# 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?)

• 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. – PhysicsDave Dec 5 '18 at 15:32
• @PhysicsDave I've got a reliable and fairly accurate measuring module. – kristjank Dec 6 '18 at 17:06
• Try and get a spec online of its spectral response for the CCD, PMT, InGAs, or vanadium, etc., detector that you have. – PhysicsDave Dec 6 '18 at 18:22

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 $$\frac{A (\omega)}{R (\omega)}=\frac{D_A (\omega)}{D_R (\omega)}.$$