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This question already has an answer here:

I am curious about the ability for a lens to start a fire without using sunlight. Is it possible for alternative light sources (LED, incandescent, etc.) in combination with a very large Fresnel to start a fire? Or more appropriately how much energy could a very large Fresnel lens concentrate into one point using everyday light sources?

In considering different types of light sources, which would be best for generating the most heat? Luminous efficacy and the size of the light source seem the most obvious but how would the color of the light affect the lens' ability to generate heat?

Will a high focal length and large aperture lens (such as a giant spot Fresnel) even be able to focus the amount of light sources necessary into a small enough point to generate enough heat?

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marked as duplicate by Floris, RedGrittyBrick, Kyle Kanos, user36790, Jon Custer Nov 15 '16 at 13:44

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    $\begingroup$ what-if.xkcd.com/145 (I love xkcd!!!). $\endgroup$ – userLTK Nov 15 '16 at 2:06
  • $\begingroup$ Well, the wire in an incandescent bulb burns very nicely. You just have to put it in an oxygenated atmosphere. Does that count? :D In any case, the light source in an incandescent bulb works exactly the same way as in the Sun, so of course it's possible. More interesting light sources are LEDs, lasers, moonlight, fluorescent tubes and similar. $\endgroup$ – Luaan Nov 15 '16 at 11:41
  • $\begingroup$ If you have enjoyed motion picture films then you have seen the means. If the film stops with the light on and the shutter open the heat on a single frame promptly melts and chars the film. In the days of celluloid film this was a huge fire hazard in theatres and caused a shift to acetate base films as soon as the material was mature enough. $\endgroup$ – KalleMP Sep 8 '18 at 21:13
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When you use a lens (or a focusing mirror) to increase the "power" of a light source - whether it be the sun, or another light source - what you are really doing is making the light source "look bigger". Which is a lot like "being closer" to the source.

For example, if you say the sun normally is a disk that spans about 0.5° in the sky, then if I have a lens that makes it look like it's 5° across, the area is about 100x larger (diameter squared), and the energy flux will feel roughly "like the power of 100 suns". I say "roughly" because there are inefficiencies in lenses and mirrors.

The same principle can be applied to any other light source - when you make the light "seem bigger", the net effect is the same as that of "getting closer" or "having more lights". This means that the question of whether you can use a lens to cause something to catch fire really comes down to this: if you got really close to your light source, would the thing you are trying to set fire to catch fire? If the answer is "yes" (like it is for the sun), then you can use a lens to "make it look like you are that close". If the answer is "no", then a lens won't help.

Specifically, it is not possible, with lenses / mirrors, to make an object hotter than the light source you are focusing. This is explained in more detail in this earlier answer

One thing to note - light bulbs have a minimum distance to the filament (the bulb has a certain size) both to keep the glass from getting too hot, but also to lower the risk of things catching fire because they are too close to the filament. However, it is quite easy, with a lens, to "make it look like you are really close to the filament". In other words, a decent lens (that is, a lens with sufficiently low f number1) should be able to focus a filament of an incandescent bulb onto a piece of paper such that it can burn a hole in it.


1 The f number is the ratio of the focal length of a lens to its diameter. The lower the number, the bigger the lens - and the more light per unit area it collects.

