I am making a spectrometer as my school project. It's main application will be analyzing the light emitted by various light sources. It measures visible, near IR and UVA/UVB radiation. I need it to be relatively weatherproof, so it needs to have some form of a window to protect the measuring modules from water, dust, debris etc.

I am concerned that passing light through a transparent material would cause a change in the spectrum. Are my concerns true, or are they false or too small to make a difference? Additionally, what would be the best material to use as a window? (the only material available to me right now are clear plastic sheets designed for use with printers, would that be at least marginally OK?)

  • $\begingroup$ Also you will need a detector, like a camera chip for visible but it very much has a spectral response as do IR and UV ones, which are expensive. $\endgroup$ – PhysicsDave Dec 5 '18 at 15:32
  • $\begingroup$ @PhysicsDave I've got a reliable and fairly accurate measuring module. $\endgroup$ – kristjank Dec 6 '18 at 17:06
  • $\begingroup$ Try and get a spec online of its spectral response for the CCD, PMT, InGAs, or vanadium, etc., detector that you have. $\endgroup$ – PhysicsDave Dec 6 '18 at 18:22

Any material the light interacts with will indeed change your spectrum, even if only a little. But a window is not the problem—pick a UV fused silica, which will transmit your entire range of interest and has a relatively low refractive index.

What you need to realize is that your entire spectrometer has a “spectrum”, which represents the spectral efficiencies of all of its components, which will not be the same for every color. Your detector has a non-flat spectrum $S_D(\omega)$, as does your dispersing element (a grating?) $S_G(\omega)$, your window $S_W(\omega)$, etc. These combine to make your spectrometer spectrum: $$S_s (\omega) = S_D(\omega) S_G(\omega) S_W(\omega)... $$ What you actually measure, $D(\omega)$, will be related to the incident spectrum $I(\omega)$ as $$D(\omega) = I(\omega) S_s (\omega).$$

So if you want your measurements to be absolutely accurate, you need to either (a) calibrate your spectrometer with a known source to find $S_s (\omega)$, so you can divide it out, or (b) do relative measurements, where you are comparing a sample spectrum $A (\omega) $ with a reference spectrum $R (\omega) $, both measured with the same spectrometer configuration. Then $$\frac{A (\omega)}{R (\omega)}=\frac{D_A (\omega)}{D_R (\omega)}.$$

  • $\begingroup$ Thank you for now, but this spectrometer doesn't need to be extremely accurate. I'm just making sure that use of common materials won't make my device useless. Thank you again for a quick and meaningful answer! $\endgroup$ – kristjank Dec 6 '18 at 17:05

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