Which transparent material alters light the least? 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?)
 A: 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
$$\frac{A (\omega)}{R (\omega)}=\frac{D_A (\omega)}{D_R (\omega)}.$$
