Can you start fires with moonlight?

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Offline Luke

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Can you start fires with moonlight?
« on: 24/10/2011 00:01:02 »
Luke McNeill  asked the Naked Scientists:
   
Dear Nakeds

Everybody knows you can use a magnifying glass to focus sunlight to burn things, and the larger the lens, the more intense the heat. How large of a lens would you need to use to focus light from the full moon to burn paper?

Thanks,
Luke McNeill
Arlington, MA, USA

What do you think?
« Last Edit: 24/10/2011 00:01:02 by _system »

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Offline Airthumbs

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Can you start fires with moonlight?
« Reply #1 on: 24/10/2011 00:18:21 »
I know people peer at the moon through pretty powerfull telescopes and don't seem to get any damage to their retina so maybe you would need an absolutely huge magnifying glass  [:o]
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Offline Soul Surfer

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Can you start fires with moonlight?
« Reply #2 on: 24/10/2011 08:53:51 »
No I don't think that you could, because of the limitations of conventional geometrical optics, because it is not possible to concentrate an image of an object to a higher radiation intensity than the object itself.  The sunlit surface of the moon gets well above the boiling point of water but not as high as the ignition temperature of  tinder so you might be able to make a cup of tea but not start a fire.

By the way looking at the full moon at low magnification in a medium sized telescope is quite painful and not recommended.   I speak fro experience with my own 8 inch (200mm ) Schmidt Cassegrain Reflector.
« Last Edit: 24/10/2011 08:56:52 by Soul Surfer »
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Offline syhprum

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« Reply #3 on: 25/10/2011 06:41:06 »
The subject of whether the concentrated spot of light from an optical system could exceed the temperature of the source of radiation was discussed at length recently on this forum.
The concensus of opinion was that it could which rather surprised me, if this is so it must apply to moonlight as well as sunlight although I cannot visualise an optical system that would do it.

SS "because it is not possible to concentrate an image of an object to a higher radiation intensity than the object itself" 
« Last Edit: 25/10/2011 06:42:50 by syhprum »
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Offline Geezer

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« Reply #4 on: 25/10/2011 08:06:59 »
because it is not possible to concentrate an image of an object to a higher radiation intensity than the object itself.

That's a great point. It sounds very likely to me, but I've really no idea why! Is there a theorem that explains why that is the case?
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Offline JP

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« Reply #5 on: 25/10/2011 17:13:32 »
I'm with you Geezer.  It sounds likely, but I don't understand why it has to be the cast.  Initially, it would seem that the second law of thermodynamics requires it.  After all if you form an image of a single point on a black body, that point will form a slightly larger blob (assuming it's diffraction limited), which means that the temperature is lower than that of the point source.  If it somehow was a higher temperature, then it would radiate back to the source point, heating it up, and you could use it to create a perpetual motion machine.

Where I get confused is what happens with an extended source.  In that case you have many points being imaged, and although each point-to-point image has to lose temperature, if your magnification is less than one (meaning your image is smaller than the object), then you're concentrating this energy, and I can't come up with a simple argument of why you can't increase the temperature this way.  This is a lot like compressing a gas to raise it's temperature, although in the case of a gas, the compression requires mechanical work.  In the case of light, it just requires a lens.  I suspect there is a good argument of why an imaging system can't do this, even with magnification, but I can't come up with it off the top of my head.

I also recall seeing an article by Prof. Roland Winston (U. of Chicago), which I cited in the previous debate over this.  He claimed that by using a particular nonimaging system, he was able to focus sunlight to heat an object up to hotter than the temperature of the sun.

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Offline Pmb

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Can you start fires with moonlight?
« Reply #6 on: 25/10/2011 21:34:17 »
I'm with you Geezer.  It sounds likely, but I don't understand why it has to be the cast.  Initially, it would seem that the second law of thermodynamics requires it.  After all if you form an image of a single point on a black body, that point will form a slightly larger blob (assuming it's diffraction limited), which means that the temperature is lower than that of the point source.  If it somehow was a higher temperature, then it would radiate back to the source point, heating it up, and you could use it to create a perpetual motion machine.

