I found this article at SPIE, thought it would be a good reminder to keep the mind on the physics rather than blindly following the nice mathematical formulas.

The following excerpt gives the physical picture of two-wave interference by focusing on the detection :

"The detecting dipole molecule tries to undulate in response to both superposed EM stimulations. Dark fringes are found where the fields are 180° out of phase because the detecting dipoles cannot undulate in two opposing directions at the same moment. They are not stimulated and hence cannot absorb energy from the fields. The field energy (photons) are not absent from these locations. They simply cannot be absorbed. Bright fringes are found where the two fields stimulate the dipole undulations in phase and in the same direction, maximizing the absorption of energy. The operation or ‘summation’ of amplitudes and phases, represented by the superposition equation, is carried out by the detecting dipoles, not by the two fields themselves. Superposition fringes are locally created by the subnanometer-size detecting molecules!"

Hi guiding_light,

Thanks for that link - a very nice summary article. You should post it over at the "Problem with the double slit experiment" thread.

Besides the paragraph you quoted, I also found this insightful:

"Electrons are indivisible, and their binding energies are quantized. Thus their response to light can only be registered as discrete quantum events irrespective of whether light energy constitutes indivisible photons. Light beams always travel with a finite velocity carrying a finite flux density delivering energy in time. Any registration of light energy patterns will therefore always be built up as accumulated discrete quantum events whose temporal rate will be proportional to the incident flux rate. A low photoelectron count rate is insufficient proof of the indivisibility of photons."

And I particularly like his italicized final statement:

"There is no ‘interference of light,’ there are only ‘superposition effects due to light."

~Jim

Hi Guiding Light,

Of spatially mapped interference

Looking at this plot of photon counts
http://www.teachspin.com/instruments/two_s...combiplot2.gif

( taken from here http://www.teachspin.com/instruments/two_s...periments.shtml )

Loosely we see that the constructive interference zones represent areas of twice the power in half the space .. leading to a peak photon count of 4 times the number of photons counted with just one slit open. The photons aren't 'hidden' in the dark areas .. they've upped stakes and moved into the bright areas.

With both slits open the total number of photons is the same (checked by graphical count .. within 5%)

I checked through the conservation of energy here .. as best I could.
http://forum.physorg.com/index.php?showtop...ndpost&p=160646

My conclusion .. in the case of the two slit experiment is that there can be no photons to detect at points of total destructive interference.

And of temporally mapped interference .. ????

If two 1 Watt laser of slightly different frequencies are directed onto a black (body) surface is the total power delivered 2 Watts ?

Assuming yes ..

Taking a slow beat (say 1Hz) .. for half a second we have destructive interference and in the other half second we have twice the amount of energy delivered in half the time ??? To me it would seem the photon count MUST increase by a factor of four at the time of peak constructive interference .. unless there is a free source of photons then the photon count MUST fall to zero during total destructive interference.

If 'no' then I have no point to make.

Best wishes,

-C2.

Hello C2, et al.

It would be interesting to see what the photon count would be if the slit plate was removed. I say this because two slits have twice the open area of a single slit. Would a single slit with the same area as the double slit have the same photon count?

Hi Montec,