Say I have one LED in a flashlight, it produces some light.

Now, if I have 3 of those LEDs in the same flashlight, each LED receiving the same amount of energy than the single LED was receiving in the model described above (thus producing, individually, the same amount of light), will I get (perceive) more light? Will it be brighter? (considering the 3 LEDs very close to each other, same direction, etc.)

Quickly said: does the perceived light "sum up"?

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    $\begingroup$ Ask yourself what brightness is in term of properties of light (number of photons). Understanding this, you should see the answer. $\endgroup$ – Will Jun 27 '13 at 0:43
  • $\begingroup$ Does it have to be a flashlight and LEDs ? It is the same answer as having a 100 Watt bulb in a room, or 3 ,100 Watt light bulbs in the same room. $\endgroup$ – anna v Jun 27 '13 at 4:24
  • $\begingroup$ @annav yes, it can be a light bulb for all I care, and the answer would be the same I think. But what is this answer? $\endgroup$ – Totor Jun 27 '13 at 10:15

Remember that the intensity is proportional to the square of the electric field. So if the three electric fields add up, then the combined intensity will increase not by a factor of 3, but a factor of 9. However, in practice the three electric fields are unlikely to be in phase with each other. Further they are also unlikely to point in the same direction. In addition, the possibility also exists that they are all of same magnitude and point at $120^{o}$ in which case they would exactly cancel each other out. So the intensity could actually be anything between 0 to 9 times the individual intensities. This is with regard to the instantaneous intensity, the intensity at a particular instant of time. The intensity which we actually perceive as humans would be the time averaged intensity( The human eye cannot perceive fluctuations more than 20 Hz and visible light frequency is much higher than that.) So we would have to compute the time averaged intensity(square of electric field) over one time period. $$I = \frac{1}{T}\int_{0}^{T}(\vec{E_1(t)} + \vec{E_2(t)} + \vec{E_3(t)})^2dt$$ Because of their dependence on sines and cosines only terms which have squares of these functions would contribute and rest would be zero and so this would work out to be exactly 3 times the intensity of an individual lamp. So it turns out that they do add up - but this cannot have been concluded a priori the math would need to have been worked out.

  • $\begingroup$ The light from the different LEDs isn't coherent and so won't interfere. There will be three times as much light, though how the eye percieves this is anyone's guess. $\endgroup$ – John Rennie Jun 27 '13 at 8:20
  • $\begingroup$ Why not E1^2 + E2^2 + E3^2 if the waves don't interfere with each other? Or can't you, as Will suggests, consider light as a bunch of photons rather than a wave? (btw, can you really "cancel" some light by emitting the perfect opposite-phased "signal"?) $\endgroup$ – Totor Jun 27 '13 at 10:40
  • $\begingroup$ "but this cannot have been concluded a priori the math would need to have been worked out." My apriori argument is that energy is additive. 3 one hundred watts each light bulb give the same light as one three hundred watts light bulb. $\endgroup$ – anna v Jun 27 '13 at 11:42
  • $\begingroup$ @Totor Yes, you can "cancel" some light, though energy must still be conserved. So if you interfere two light waves, the result is some bright spots and some dark spots - you can have perfect destructive interference locally, but not globally. $\endgroup$ – Kyle Oman Jun 27 '13 at 13:09
  • $\begingroup$ @annav so it is not additive? I mean, if 3 light bulbs give the same light as one, you get no benefit as having 2 more light bulbs, right? $\endgroup$ – Totor Jun 27 '13 at 13:42

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