Where I get confused is what happens with an extended source.  In that case you have many points being imaged, and although each point-to-point image has to lose temperature, if your magnification is less than one (meaning your image is smaller than the object), then you're concentrating this energy, and I can't come up with a simple argument of why you can't increase the temperature this way.  This is a lot like compressing a gas to raise it's temperature, although in the case of a gas, the compression requires mechanical work.  In the case of light, it just requires a lens.  I suspect there is a good argument of why an imaging system can't do this, even with magnification, but I can't come up with it off the top of my head.

I also recall seeing an article by Prof. Roland Winston (U. of Chicago), which I cited in the previous debate over this.  He claimed that by using a particular nonimaging system, he was able to focus sunlight to heat an object up to hotter than the temperature of the sun.
If you store the energy and release it all at once with lenses toward the target.

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Offline JP

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« Reply #7 on: 25/10/2011 21:56:41 »
Well if you store it, it's obviously going to be able to make something hotter than the moon's surface.  The question I have is about heating a black body using only light emitted from another black body along with optical components (lenses, mirrors, etc.) 

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Offline CliffordK

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« Reply #8 on: 25/10/2011 22:06:00 »
If you have a point source, such as a lightbulb, then it can not generate any more heat than the source itself, although, you could perhaps generate a method to capture and store the energy.

There are a few problems with extending that analogy to the Moon.

First of all, most of the light we see is not radiated heat, but rather reflected light.  Consider the moon as being a large mirror as a component of your solar collector.  Certainly the mirrors in your solar collector don't have to heat up as much as the target.

The other point is that the moon is not a point source, but rather a diffuse source with a diameter of about 3,400 km in diameter.  If you could capture 99% of the light reflected from the moon's surface, it would be like having a solar collector 3,400 km in diameter, reflecting that light onto a point-source, it would get hot.

The problem is that we can't collect 99% of the light reaching and being reflected from the moon.

Anyway, I'd say you would need an awfully big solar collector.  I wonder what you'd get from one of the large commercial solar arrays.

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Offline yor_on

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Can you start fires with moonlight?
« Reply #9 on: 26/10/2011 02:41:14 »
Assuming that you can concentrate/direct photons by some lens to hit a specific spot it should be possible? It's not as the photons from the moon has a 'cold energy' :) Is it?

"After all if you form an image of a single point on a black body, that point will form a slightly larger blob (assuming it's diffraction limited), which means that the temperature is lower than that of the point source.  If it somehow was a higher temperature, then it would radiate back to the source point, heating it up, and you could use it to create a perpetual motion machine." is a good argument though.
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Offline imark70

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Can you start fires with moonlight?
« Reply #10 on: 31/10/2011 05:44:40 »
I've heard that the answer's 'no'. The proof offered is that we can point an instrument at an object in space and tell it's temperature. That is to say, we can measure the heat the object radiates and the moon is not hot enough.

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Offline Bored chemist

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Can you start fires with moonlight?
« Reply #11 on: 31/10/2011 07:10:10 »
You can't do it with a magnifying glass or other conventional optics. In principle it's possible using non-imaging optics.
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Offline Geezer

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« Reply #12 on: 31/10/2011 07:54:53 »
You can't do it with a magnifying glass or other conventional optics.

Yes, but why is that?
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Offline Geezer

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« Reply #13 on: 31/10/2011 15:58:00 »
Oh, I think I might have figured it out.

It's a matter of focus and color (wavelength). When you focus the light through an optical magnifying thingy, you create an image of the object on a surface. When it's properly focused, the colours of the object are as true as they can be for the particular optical system being used.

This applies to visible and also to invisible infrared (IR) light. The maximum intensity (temperature) of the IR occurs when it is properly focused on the surface, and assuming there is no attenuation of the IR through the optical system, the temperature of the image is the same as the temperature of the object. If it was not, the colors would look all wonky.

