Is there any optical component in existence that uniformizes randomly pointing rays?
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What do you mean by "uniformize"? The answer to your question depends on your meaning. – Colin McFaul Dec 27 '13 at 21:35
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To make uniform: http://www.thefreedictionary.com/uniformize Make every ray have the same direction. – Vial Dec 28 '13 at 01:58
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1With the diagram it's clear enough and it's a great question because it gets right to exactly how the second law of thermodynamics applies to optics. I've added the thermodynamics and entropy tags accordingly. I also added geometric optics. – Selene Routley Dec 28 '13 at 05:37
3 Answers
To add to Carl Witthoft's answer: your proposed device would violate Conservation of Optical Extent aka Optical Étendue unless it were an active device (i.e. one needing a work input to "uniformise" a given quantity of light).
The law that optical extent can only be held constant or increased by a passive optical system is equivalent to the second law of thermodynamics for light, because the optical extent of a light source is its volume in phase space.
The optical extent $\Sigma$ for the light radiated from a surface $S$ is:
$$\Sigma = \int_S \int_\Omega I(x) \cos(\theta(x, \Omega)) \,{\rm d} \Omega\, {\rm d} S$$
where we integrate the intensity $I$ at each point $x\in S$ over all solid angles $\Omega$ taking account of the angle $\theta$ each component of the radiation from point $x$ makes with the surface's unit normal. Then we integrate this quantity over all points on the surface $S$.
So, the $\Sigma$ for your output would be nought, whilst it would be large for your input, so no passive imaging device can do what you ask.
So, another way of putting Carl's answer would be that the proposed device would have to "forget" the state encoded in the input light's wavefront direction at each point. Thus your proposed device, if at all possible, would needfully be an active device, needing work input of $k_B\,T\,\log 2$ joules for each bit of light state forgotten in accordance with the Landauer Principle form of the second law of thermodynamics. I say more about this in my answer here.

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1Very well done. Guess I have to go off and learn how to stick LaTex into my answers :) – Carl Witthoft Dec 28 '13 at 13:46
No. Not that you've defined "uniformize" very well here, but it appears you're asking for every ray from every source point to be directed in some parallel direction. Leaving aside self-diffraction effects, just stop and think: how would any material or device be able to accept a bundle of rays from all angles at a given point on the device, and emit them all in a parallel bundle? This would be the equivalent, say, of a mirror which reflected every incoming ray in the same direction regardless of the incoming angle.
Now, I could cheat really badly and propose a lasing medium which is pumped with all those incoming rays (i.e. absorbs everything) and emits energy in a collimated beam, but that's clearly not what you are getting at -- I hope!

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How about using a diaphragm and a narrow-band color filter? Whatever passes through will be close to a monochromatic plane wave. – Maxim Umansky Dec 27 '13 at 22:31
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"This would be the equivalent, say, of a mirror which reflected every incoming ray in the same direction regardless of the incoming angle." Actually, these already exist, they're called retroreflectors. – DumpsterDoofus Dec 28 '13 at 01:12
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2@DumpsterDoofus No this is not the same thing. A cataphote simply flips any ray over so that it goes back whence it came. So that ray's state is still encoded in the output: same direction, opposite sense and so the operation is reversible (bijective). Sure, the cataphote is "direction independent" - it does the same thing to all rays, but the output of that operation still lets you tell those rays apart. Crucially nothing is forgotten and so the process can go forward passively and still be in keeping with Landauer's principle. – Selene Routley Dec 28 '13 at 01:31
No lossless optical system can form an image (of some radiation source) that is "brighter"; (more strictly higher "radiance" or higher "luminance" in the case of photometric rather than radiometric measure ), than the source. An early proof of this was given by the well known 19th century thermodynamicist , Rudolph Clausius, based on the second law of thermodynamics.
If such a system existed, and the source was an area of a blackbody radiator, at some Temperature T kelvins, the higher radiance (Watts per steradian per square meter (source area) ) image, applied to a receiving black body absorber, would drive that body to an equilibrium Temperature greater than T kelvins, and the second law forbids that.
