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If I would set up electromagnetic emitter that would emit one wave, but while it would be emitting it I would spin emitter 180 degrees at such speed that one half of wave would be emitted in that time (or it could be 1 degree, but the speed of turn would have to match the half of wave's oscillation) and then would set up a lot of sensors on the path of the wave that are able to measure strength of electromagnetic field (so that it would hit multiple sensors with that partial wave, sensors can be placed as far away as it needs be to allow the small turn in angle allow to have bigger affect). Would all the sensors show the value of the field going up to the peak and back down to zero, or would first sensor absorb the wave and all other sensors would show 0 values.

Update: Reason I am asking is that I can't wrap my head around - for sensor to register something it has to be affected by electromagnetic field - to affect sensor some energy has to be used, if wave would affect all the sensors it would imply that it has infinite energy as increasing the amount of sensors would still 'do to the work' register a value, but there has to be more 'work' done with each additional sensor, so in my understanding 1st sensor (or some other) would have to take all the energy of the wave and rest would not get anything.

Please forgive me if question is unclear or silly, I am not a physicist. If you need additional information please ask I will do my best to clarify.

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The velocity of light is 299792458 meters per second. That is the velocity of displacement of a peak to peak as the wave passes.

emradiation

the wave in slow motion

It is not clear what you want to do with your experiment but it is not possible to carry it out the way you envisage. There is no technology that could spin anything that fast.

Now in your title you ask

Would photon be absorbed on first sensor?

A photon is an elementary particle , it does not have a wave structure in space by itself. It helps build up the classical electromagnetic wave with zillions of other photons.

A photon either interacts, with some probability (given by the square of the wavefunction of the solution of the QM boundary problem) or passes through unscathed. It is not divisible.

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  • $\begingroup$ "It helps build up the classical electromagnetic wave with zillions of other photons". Can you please elaborate this point? $\endgroup$
    – user22180
    Jul 2, 2014 at 20:20
  • $\begingroup$ @user22180 Have a look at my answer to this question physics.stackexchange.com/questions/105417/… $\endgroup$
    – anna v
    Jul 3, 2014 at 3:23
  • $\begingroup$ Thanks for the answer, I have replaced title of the question and edited the text a bit to better represent what I am looking for. My question is that for sensor to register something it has to be affected by electromagnetic field, to affect some energy has to be used, if wave would affect all the sensors it would imply that it has infinite energy as increasing the amount of sensors would still do to the work, but there has to be more 'work' done with each sensor, so in my understanding 1st sensor would have to take all the energy of the wave and rest would not get anything, or I am being silly $\endgroup$
    – Matas Vaitkevicius
    Jul 3, 2014 at 8:50
  • $\begingroup$ well, it does not work that way. Sensors have a given area that responds either to individual photons ( as in photomultipliers) or the the summed up energy of individual photons. As I said , you cannot turn the beam around in the way you imagine, half wavelength, it is not physically possible because of the tiny, nanometer wavelengths and the very high velocity of light. Suppose you could, which is impossible , the photons making up the front wavelength would hit the frontal detectors and the ones turned would hit the 180 deg one. 1/2 1/2 by construction ( which is impossible really) $\endgroup$
    – anna v
    Jul 3, 2014 at 9:21

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