Unless you are using a mirror, the colors do have slightly different focal distances. During my pyromanic youth, I seem to remember that the highest temperatures could only be achieved with a magnifying glass when the image of the sun focused on the piece of wood I was igniting was a bit blurry. I never understood why, but maybe I do now.
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Offline JP

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« Reply #14 on: 31/10/2011 17:22:33 »
I'm not sure the spectral argument holds water, though.  It's emitted spectrum that tells you the temperature of a black body, not absorbed spectrum.  If I pumped light with a solar spectrum onto a point, then the equilibrium temperature of a black body at that point should be determined by the rate of energy flowing into it and it's surface area.

For example, if I project a magnified image of a light bulb onto a black body screen, that screen will be at a much lower temperature than the light bulb itself, since the surface area of the screen is much larger. 

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Offline Geezer

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« Reply #15 on: 31/10/2011 17:42:59 »
Er well, OK then. What I obviously intended to say is that the temperature of the image is only the same as the temperature of the object when the image is focused onto a teensy weensy point.
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Offline JP

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« Reply #16 on: 31/10/2011 19:54:53 »
Er well, OK then. What I obviously intended to say is that the temperature of the image is only the same as the temperature of the object when the image is focused onto a teensy weensy point.

Isn't that begging the question?

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Offline Geezer

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« Reply #17 on: 31/10/2011 20:01:57 »
Er well, OK then. What I obviously intended to say is that the temperature of the image is only the same as the temperature of the object when the image is focused onto a teensy weensy point.

Isn't that begging the question?

Don't think so [???]

It just means that the temperature has an upper limit. As the area of the image approaches zero, the temperature of the image approaches the temperature of the object.

I think I shall name this "Geezer's Conjecture."
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Offline JP

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« Reply #18 on: 31/10/2011 20:22:56 »
I happen to believe in the Geezer conjecture, but I don't see how it's been proven.

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Offline Bored chemist

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« Reply #19 on: 31/10/2011 20:44:53 »
I can't remember whose "law" it is, but you can't get the thing you are heating (hereafter called the ant) hotter than the thing heating it Hereafter called the sun).
The problem is, in essence, that you need a really big magnifying glass, but that will have a long focal length (because there's a limit to the radius of curvature) so the image will be big so the power density will be small so the rise in temperature will be small.

However, with non-imaging optics this doesn't apply. The conceptually easiest one is just a silvered funnel with the wide end pointed at the moon. The light that goes in the big end has to come out of the small end . If the thing is big enough the output power density can be made (practically) arbitrarily large so the temperature rise can be big.
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Offline CliffordK

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« Reply #20 on: 31/10/2011 21:37:30 »
Keep in mind that the majority of the light we see from the moon is reflected light, and not emitted light from the intrinsic heat of the moon (we don't see the IR bands).

You should be able to start a fire using the sun and a mirror, despite the mirror not getting red hot.

[attachment=15481]

Likewise, if the moon is acting at the mirror, one should still be able to capture the sunlight and start the fire.

I suppose the task gets far more difficult if one replaces the mirror with a white wall which acts as a diffuser.  Still, it should be possible to capture the sunlight reflected by a white wall.

I would think you could get some empirical results from a large solar collector. 



This one is actually owned by the University of California, Davis, so perhaps they would be willing to run an experiment.  You would have to find a way to collect night-time moonlight without contaminating the experiment with residual daytime heat.
[Oops, never mind, apparently someone had the bright idea to tear down that one.  There are other ones though.  Perhaps the Spanish PS20]

I'm not sure about the limitations of refraction.  There is a limit to the precise focus that a telescope uses.  However, in this case, one would be perfectly happy with multiple lenses, or minor perturbations in the focus.  But, still, when considered a large collector (1 sq km?) it would be much easier to make it with mirrors.
« Last Edit: 31/10/2011 21:48:30 by CliffordK »

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Offline Geezer

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« Reply #21 on: 31/10/2011 23:30:33 »
Maybe it's just because it's Halloween (here anyway) but it strikes me there's something very dodgy about the idea that you can ever concentrate the energy to make something hotter than the source. On the other hand, as long as you don't try to take more energy out than went in, why not?

