lenses and the 2nd law

Focus the sun's energy coming through the earth's atmosphere. The achievable temperature at the focal point is is always less than that of the source, in this case, the surface of the sun.

"RichD"

.... Phil

Yes. (When talking waves, "dissipating" usually means "being absorbed", that energy is being lost from the wave, and isn't usually used for the drop in intensity due to a wave spreading.)

No, because ...

... the spreading of the radiation from a source doesn't mean that the temperature of that radiation falls. I also wouldn't call it more "disorderly" as it spreads. (You need to be careful about "disorderly". Yes, you can give the 2nd law in terms of "disorder", but that is for a specific technical meaning of "disorder".)

The converse of the above is that there is no temperature rise of the radiation.

Consider a radiating object O of temperature To, enclosed in a box of lower temperature Tc. If we have another object Sin the box, it will reach a temperature between Tc and To. If S is a long way away from the radiating object O, the temperature will be close to Tc. If S is close to O, its temperature will be closer to To.

Object S doesn't know how close it is to O; it doesn't directly interact with O. It only interacts with the radiation from O and from the surrounding box. Think about a bunch of rays spreading out from S. These are the lines along which S will interact with its surroundings via radiation. What matters is whether the rays will hit O or the surroundings. What a lens (or paraboic mirror) will do (if in the correct position) is deflect these rays so more of them hit O than without the lens. The effect is the same as moving S closer to O. So the temperature of S gets closer to that of O. The best you can do is completely surround S by radiation from O (e.g., by surrounding both S and O with a spheroidal mirror), in which case S looks like it's surrounded by O. Same effect as being in an oven at To; the temperature of S will be To at equilibrium.

The energy flow is only from hot to cold, so the laws of thermodynamics are perfectly happy, with no work required.

If you look at an object through a lens, it is likely that you will see an image of the object. If you do, it will not (except in rare circumstances) be the same size as the object or in the same location as the object. Nevertheless, the image will be the same brightness as the object except for some losses such as reflection at glass surfaces.

In particular, note that as you stop down a lens, illumination by the image formed in the focal plane diminishes. That is one means of exposure control. However, if you observe the image by eye, it is as bright as the source irrespective of reducing the effective diameter of the lens with a variable iris. The brightness remains the same, but your field of view is restricted so that you may not be able to see all of the object. When you go through all of the geometry, the lens cannot collect all of the light emitted by a source or produce a temperature higher than that of a source.

Bill

sometimes it does.

if aligned correctly it does.

no.

yes (not sure what you mean)

no

the nearest thing light has to temperature is frequency and that's not reduced by dispersal.

Look up conservation of brightness

Brightness and Flux Density

No, not unless the focus could be heated to a higher temperature than the light source. When you use a magnifier to focus sunlight on a weed, the view (from the weed) is of not just 1/2 degree of the sky being hot and yellow, the rest blue and cool, but of a 20 degree segment of the sky (the full width of the lens) being hot and yellow. If you got the weed close to the sun and put a mirror behind it, it'd get the full effect, and reach the temperature of the sun's photosphere. But never a higher temperature.

Mirrors to focus sunlight from all directions onto a point would have the effect of removing radiative thermal connection to all local objects, while retaining the connection to the sun. That would make the poor weed reach equilibrium when it hit 5800K or so.

There was an article in SciAm, ages ago, about a non-imaging optical system that could heat an object to above the sun's surface temperature. Sounds fishy, except that the sun does emit a lot of UV, energy past the black-body peak wavelength, which could be useful if concentrated enough. Like, 84,000:1.

I guess one could always use a sun-powered laser to heat an object to an arbitrarily absurd temperature. One of the early "theoretical" objections to funding Townes' maser/laser idea is that it obvioulsy violated the 2nd law.

John

This is just an instance of a common situation where local entropy is reduced at the expense of increased overall entropy. It happens in humans by eating food as well as in refrigerators or heat pumps.

Part of the overall problem in this discussion is poor use of language. What does it mean to focus? Why would one expect focussing to violate the second law? Vernacular use of language easily gets you into trouble. Explain to me what is grammatical about the phrase "twice as small."

Bill

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| > A lens focuses energy, does it not? That's its | > purpose. So given a large radiating source, the | > lens concentrates a dissipating spreading wave, | > down to a small area. | >

| > Does this not violate the 2nd law of thermodynamics? | > The natural evolution is the radiation spreads, | > becoming more disorderly and lower temperature. | > The lens reverse this process, concentrating the | > energy, with temperature rise. | >

| > And the lens is a passive device, so no exogenous | > power source to do the work. | >

| > Ot it could be a parabolic mirror, same thing. | >

| > Explanation? | >

| > -- | > Rich | | |

Actually to be fair, the ratio you list (84,000:1) is the concentration ratio provided by the system which can produce irradiance levels higher than the sun. There are no tricks used in regards to using UV energy or something like that. The absorber is simply immersed in an index of refraction that is different from air. The temperature of the abosrber reached however will never exceed that of the sun as a blackbody since the absorber immersed in a higher refractive index also radiates more than one immersed in air, thus the temperatures remain equal in theory. I don't think the SciAm article lists the temperature as being higher, I think it talks about the irradiance being higher, which are two different quantities. See the following link for further details

I am not sure I understand what you mean (and I will not try to), you might however study, if you can, the relevant chapter from Clausius famous treatise.

It all boils down to the conservation of "étendue" in optical systems...

Rudolf Clausius, The Mechanical Theory of Heat etc., Eighth Memoir, On the Concentration of Rays of Heat and Light etc., London 1868.

There might be simpler accounts elsewhere, I do not know of any in English, there could be a couple in French I think.

Some people say that Clausius derivations (might) contain errors however the results are not in question.