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    $\begingroup$ No, the optical apparatus intended here makes the same amount of energy hit a smaller area, causing more energy to be absorbed and converted to heat in this smaller area. Old school physics probably assume light of a certain quality can only be generated by a source of a certain temperature - LEDs/Laser Diodes and gas discharge lamps of various descriptions are not dependent on heat for generating any kind of light you design them to make. $\endgroup$ – rackandboneman Nov 15 '16 at 10:15
  • $\begingroup$ -what you are really doing is making the light source "look bigger". Which is a lot like "being closer" to the source. Was simpler to understand $\endgroup$ – Ratna Nov 15 '16 at 10:38
  • $\begingroup$ @rackandboneman Well, this still works for Floris' rule-of-thumb - if you can make fire by moving the object closer to the light source, you can use optics to do the same thing (ignoring efficiency). The tricky part is that it really only works to the extent that the light source is a point source - and the closer you get to the light source, the worse it approximates to a point source (fluorescent tubes are an obvious example). $\endgroup$ – Luaan Nov 15 '16 at 11:48
  • $\begingroup$ I assumed you can make any lens shape - including a condenser or collimator - as a fresnel? $\endgroup$ – rackandboneman Nov 15 '16 at 13:32
  • $\begingroup$ @rackandboneman no optical apparatus chan concentrate the power per unit area to something greater than "at the surface" of the source. And while the source may not be a black body emitter, the thing you want to set on fire will approximate it. Perhaps I oversimplified - but you will need a lot of Watts per square meter in your source to begin with , otherwise optics can do nothing. $\endgroup$ – Floris Nov 15 '16 at 13:37
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You can't focus light onto a region so that it increases the temperature to a value larger than the temperature of the source. This has its origin in Fermat's principle, and since it is a variational calculus it carries to much of physics. This is primarily Lagrangian and Hamiltonian mechanics. The fundamental upshot is that the volume a system occupies in phase space of position and momentum is an invariant in conservative systems. The optical analogue of this indicates one is not able to concentrate light from a source and heat up a target material to a temperature larger than the source.

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    $\begingroup$ Sure? Temperature at the target is created by light being absorbed. And in practice, people have started fires by accident with strong (borderline illegal class 4) laser pointers that use sources that would instantly be destroyed if you brought them up to the temperature needed to start a fire. How would light carry "information" about how hot the source is (unless the quality/quantity of the light depends on the source temperature, which is not relevantly the case with a source like a LED!) $\endgroup$ – rackandboneman Nov 15 '16 at 10:08
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    $\begingroup$ @rackandmoneman There are various definitions of "Temperature" for non-thermal light sources. The thermodynamic temperature of a laser emitting medium is in fact negative (because it has been pumped such that there are more atoms in an exctited state than is possible to obtain by simple heating of the medium ). $\endgroup$ – nigel222 Nov 15 '16 at 10:33
  • $\begingroup$ @rackandboneman In the case of a laser pointer, it emits more light than what would be emitted by only thermal radiation. If you live in a hot part of the world, you may notice an appliance called air conditioner. It can make the inside of the room cooler than the outside, provided that the outside is heated more than the inside is cooled. This answer is correct if the two bodies only emit thermal radiation. $\endgroup$ – v7d8dpo4 Nov 15 '16 at 10:53
  • $\begingroup$ Substitute energy for temperature here. I wanted a short answer and skirted the issue of not Boltzmann distributed energy or nonthermalized photons. However, you can't concentrate energy from a source to an energy density than what comes from the source. At least this can't be done in a manner without adulterating conservation laws. $\endgroup$ – Lawrence B. Crowell Nov 15 '16 at 11:15
  • $\begingroup$ This "Etendue" stuff? Oops.... $\endgroup$ – rackandboneman Nov 15 '16 at 13:56
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It is very dependent on how well your target absorbs (rather than reflect) the light and converts it into heat, and how well that material conducts heat away if heated in a small spot - black, matte cardboard for example will absorb most of the energy from visible light and convert it to heat. Losses will still occur from blackbody IR radiation and surface air cooling.

In practice, one watt of energy concentrated into a few square mm (think overheating a very small resistor with 1 watt of electrical energy, with flammable vapor present) can start a fire.

LEDs can reach efficiencies in converting electrical energy to optical energy of several 10s of percents, so actual optical power is in the same order of magnitude as input power...

Ignore constraints that assume that all light is generated from blackbody radiation, unless your light source is working on the basis of blackbody radiation (an incandescent lamp does, a LED or CFL does not!).

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