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Offline CliffordK

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« Reply #22 on: 31/10/2011 23:45:54 »
In this case, the source of the REFLECTED light is the sun, and not the moon.

Even so, you would have a diffuse source (at a great distance) that you're concentrating to a single point. 

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Offline Geezer

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« Reply #23 on: 01/11/2011 00:10:40 »
In this case, the source of the REFLECTED light is the sun, and not the moon.

Yes, but's really no such thing as "reflected". Solar energy was absorbed by the Moon and energy was emitted by the Moon.
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Offline Bored chemist

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« Reply #24 on: 01/11/2011 06:59:23 »
In this case, the source of the REFLECTED light is the sun, and not the moon.

Yes, but's really no such thing as "reflected". Solar energy was absorbed by the Moon and energy was emitted by the Moon.

The same is true for light reflected by a mirror.
The important difference is that the moon is not a specular reflector. (Though, if it were, the fact that it;s not flat would make things more difficult.)
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Offline burning

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« Reply #25 on: 02/11/2011 02:09:44 »
In this case, the source of the REFLECTED light is the sun, and not the moon.

Yes, but's really no such thing as "reflected". Solar energy was absorbed by the Moon and energy was emitted by the Moon.

This is at best a dangerously misleading word game.  The are two different processes at work.  The one called reflection will produce a very different spectrum than that produced by the process called absorption and emission.  The reflected spectrum is proportional to the incident spectrum.  The emitted spectrum is proportional to the black body spectrum at the temperature of the object.*

Now, I suspect (but have not done the work to prove) that moonlight can't be focused enough, but the temperature of the moon is unlikely to be a significant limiting factor, because the spectrum of moonlight is dominated by reflected sunlight, not thermal radiation from the moon.  As Bored Chemist points out, the specular nature of the reflection is important.  The incident sunlight is getting more or less uniformly spread out over 2 π steradians, which is a considerable dilution.




* OK, there is a complication that the absorbtivity (and hence the reflectivity) of the object is a function of frequency, so it's not really simple proportionality.  However, since the absorbtivity function equals 1 minus the reflectivity function, and these two related functions are applied to two unrelated specta, the point still remains that they are measurably distinct processes.

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Offline Geezer

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« Reply #26 on: 02/11/2011 03:32:58 »
This is at best a dangerously misleading word game. are applied to two unrelated specta, the point still remains that they are measurably distinct processes.

Doesn't the question boil down to energy transmission and absorption. I could be wrong, but I don't think photons simply "bounce off" mirrors.
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Offline burning

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« Reply #27 on: 02/11/2011 04:40:12 »
This is at best a dangerously misleading word game. are applied to two unrelated specta, the point still remains that they are measurably distinct processes.

Doesn't the question boil down to energy transmission and absorption. I could be wrong, but I don't think photons simply "bounce off" mirrors.

For the moment, don't think about what's happening at the level of a single photon, and let's consider the two extreme behaviors.

An ideal absorber, i.e. a "black body," absorbs all light in the sense that any spectral information about light incident on it is lost.  The light emitted from the surface of the black body is a function of its temperature only.  True, the temperature of the body will change if the net power absorbed does not equal the net power radiated. But whether we shine 5 Watts of infrared, 5 Watts of visible, 5 Watts of UV, or 5 Watts distributed over a broad band, it doesn't matter; the spectrum coming from the surface of the body will be the same.

An ideal reflector, a "white body", can be said to not absorb, because the light coming from its surface completely preserves the spectral information of the light incident on its surface.  Indeed, if there is no light incident on it, there is also no light emitted from it.

OK, now realistic objects are neither ideal absorbers nor ideal reflectors.  But they can be understood as some combination of the two (neglecting the third process of transmittance, which is a safe omission in the case of the moon).  Some fraction of the light incident will be absorbed and thermalized; this light will potentially affect the temperature, which in turn determines what light is radiated from the surface.  Some fraction will come back out spectrally unchanged from how it was on the way in.

