I am working with a Czerny–Turner monchromator (a Bentham TMc300, to be precise), with adjustable entrance and exit slits. I started by acquiring the signal with the slits completely opened, and I got a 50 nm FWHM signal. I would like to reduce its bandwidth without changing any internal element. I therefore reduced the width of the exit slit, and the best I could get to was a FWHM of 20 nm. I thus started to reduce the entrance slit width, and to my surprise the bandwidth decreased down to 8 nm.

That's ok for what I want to do, but how comes that this happens? I keep thinking about the process, but it seems to me that the entrance slit should influence mainly the power of the light signal, but not its spectral bandwidth. I think I'm missing some fundamental optical point here (and a stupid one, probably), and I want to solve it.


1 Answer 1


Yeah, this is expected.

Perhaps the best way to understand this is to consider what would happen with the interior elements of the monochromator if your input light were truly monochromatic. Given that assumption, what really matters is the response of the diffraction grating (element D in the Wikipedia diagram below) would be basically the same as a flat mirror (oriented at some nontrivial angle, sure, but that doesn't matter) since all it's doing is taking collimated beams coming in from some angle and producing a collimated beam at some other angle:

Image source

(Heck, that's basically what the ray diagram is actually showing.)

That means, therefore, that the C-D-E combination is rather easy to analyze under those conditions ─ basically, it is a re-imaging configuration, and an illuminated entrance slit at B will produce an unmagnified real image at the focal plane F.

Now, if you increase the complexity just a little to consider light of two discrete monochromatic components with wavelengths very close to each other, then you do the same thing twice, except that now you'll have two images of the entry slit at the focal plane, and if the wavelengths are close enough then those two images will overlap, and you will be less able to resolve the two components.

However, if you then reduce the width of the entrance slit (and assuming, of course, that you're not limited by the grating itself), then those images will also become thinner, and if you make them thin enough then you will again be able to make out the two separate images, as they will be thinner than the separation between them.

So, yeah, that's basically it. This obviously needs to be extended to a continuous spectral distribution, but that extension is basically trivial.

(Though if you actually have the device in front of you, there's nothing better than actually seeing this for yourself. Take a light source that produces multiple different discrete emission lines at disparate wavelengths, remove the exit slit F, and maybe for good measure replace the entrance slit B with something more distinctive ─ say, a cutout star, or something with a recognizable shape. Then you should be able to see, with your bare eye (or possibly with a magnifying glass? depending on the instrument) multiple different images of the object at B, with the different at different locations.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.