It is my understanding that Raman Spectroscopy uses a notch filter (or sometimes an edge filter) to remove the light from the laser, so only the relevant Raman bands are present. However, if this is the sole purpose of the filter, then why not simply ignore that wavelength in the generated data and leave out the expensive filter?
1 Answer
The incident light is many many orders of magnitude brighter than the light at the Raman frequencies. If you let the fundamental in, it will bounce around whatever spectrometer and detector that you have, and inevitably some (sometimes a lot) of that light will reach the detector. Note that this is light that has scattered off of walls and baffles and what-have-you. That is, it is no longer following the path of the radiation of interest. The light fills the spectrometer and detector, and simply clouds the spectrum. One must take care to eliminate as much of this as possible. This is the reason that double and triple monochromators are used in Raman experiments.