Now if we consider what's going on at the photon level, are the photons coming out that preserve spectral information of the incident light the same actual photons that were incident in the first place?  Probably not, but so what?  Why should detailed knowledge of what is going on at the particle level invalidate the use of the word "reflection" for the process of redirecting light with its spectral information intact?  Particularly, why should we insist on giving it the same name as the process that removes all spectral information of the incident light?

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Offline Geezer

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« Reply #28 on: 02/11/2011 05:02:03 »
OK, but I suppose we should try to get back to the original question and try to exlain the answer.
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Offline Bored chemist

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« Reply #29 on: 02/11/2011 06:58:18 »
"But they can be understood as some combination of the two (neglecting the third process of transmittance, which is a safe omission in the case of the moon)."
Not if we are interested in the spectrum of the light reflected.
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Offline burning

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« Reply #30 on: 04/11/2011 03:02:26 »
OK, I've been thinking about this for a couple days, and I've finally got it.  The quantity power per unit area per unit solid angle* is one that can only decrease as light goes through any optical system.  If the system consisted of only perfect mirrors and perfectly transparent lenses, the quantity would be conserved.  This basically boils down to being a consequence of conservation of energy.

Now the power per unit area from the Sun at the surface of the Moon is essentially the same as at the surface of Earth, and we know that it isn't enough to start a fire.  The moon is a diffuse reflector, so even if it were a perfect diffuse reflector, the power per area per solid angle coming from the moon would be the incident power per area divided by 2 π steradians.

No matter what optical system we might devise to focus this light, it is impossible to devise one that subtended more than 4 π steradians from a point on the surface of the target paper.  Which means that the absolute theoretical maximum power per unit area of focused moonlight would only be twice the power per unit area of sunlight at the Earth.  I'm pretty sure that only doubling the power per unit area of sunlight would not be enough to start a fire, and any realistic value for reflected moonlight is going to be less than this amount.



*Wikipedia calls this the radiance.  In my graduate optics class it was called the intensity.  Radiance is probably better, since a lot of different quantities get saddled with the name "intensity" by someone or other.

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Offline Geezer

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« Reply #31 on: 04/11/2011 05:46:35 »
and we know that it isn't enough to start a fire.

I suspect you didn't really mean to say that [:D]
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Offline willyp00

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Can you start fires with moonlight?
« Reply #32 on: 04/11/2011 09:14:04 »
Luke McNeill  asked the Naked Scientists:
   
Dear Nakeds

Everybody knows you can use a magnifying glass to focus sunlight to burn things, and the larger the lens, the more intense the heat. How large of a lens would you need to use to focus light from the full moon to burn paper?

Thanks,
Luke McNeill
Arlington, MA, USA

What do you think?
If the lens were big enough and the conditions were perfect, yes. What is the directed power of reflected moonlight at us? There must be some langleys or watts out there

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Offline willyp00

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« Reply #33 on: 04/11/2011 09:18:11 »
In this case, the source of the REFLECTED light is the sun, and not the moon.

Yes, but's really no such thing as "reflected". Solar energy was absorbed by the Moon and energy was emitted by the Moon.

This is at best a dangerously misleading word game.  The are two different processes at work.  The one called reflection will produce a very different spectrum than that produced by the process called absorption and emission.  The reflected spectrum is proportional to the incident spectrum.  The emitted spectrum is proportional to the black body spectrum at the temperature of the object.*

Now, I suspect (but have not done the work to prove) that moonlight can't be focused enough, but the temperature of the moon is unlikely to be a significant limiting factor, because the spectrum of moonlight is dominated by reflected sunlight, not thermal radiation from the moon.  As Bored Chemist points out, the specular nature of the reflection is important.  The incident sunlight is getting more or less uniformly spread out over 2 π steradians, which is a considerable dilution.




