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Light has always been a mystery for me. When we do a simple experiment of holding a lighter in one hand and a light source behind it, it shows no shadow on the wall at all. Why is this happening? Is it because the flame is pure energy and doesn't have matter in it, or is it due to light passing through the flame and not getting reflected? If it is, why isn't the light getting reflected or retracted?

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Imagine that you have a titanium screen. It's thin, so it's difficult to see. You can use a torch to heat some areas to incandescence. These areas become easily visible.

Now if we take a very bright light source and shine them at these lit areas, we won't see a shadow that gives us any information about the pattern of the light. Both the cool and hot areas interact with the bright light beam in the same way. The grid itself may be visible, but the pattern won't reflect anything about the relationship between the dark and light regions.

The flame is the same thing. It contains matter (such as fuel or soot particles), but similar matter is also present outside the flame. The imaging light may be slightly attenuated by these particles, but there is no sharp division in their location that corresponds to the shape of the flame. This lack of sharp division will prevent imaging.

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There is nothing in the flame to absorb or reflect light (except a very slight amount of smoke). There is some slight refraction--but you'll only notice it when the shadow is far from the flame and the light source is very intense (a point source). When you look at the flame, you don't see through it because the flame is so bright that your eyes only notice the flame. If your eyes and brain had a wide enough dynamic range then you could see both the flame and the things behind it.

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  • $\begingroup$ doesn't sound right to me. If a flame can emit light, then it can also absorb light. $\endgroup$ – Marty Green Dec 17 '15 at 21:13
  • $\begingroup$ @MartyGreen I'm sure the molecules in the flame do absorb light, but not in the visible range. $\endgroup$ – Asher Dec 17 '15 at 22:23
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    $\begingroup$ If they emit light in the visible range, then they must also absorb light in that range. That's how it works. If it were different, you could construct a closed system where heat flows from low to high temperature. $\endgroup$ – Marty Green Dec 17 '15 at 22:36
  • $\begingroup$ @MartyGreen: Glass glows when its hot enough, but is otherwise transparent. However, there is the issue of whether an excited atom has the same optical behavior as it does when its not excited. $\endgroup$ – Digiproc Dec 18 '15 at 10:15
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The flame has to have matter in it, because it is producing light itself. According to EM theory the resulting field observed (at a wall, for example) is sum of the field due to primary source and the flame. Some scattering and absorption does happen in the flame and this should lead to decrease of intensity in the region shadow would be expected. If you can't see this decreased intensity, the simplest explanation is that it is too small a difference to be detected by your eye.

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Everyone's trying to be all complicated, but it's an extremely simple answer. A shadow is nothing more than the result of light being blocked by something or another. But a flame is a light source itself. If you take a lamp bulb and a flashlight, and shine the flashlight on the lit lightbulb, the lightbulb won't produce a shadow, either.

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I think the conundrum comes from the mistaken idea that every "thing" that is visible is a "material" whose solidity is proportional to its visibility. It's an almost subconscious idea built up from a vast amount of inductive experience with reflective surfaces. But it breaks down with light-emitting gases/smokes (and plasmas).

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Try running your lighter without lighting it. How much shadow do you expect?

How much more shadow do you expect if it's lit? Probably more because there will be a little tiny bit of smoke, but not much more.

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We usually see shadows thrown by buildings and people walking around in streets on a bright sunlit day.

Obviously, what is happening here, is that light is being blocked from travelling by matter in its line of flight and this gives a shadow.

Not all matter will do. For example, a window is made up of matter but it is transparent to light and let light through and we see no shadow thrown by a window.

Air, too, is made up of matter, but it is transparent to light and we see no shadows thrown by air too.

A lighter flame is is essentially hot gas and like air is transparent to light. However, unlike air, it is giving off light as it combusts and so - rather than throwing a shadow - it throws up a spot of light on a sheet of paper held close to it, even in a sunny day.

However, combustion need not be so clean and may be 'sooty'; for example, certain cheap oil-lamps. And this soot is carried up by the hot gas. This soot can stop light and so here you will see a shadow, not of the flame itself, but of the soot borne aloft by the hot gas.

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I looked at https://www.youtube.com/watch?v=S3p-jzyZXJM , and my impression is there is some slight partial shadow of the lighter flame. There is a similar question at Shadow of fire doesn't exist , and the photographs in the answers there clearly show that candle flame does cast partial shadow. Maybe the intensity of light is higher for a lighter than for a candle, so I suspect you will have a clear partial shadow of lighter flame if you use a much brighter light source, especially if the spectrum of the light from the source is close to the spectrum of the light from the lighter flame.

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