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In Physics class we were doing the two slit experiment with a helium-neon red laser. We used this to work out the wavelength of the laser light to a high degree of accuracy. On the piece of paper the light shined on there were patterns of interference, both constructive and destructive. My question is, when the part of the paper appeared dark, where did the energy in the light go?

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==OUT OF TOPIC== Can you confirm that the 3 slit experiment does not produce any additional interference patterns? – ja72 Jan 31 '11 at 18:39
up vote 7 down vote accepted

It goes to the brighter strips. In these regions there is a constructive interference, which is actually brighter than would be with just one source.

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This sounds like the right answer.. I suppose the energy is somehow transferred between the peaks and troughs? – Thomas O Jan 24 '11 at 13:41
Yes, that's right. – Rafael Jan 24 '11 at 15:46

Try integrating the power in the resulting interference pattern; you will find that energy is indeed conserved. In the case where the slit is illuminated by a plane wave and observed in the far-field, this is very easy: the interference pattern is simply the Fourier transform of the pattern of slits, and conservation of energy is then given by Parseval's theorem!

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Energy is indeed conserved. The link below gives a nice explanation...

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Answers consisting only of a link are discouraged. – dmckee Jul 23 '12 at 1:40

It goes where the photons go. Best explanation I've seen is in Feynman "QED: the strange theory of light and matter". One does not have to unlearn anything. It is even more interesting after reading his more technical works.

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