* OK, there is a complication that the absorbtivity (and hence the reflectivity) of the object is a function of frequency, so it's not really simple proportionality.  However, since the absorbtivity function equals 1 minus the reflectivity function, and these two related functions are applied to two unrelated specta, the point still remains that they are measurably distinct processes.
Reflection is absorption and emission.

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Offline Bored chemist

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« Reply #34 on: 04/11/2011 11:03:32 »
Luke McNeill  asked the Naked Scientists:
   
Dear Nakeds

Everybody knows you can use a magnifying glass to focus sunlight to burn things, and the larger the lens, the more intense the heat. How large of a lens would you need to use to focus light from the full moon to burn paper?

Thanks,
Luke McNeill
Arlington, MA, USA

What do you think?
If the lens were big enough and the conditions were perfect, yes. What is the directed power of reflected moonlight at us? There must be some langleys or watts out there

Actually, someone (under the name of Burning) has just shown that, even if the conditions were perfect, the answer is no.
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Offline syhprum

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« Reply #35 on: 04/11/2011 12:16:03 »
There is no doubt that it cannot be done with imaging optics but the jury is still out as it whether it can be done with non imaging optics
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Offline JP

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« Reply #36 on: 04/11/2011 13:34:18 »
*Wikipedia calls this the radiance.  In my graduate optics class it was called the intensity.  Radiance is probably better, since a lot of different quantities get saddled with the name "intensity" by someone or other.

The radiance is power/(unit area*unit solid angle)

Radiant intensity is power/(unit solid angle)

Irradiance is power/(unit area).

Intensity is most often associated with irradiance, since you're measuring energy incident upon a pixel (a tiny unit of area) integrated over some time.

But good explanation.     

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Offline Geezer

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Can you start fires with moonlight?
« Reply #37 on: 05/11/2011 05:12:00 »
Intensity is most often associated with irradiance, since you're measuring energy incident upon a pixel (a tiny unit of area) integrated over some time.   

Very nice, but still a load of old moonshine.

A simple magnifying glass could easily start a fire on the Moon (if there was some oxygen) using the light from the Sun. The Earth and the Moon are essentially the same distance from the Sun.

The reason the light from the Moon cannot start a fire on Earth is simply because the Moon has "soaked up" all the IR from the Sun. Basically, the Moon is a lousy reflector. If it was not, it would look like a second Sun.
« Last Edit: 05/11/2011 05:35:28 by Geezer »
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Offline syhprum

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« Reply #38 on: 05/11/2011 08:15:13 »
Has anyone checked the output of a solar panel when illuminated by Moonlight I would expect it to be pretty low
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Offline burning

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« Reply #39 on: 05/11/2011 15:23:55 »
and we know that it isn't enough to start a fire.

I suspect you didn't really mean to say that [:D]

That value of watts per square meter is not enough to start a fire.  Have you ever put a piece of paper out under the sun and see it ignite?  Yes, you can magnify the direct light of the sun and start a fire, but then you wouldn't have the incident power per unit area, you would have a greater one, due to the magnification.


(Edit: Just replacing quote I was responding to that got lost in pre-post editing.)
« Last Edit: 05/11/2011 15:54:37 by burning »

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« Reply #40 on: 05/11/2011 15:54:10 »
Intensity is most often associated with irradiance, since you're measuring energy incident upon a pixel (a tiny unit of area) integrated over some time.   

Very nice, but still a load of old moonshine.

A simple magnifying glass could easily start a fire on the Moon (if there was some oxygen) using the light from the Sun. The Earth and the Moon are essentially the same distance from the Sun.

The reason the light from the Moon cannot start a fire on Earth is simply because the Moon has "soaked up" all the IR from the Sun. Basically, the Moon is a lousy reflector. If it was not, it would look like a second Sun.

If the Moon were a perfect specular reflector (and flat and oriented correctly) it would look like a second Sun.  If it were a perfect diffuse reflector, it most certainly would not.

