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I was trying to design a setup to measure a laser beam's wavelength and (hopefully) its intensity (\$W/m^2\$). I mainly plan to work in the visible spectrum with intensities between 200 and 900 \$W/m^2\$ and a beam diameter of around \$5\, mm\$.

I thought of using a pair of different photodiodes to measure their reverse currents when illuminated. Even though these reverse currents depend on both wavelength and intensity, I think I would be able to find a single pair of these quantities that would match the readings using their datasheets. Would you consider this a legitimate and accurate way of doing this task, or would you rather not do it this way?

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    \$\begingroup\$ I would not expect that to be very accurate, but it would be straightforward to try simulating for two photodiodes to see how well you can back out the wavelength. Your measurement of power will similarly have a lot of uncertainty unless you have some way to calibrate the photodiodes. \$\endgroup\$
    – user1850479
    Commented Nov 14, 2023 at 2:22
  • \$\begingroup\$ And dark current scales exponentially with temperature, of I remember correctly. Photonic efficiency: not quite sure about the temperature dependency oh that, would be something worth reading literature on. My suspicion is you'll want to make sure the temperature is somewhat stable, but since phone cameras give mostly the same amount of SNR, this won't be too critical \$\endgroup\$
    – Marcus Müller
    Commented Nov 14, 2023 at 2:43
  • \$\begingroup\$ Thanks for the answers. There is a possibility to calibrate the photodiodes with two lasers I have well characterized. I thought of the temperature dependence as well, I guess I should try to get the dark current - temperature and photonic efficiency - temperature curves from the datasheets and take that also into account. I also thought I could implement these adjustments using a logic made in Arduino so that it isn't so burdensome everytime I use a different laser. This last part may prove to be very difficult so maybe there is a better way of doing all this. \$\endgroup\$
    – Arturo Ruiz
    Commented Nov 15, 2023 at 20:59

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