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Most photo detectors uses the wave aspect of light while measuring the intensity of any incident radiation , thus yielding a value in $\text{W/cm}^2$ , also there are photo detectors that shows the same result in photons incident per unit square of the surface . My question is whether does this agrees with the uncertainty principle in terms of the uncertainty among the wave and particle nature of light .

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    $\begingroup$ I'm unclear on exactly what distinction you are trying to make here. In particular, your first sentence is not true for a semiconductor detector. Please clarify. $\endgroup$ – Jon Custer Aug 5 '19 at 13:09
  • $\begingroup$ "Particle" and "wave" are different ways of understanding/explaining light. Maybe there is a photo detector out there whose action is best explained by talking about waves, but even if that's true, I would not say that the detector uses the "wave aspect." $\endgroup$ – Solomon Slow Aug 5 '19 at 14:24
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You can only absorb energy from the electromagnetic field in units of the photon energy. This is why we call the photon the quantum of electromagnetic energy.

Any photodetector that works by absorbing energy from the field it's measuring thus only absorbs that energy in discrete quantities, aka photons.

That means that most of the common photodetector types, including semiconductors, tubes, and bolometers, are all interacting with the field by absorbing photons from the field, not by somehow measuring the intensity without absorbing any energy.

If you have a sensor (other than a bolometer) that displays its result in W/cm2, that's most likely done by simply scaling the measured photon flux according to the expected wavelength being measured.

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  • $\begingroup$ Well then , if I am using a standard photo meter to measure the intensity of light directly in one case , while in another, I am using the data from a photoveltic cell to record the photo current (as photo current depends on the number of photons incident on the surface of the cell per unit time) and hence derive the intensity from it , will both the measurements yield same result. $\endgroup$ – rajarsifoxy Aug 6 '19 at 2:21
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    $\begingroup$ As far as I know, a "standard photometer" (as used for photography) is a photovoltaic cell. But there are several possible ways to make a photometer, and I don't know which one you mean by "standard". In any case if you have two detectors, and you calibrate both for the spectrum of the light you're measuring, then they should give the same results. $\endgroup$ – The Photon Aug 6 '19 at 4:18

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