For an observer on Earth or the Moon, the sun subtends a very small solid angle (about 6 x 10-5 steradians).  After reflecting off a perfect diffuse reflector located where the Moon is, that light would be spread out over 2π steradians, i.e. a reduction in radiance (power per unit area per unit solid angle) of five orders of magnitude.  Since the Moon is coincidentally about the same solid angle as the Sun when viewed from the Earth, that means that the unfocused power per unit area from the Moon is at best (assuming perfect reflectivity across the whole spectrum) five orders of magnitude less than the unfocused power per unit area from the Sun.

Now, it may be that if we replaced the moon with a flat specular reflector, but an imperfect one which had the same spectral reflectivity of the moon, that we would indeed lose too much power at frequencies that paper absorbs efficiently to be able to focus the moonlight enough to start a fire.  I can't speak for anyone else, but it was never my intention to say that diffuse nature of reflection from the Moon was the only factor.  I do assert, however, that understanding the consequences of diffuse reflection is sufficient in and of itself to answer the question with a No.

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Offline Bored chemist

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Can you start fires with moonlight?
« Reply #41 on: 05/11/2011 16:08:18 »
The amount of radiation absorbed (rather than reflected) by the moon is a bit of a red herring, It heats up the moon until the moon re-radiates it as pretty much black body radiation so, with the right sort of lens (or a good mirror) not much energy is lost. The big issue is, as Burning says, that the light is spread out so much.
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Offline syhprum

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« Reply #42 on: 05/11/2011 16:19:25 »
The ignition temperature of paper is reputed to be 451░F (233░C) has this been verified ?
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Offline JP

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« Reply #43 on: 05/11/2011 16:27:27 »
The ignition temperature of paper is reputed to be 451░F (233░C) has this been verified ?
Yes.

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« Reply #44 on: 05/11/2011 17:11:04 »
The ignition temperature of paper is reputed to be 451░F (233░C) has this been verified ?
Yes.
No.
I always rather doubted this so I checked.
I got some lead, melted it and heated the liquid to about 400C (as measured with a thermocouple thermometer).
I then poured it into a folded  piece of paper.
The paper didn't even smoke much.
My thermometer might not be accurate, but the melting point of lead is about 327C or 621 F so this paper was certainly hotter than that.
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Offline syhprum

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« Reply #46 on: 05/11/2011 19:21:21 »
I am a little surprised that Ray Bradbury did not do some research before settling on a title for his novel, it is obvious to most people that 451░F or even 451░C is much too low. 
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« Reply #47 on: 05/11/2011 19:27:17 »
I am a little surprised that Ray Bradbury did not do some research before settling on a title for his novel, it is obvious to most people that 451░F or even 451░C is much too low. 

Check the link above.  451 C is not too low, although F is.

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Offline Geezer

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« Reply #48 on: 06/11/2011 00:30:44 »
The amount of radiation absorbed (rather than reflected) by the moon is a bit of a red herring, It heats up the moon until the moon re-radiates it as pretty much black body radiation so, with the right sort of lens (or a good mirror) not much energy is lost. The big issue is, as Burning says, that the light is spread out so much.

I suppose that's true, although it probably takes a little time to warm up after sunrise.
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« Reply #49 on: 06/11/2011 00:51:05 »
and we know that it isn't enough to start a fire.

I suspect you didn't really mean to say that [:D]

That value of watts per square meter is not enough to start a fire.  Have you ever put a piece of paper out under the sun and see it ignite?  Yes, you can magnify the direct light of the sun and start a fire, but then you wouldn't have the incident power per unit area, you would have a greater one, due to the magnification.


(Edit: Just replacing quote I was responding to that got lost in pre-post editing.)

Right, so it's just a matter of magnification. In that case, unless the spectrum of moonlight does not allow it, it should not be impossible with moonlight either, although it may be completely impractical.

Anybody know what the spectrum of moonlight looks like? I was going to say it looks as if it's not emitting much towards the red end, but, on reflection, I think that's not right. I imagine the light from the Sun must appear white on the Moon.
« Last Edit: 06/11/2011 01:46:41 by Geezer »
There ain'ta no sanity clause, and there ain'ta no centrifugal force Šther.