Ergo, no such system exists.
Also, since lossless optical systems are reversible, any lossless optical system that formed an image that has lower radiance (luminance) than the source (less "bright"), could simply be reversed; object and image, to get a "brighter" image. Hence no such system exists either.
So "brightness" (ugly word) or Radiance, (luminance) is an INVARIANT property of any radiation source. In paraxial optics, this is usually referred to as "the Lagrange invariant" nhu, or NHSinU for non paraxial imaging systems; n (N) being the medium index of refraction (relative to vacuum), h (H) being the height of the source object or image , and u (sinU) referring to the ray angle.
The quantity (nhu)^2 or (NHsinU)^2 , or an integral form of that is what Rod Vance referred to as "etendue", which is a $5 French word for "throughput."
Although simple proofs of this expression usually relate to "imaging" optical systems, the exact same restriction applies equally to non-imaging optical systems, such as CPCs (compound parabolic concentrators ), which are sometimes used in solar energy concentration.
Although often applied to sources and their images, the invariance of etendue also applies equally everywhere in the radiation propagation path. You can exchange luminous area for luminous solid angle, but their product is fixed.
Although radiant flux from say the sun spreads out over a large area on the earth, (surface irradiance in Watts per square meter), when perceived by the human eye, the eye forms an image on the retina, that is as "bright" (luminance or radiance) as the surface of the sun, less losses in the eye. Which is why, you shouldn't look at the sun.
Professor Roland Winston, formerly at the University of Chicago, and Argonne National Laboratory, but now at University of California Merced , and colleagues, made a non imaging solid (YAG) CPC concentrator that gave a spot five times "brighter" than the sun, but that spot was inside the YAG medium, of index about 2.7, and there is no way to get it out of the crystal because of TIR. (total internal reflection). The second law is not violated.
The student can find rigorous treatment of etendue ideas in Born & Wolfe, and other standard texts.
The word "brightness" is not any SI quantity or measure; you can't put any number to it.
When used casually by physicists (we should know better), it invariably refers to the radiometric quantity "radiance" measured in Watts per steradian per square meter, or to the photometric quantity "luminance" measured in Lumens per steradian per square meter, which is a psycho-physical response of the human eye to radiant energy in the 400-800 nm "visible" octave. Also people dealing with electron or ion sources, such as thermionic cathodes or field emission sources used in plasma displays, or in particle accelerator physicists, use "brightness" as a colloquial expression for the electrons or ions per steradian, per square meter of their sources, or even coulombs (or Amps) per steradian, per square meter.
We all know what we mean; but we should be more rigorous and use proper terminology, when dealing with students or the lay public, who usually encounter "brightness" in Shakespearian poetry. Brightness is never used as an alternative to radiant or luminous intensity (Watts or lumens per steradian) or for surface irradiance or illuminance (Watts or lumens per square meter), or for radiant or luminous emittance, also measured in Watts or lumens per square meter for a source.
Intensity is strictly defined for a point source, hence Watts or lumens per steradian, with NO reference to size. So intensity (radiant or luminous) DOES NOT change with distance from the source. However, real sources have a finite area or diameter, so close to the source, measurement of intensity becomes a problem. So long as we observe or measure, at a distance at least ten times the source diameter, we get results with little error. Actually about 1/2% at ten times source diameter.
Another peeve, is that some think lasers don't follow the inverse square law for wave front irradiance at a distance. They do, so long as you are several times the Raleigh range distance from the beam waist for the TEM00 fundamental mode.
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+1 You need to add the Stefan Boltzmann law to your second paragraph to make the argument complete. The second BBR radiates back, so you need to take account of the radiation in two directions. The second BBR's temperature stops rising when the power in = power out: the point is that the Stefan Boltzmann law shows that this second temperature is higher than that of the first BBR (owing to the assumed lower area and higher radiance), thus violating the second law. – Selene Routley Dec 30 '13 at 22:53