When trying to intensify a beam of light by refracting it through a lens, (as in a lighthouse fresnel system or similar railroad style switch lamp from years past), is the beam intensity increased by moving the lens closer to the beam? That is, refracting it as close as possible to the light source? Or of a similar condition, would an old kerosene barn lantern be brighter if the globe was an inch away from the flame instead of 3 to 4 inches?


  • $\begingroup$ Well you can 'locally' intensify a light source but only at the expense of de-intensifying it elsewhere. But this really comes under a general topic of introductory optics and seems to broad for an answer to be given. Can you clarify/specify exactly what you are asking? $\endgroup$ – Nic Jun 8 '11 at 14:12
  • $\begingroup$ Very simply, I am asking if you moved the fresnel lens in a lighthouse, closer to the source of the flame, would it "concentrate" the refracted light making it appear brighter? $\endgroup$ – T Eizinger Jun 9 '11 at 0:35
  • $\begingroup$ T Eizinger, I deleted Georg's comment and your response to keep the comment thread clean. Georg can be rather harsh sometimes but I hope you won't let him get to you. This is not a bad question at all :-) $\endgroup$ – David Z Jul 22 '11 at 18:42

Surprisingly, the maximum intensity one can obtain from a light source does not depend on the distance to the light source nor on the distance between the light source and the imaging equipment. Instead, it depends on the distance between the imaging equipment and the object being heated (assuming that the imaging system is designed to be the best possible for these distances). What you want to do is "refract as close as possible" to the object being heated.

By thermodynamics, it's not possible for the light emitted by an object to heat up another that is already hotter than the first. For example, this means that we can't use an imaging system (or a non imaging power concentrating system such as a collection of mirrors) to use sunlight to heat a spot on the earth to a temperature higher than that of the sun.

So when you use an imaging system to magnify the effect of the sun, you do not get an image that is hotter than the sun. The best you can do is to make the image of the sun appear larger. And indeed, this is what a magnifying glass does. If you look through one (from the distance appropriate to frying an ant), you will see that objects in the direction of the image appear larger than they actually are. So when you apply a magnifying glass to an ant, what the ant sees is that the sun has increased in size.

The best you can do is to make the sun appear over all the sky, plus, with a mirror, to appear in the direction of the ground as well. Alternatively, you can arrange for the bottom side of the ant to be an insulator. (Uh, even concrete is a pretty good insulator if you try this at home.) This will heat the ant up until it approaches the temperature of the sun (with optimum optics and insulation). At that time the ant will radiate as much heat back to the sun as the sun radiates to the ant. And so the heat flows will balance and the ant will cease heating up. (All this assumes black body type radiation which is generally an okay approximation for this sort of thing.)

  • $\begingroup$ Thank you, Carl. An excellent and clear explanation of my inquiry. I do so appreciate it. $\endgroup$ – T Eizinger Jun 10 '11 at 0:29
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    $\begingroup$ Thanks! From a theoretical point of view, given a large enough telescope, one could heat an ant up to the temperature of a distant star. That's about as far away as you can get. (And a nice calculation: just how big would the telescope lens have to be.) $\endgroup$ – Carl Brannen Jun 10 '11 at 4:12

A point source of light radiates that light spherically outward, and the total power radiated by the light remains the same at all distances. As one gets further away, there is a greater total surface area, and the intensity of the light (power/area) is reduced.

Lenses don't "intensify" light, they only change its direction (and possibly attenuate it). Rays of light that would have been diverging from each other can be made parallel (to form a beam) or converging (to focus the light in one spot). There is still the same total amount of light, it has just been concentrated into a smaller area.

Making a beam requires that the rays of light exiting the lens have to be parallel or nearly parallel. This requires that the light source be at the focus of the lens. If the lens were moved relative to the light source, it would no longer form a beam and instead either converge or diverge.

Substituting a far-away lens for a nearer one with the focus in the same position would not necessarily intensify the beam. The light source radiates the same amount of power per unit of solid angle, and a larger, far-away lens could do the same work of a smaller, nearby lens.

  • $\begingroup$ Dan. Thank you for your explanation. It is what I was looking for and I appreciate the time you spent to explain it. $\endgroup$ – T Eizinger Jun 10 '11 at 0:30

umm, as long as Light/Plasma is defused by Plastic/Glass in any way/form; our eyesight will not recognize/perceive an illumination of anything. TV Shows have presented the false impression of refractive amplication for lighting and lasers being selfish amplifiers. What SciFi Programs HAVE gotten-right is that an underground tunnel/cavern can be illuminated with shiny surfaces bouncing available directed lighting via making a circle of reflections; what they are not explaining is that their eyes were already used to very-very low brightness from flashlights/candles and with the sun being reflected onto objects once our eyes are used to such low intensities seems to perceive the room as being completely illuminated by such relayed intensities. ( an old-fashioned flashing-light/flash-bulb/strobe-flash from a camera taking a picture in the darkness of moonlight is about the only explanation that everybody, whose seen such a dramatic sudden lighting, would grasp/understand as being a lights amplifier. It Is, however,Just An Illusion. ). JW USAF Retired.


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