's what you meant. I think you're asking about the one-lens problem: Just a single glass lens and no human eyes. |
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Sep 11 |
comment |
simple test/measurement to quantify water opacity Do you care whether the light is absorbed versus scattered by the colorant? |
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Sep 11 |
comment |
total intensity measurement (of the whole visible light spectrum; from 400 - 800 nm) with a powermeter PM100? You also need to know the response curve of the photodiode. (amps per watt of light-power, as a function of wavelength.) If this curve is very flat in your range, you can use the 633nm reading directly. Otherwise, use the 633nm value on the response curve to figure out how many amps are coming out of the detector, then that has to equal the integral of (spectrum * response curve) [integrated over wavelength] :-) |
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Aug 28 |
answered | References for real life applications on advanced EM |
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Aug 23 |
awarded | Critic |
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Aug 23 |
comment |
Reciprocal Lattices This is just wrong. For example, the reciprocal space of a 2D triangular lattice is a 2D triangular lattice. Not hexagonal. |
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Aug 19 |
revised |
Physics of Focusing a Laser fix imprecise terminology about gaussian beams |
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Aug 19 |
revised |
Physics of Focusing a Laser edits to "in practice" section--explain cylindrical lens thing and add telescope note |
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Aug 19 |
revised |
Physics of Focusing a Laser typos in inequality formulas |
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Aug 19 |
answered | Physics of Focusing a Laser |
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Aug 3 |
answered | How beam focusing looks like in electron microscope? |
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Aug 3 |
comment |
The Ozma Problem @ANKU -- Nope. I have a strong impression that it should be possible, but I don't enough to spell out the procedure. I suppose I could be wrong. Maybe someone more knowledgeable will help... :-) |
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Aug 3 |
comment |
Why are lasers inefficient? @Martin Beckett -- What kind(s) of green lasers have high wallplug efficiency? |
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Jul 30 |
awarded | Commentator |
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Jul 30 |
comment |
Calculating diffraction-limited resolution for a lens setup I want to add a couple things to this nice answer: First, "the f-number at the image plane" is NOT the same as the f-number of the last lens of the system. I don't want Colin to be misinterpreted! This is just another way to discuss the "angle of the converging cone of light". Second, for the exact relation that Colin alluded to between convergence/divergence angle and diffraction-limited spot size (a.k.a. "beam waste"), you can use the Gaussian beam approximation. :-) |
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Jul 30 |
comment |
Why are lasers inefficient? In fact this makes red laser diodes the #1 most efficient way we know of to turn electricity into visible light energy! More efficient than fluorescent, more efficient than LEDs, etc. If only green and blue lasers were as efficient as red, then lasers would be the best energy-efficient way to light your home! (Of course you would need diffusers!) |
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Jul 30 |
answered | The Ozma Problem |
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Jul 30 |
comment |
What happens to light in a perfect reflective sphere? @lurscher -- Yes, optical diodes cannot be perfect in reality, just as walls cannot be perfectly reflecting in reality. This is a silly hypothetical. Your statement about inelastic scattering is I think misleading: A photon reflecting off of a moving mirror will be redshifted if the mirror is moving away from the photon, or blueshifted if the mirror is moving towards the photon. I don't think it makes sense to call this "inelastic scattering", even though there is a change of frequency in some reference frame. At least it's not the usual "inelastic scattering" people think of. |
