If you light a candle it starts emitting photons, right? And photons travel at the speed of light. But how come they can't light the whole of a dark room? It means that the photons are not reaching the dark corners of the room, but that is impossible if they travel at 300 thousand kilometres per second?
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34$\begingroup$ Clearly you haven't heard of the theory of dark suckers: travel-watch.com/mario/dark_suckers.htm ;) $\endgroup$– Mikael FremlingAug 18, 2017 at 8:44
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48$\begingroup$ Put a mirror in a dark corner. Can you see the candle in the mirror? What does this tell you about the supposition of your question? $\endgroup$– Eric LippertAug 18, 2017 at 16:39
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5 Answers
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It might well be that the photons reach the corners of the room but that doesn't mean that you can see the corners. Why? For you to see something, photons have to be emitted or reflected by the object and afterwards reach your eye. Also, it is not enough that one photon reaches your eye, but several photons are necessary for you to see the object (e.g. here).
So why does a single candle not suffice to light the whole room? While the candle emits photons in (nearly) all directions, the number of photons it emits is too small. There are simply not enough photons being reflected back into your eye to see the whole room.
The reason that you can see objects close to the candle but not far away is the inverse square law.
(source: http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/imgfor/isq.gif)
The candle emits (approximately) an equal number of photons in each direction. If an object is very close to the candle nearly all photons emitted in the direction of this object will hit it and be reflected. However, if the candle is far away, the number of photons per unit area gets smaller and smaller because the number of photons emitted in this direction is constant but the area becomes larger. Thus, an object far away is hit by much less photons. Then, the photons are either absorbed or reflected in all directions and while some travel into the direction of your eye, the inverse square law becomes important again. Hence, in the end, only very few photons are reflected by the corner and reach your eye and you can't see it.
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45$\begingroup$ The final piece you're missing is the sensitivity range of the eye. You can see a single photon. You can see a sunlit snowfield. But you can't see both at the same time. The eye adjusts its sensitivity range for the environment; the sensitivity for seeing the bright candle flame can't also see the dim reflections from the corners of the room. $\endgroup$– MarkAug 18, 2017 at 20:00
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1$\begingroup$ Thanks for adding this! This fact is (implicitly) mentioned in the article of Baez I linked to (cf. photopic and scotopic vision), but it probably would have been better to include it in the answer. $\endgroup$– VirftAug 19, 2017 at 7:16
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The light does reach every corner of the room, because from every corner you can actually see the candle light.
To light the surface of the walls, the light needs to be reflected and then hit your eye again with enough energy so that your receptors register it. Reflection however drains energy from the beam, and irregular surfaces require many reflections (on average) till the beam ejects at an angle that aims towards your eye.
If you cover the walls of that room with a flat, non-absorbing surface like a mirror, you would see the reflection of the candle clearly. So the light obviously traveled to the corner of the room and back to your eye.
If you use an irregular, absorbing surface like black cloth, the beam will bounce back and forth inside that surface till all its energy is consumed/transformed. As in this case the remaining energy of the beam that is able to reach back to your eyes is very low to none, the corners appear dark even though they are hit by the same intensity as in the case with the mirrors.
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$\begingroup$ Note that the answer has been edited to address the issues raised by @OrangeDog. However, it should be pointed out that reflection and intensity loss can be described in the photon picture as well as the wave picture. $\endgroup$– garypAug 18, 2017 at 14:41
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$\begingroup$ @garyp yes, just not in the way they tried to. It's still a bit iffy, as they apparently just replaced "photon" with "beam", so now we have a beam bouncing around inside a black surface, which also makes no sense. $\endgroup$ Aug 18, 2017 at 14:43
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$\begingroup$ @OrangeDog Yes, I agree. The ideas in this answer are ok, but they are not expressed clearly. $\endgroup$– garypAug 18, 2017 at 14:54
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$\begingroup$ I am open for suggestions, but as said: I am trying to keep the answer simple, and simple doesn't necessarily mean correct in every detail. $\endgroup$– TwoTheAug 21, 2017 at 7:14
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$\begingroup$ Simplicity must not include incorrect claims, unless specifically noted so. You can easily talk about photons and intensity loss in same sentence, but you should not claim that 'photons lose energy' (which doesn't happen in linear optics, although can be said for nonlinear effects, where photon can be absorbed and re-emitted with different energy - simplification of course). You could say that 'some photons can be lost', which is simplified again, but not incorrect. $\endgroup$– ArvoAug 21, 2017 at 10:46
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Every source of light emits certain amount of photons of certain wavelength. Every photon travels at the same speed in the medium.
But to light the whole area, what actually matters is the luminous intensity of the light source and not the speed of light.
If the light source has high luminous intensity, it emits more photons in a solid angle at a given time than a light source with low luminous intensity.
As candles don't have enough luminous intensity to light the whole room, still some parts of the room remain dark.
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In order to "light the dark corners of the room" several things need to happen:
The candle needs to emit photons. This it certainly does and at a fairly large rate, something like 10^18 (!?) photons per second or so. A large part of these photons are however not in the visible spectrum, but infrared, so you cannot see them.
As you move away from the candle, the number of photons that hit a certain area diminishes proportional to one over the distance squared (surface area of a sphere). So only a small fraction of these photons will end up in the corner of the room that is several meters away as most photons will end up lighting other parts of the room.
Now, at the wall/corner some of the photons will get absorbed (lost), some will get reflected/re-emitted, but for typical walls this would be diffuse reflection in all kind of directions. In order for you to see the corners, only part of the reflected light (only that part which gets reflected into your direction) can be used for observation. And as with the candle lighting the corner, also for the reflected light your eye's distance from the corner will matter with the same proportionality ($\propto 1/r^2$). So ideally you'd be close to the corner, but then you run into the risk of throwing a shadow on the corner.
Lastly the light hits your eyes and needs to be detected which seems to be possible with a threshold of only around 10 photons.
In the discussion above I neglected (at least) light absorption in the air and the eyes (before hitting the receptors). Also I neglected second order processes (multiple reflections), but I believe both of these are minor compared to the factors mentioned.
answered Aug 18, 2017 at 8:04
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They are reaching the corners but are then absorbed by the material. The material must re-emit light to hit your eyes, before you see it and think of it as lit. If there simply arrives too few photons to a specific point, then not enough light is re-emitted from that point to your eye and your eye won't register this as lit.
