<p>What happens when you double the width of the entrance slit of a spectrograph, in turn letting in twice as much light, would there be an improvement in the Fourier transform of the measured spectral profile d(f) following to higher frequency f?</p>

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2021-03-31T01:02:27.493Z
The slit is a cut in the transverse spatial domain, the spectrometer (assuming a dispersive-type (e.g. grating or prism)) spreads wavelength into the transverse spatial domain. Bigger slit smears out the spectrum, so no, there would be the opposite of an improvement.
uhoh
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2021-03-31T03:05:57.103Z
Thanks for the reply. @uhoh ; would this mean a more noise towards higher frequencies in the Fourier domain then?
Jay D
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2021-03-31T04:46:41.103Z
Well I think it is best if you first add more information. Are you talking about a conventional spectrometer that uses a diffraction grating, or a Fourier Transform spectrometer? Is the noise you are talking about just statistical (e.g. number of photons) or is there additional noise in the light source of some kind, or is it detector noise? Presumably data will be less noisy if you have a bigger aperture and more light and the resolution will be poorer, but it's hard to make more general statements without additional details.
uhoh
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2021-03-31T06:23:36.797Z
A conventional one using a diffraction grating; the noise would come from the background  like heat, noise, stray light, electronic noise etc.
Jay D
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2021-04-03T05:06:18.900Z
@Jay D I don't know how these things are actually constructed, but I'd be very surprised if a spectrograph for a telescope didn't already utilize all the available light from a star.  So I'm not sure where 'slot-width' would come in.
Roger Wood
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2021-05-07T06:45:22.893Z
What do you mean by an improvement to the FT?
ProfRob
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2021-05-07T06:47:34.547Z
@RogerWood almost all spectrographs have slits defining the illumination of the dispersive element (and as a result, do not use all the available light from a star).
ProfRob
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2021-05-09T02:34:52.517Z
@ProfRob I'm very very dubious about your statement. Do you have a reference to support that?  Spectrographs of low magnitude objects are invariably starved of photons. No-one would intentionally throw any away. I don't know the technology at all, but I can imagine there might be a physical slit involved in trimming an optically shaped beam. In that case a few photons might not make it through - rather like illuminating a parabolic antenna/mirror and some photons fall off the edge..
Roger Wood
CC BY-SA 4.0
2021-05-09T07:18:51.197Z
@RogerWood you're dubious? 
Here are two links for 8-m telescopes, though why you couldn't find this I don't know. https://www.gemini.edu/instrumentation/gmos https://www.eso.org/sci/facilities/paranal/instruments/uves/inst.html
ProfRob
CC BY-SA 4.0
2021-05-09T07:24:35.990Z
@RogerWood if you type "spectrograph slit" into Google you'll find plenty of other stuff to read. Slitless spectrographs do exist. They either have fibers defining the entrance (and light is lost when light enters the fiber) or they are used in a slitless mode to do spectrophotometry.
ProfRob
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2021-05-09T20:00:05.940Z
@ProfRob Thanks, That's helpful. I've read up some of the references now. But my understanding remains the same.  The slit is placed in the image plane to remove objects above and below (perpendicular to) the slit whose spectra might overlap with the target object. Ideally the slit will admit ALL the light from the target and NONE from the unwanted neighbors. I'm sure there are many circumstances when this ideal cannot be achieved. http://slittlefair.staff.shef.ac.uk/teaching/phy217/lectures/instruments/L16/index.html
Roger Wood
CC BY-SA 4.0
2021-05-09T20:28:56.200Z
"The second function of the slit is to provide stable spectral resolution. This can be understood by noting that a spectrum is essentially an infinite number of images of the telescope focal plane, each shifted slightly in wavelength. Hence, without a slit present, the spectral resolution of an object would be defined by the width of the object. For a point source like a star the would be defined by the seeing. However, since seeing varies with time, the spectral resolution would then vary with time." at least read the notes you are quoting at me.
ProfRob
CC BY-SA 4.0
2021-05-09T20:30:47.190Z
.It isn't "ideal" to admit all the light because you make your spectral resolution worse and time variable. It is positively disastrous for radial velocity measurements. That's why almost **all** astronomical spectrographs use a slit or a fiber (and I've observed with plenty of them).  @RogerWood. ps Contrary to what Stuart Littlefair says, it isn't just the seeing width, the longer term differential image motion and tracking errors would destroy the spectral resolution even more
ProfRob
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2021-05-10T01:34:48.067Z
@ProfRob The spectral resolution of a point source will highest when all the light is admitted and the resolution is determined by the primary.  Theoretically, passing the light through an unneccesarily narrow slot will both lose photons and make the spectral resolution worse!  But I can understand that if the image is dancing around, any motion perpendicular to the slot would also appear as a loss of resolution. Then, as your say, I can well believe it's better to compromise and include a slot.   BTW, I appreciate your taking the time to educate me!
Roger Wood
CC BY-SA 4.0
2021-05-10T06:42:48.957Z
@RogerWood that isn't correct.
ProfRob
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2021-05-10T20:31:01.517Z
@ProfRob. Yes, thanks.  Stopping down the slot will not degrade spectral resolution, since the slot is in the image plane.
Roger Wood
<telescope><light><spectroscopy><wavelength><instruments>
What happens to the Fourier Transform of the Measured Spectrum?
Jay D
# What happens to the Fourier Transform of the Measured Spectrum?

(Asked by: username_0 on 2021-03-31)

What happens when you double the width of the entrance slit of a spectrograph, in turn letting in twice as much light, would there be an improvement in the Fourier transform of the measured spectral profile d(f) following to higher frequency f?

**Tags:** telescope, light, spectroscopy, wavelength, instruments
> **Comments:**
> **username_1**: The slit is a cut in the transverse spatial domain, the spectrometer (assuming a dispersive-type (e.g. grating or prism)) spreads wavelength into the transverse spatial domain. Bigger slit smears out the spectrum, so no, there would be the opposite of an improvement.  
> **username_0**: Thanks for the reply. @uhoh ; would this mean a more noise towards higher frequencies in the Fourier domain then?  
> **username_1**: Well I think it is best if you first add more information. Are you talking about a conventional spectrometer that uses a diffraction grating, or a Fourier Transform spectrometer? Is the noise you are talking about just statistical (e.g. number of photons) or is there additional noise in the light source of some kind, or is it detector noise? Presumably data will be less noisy if you have a bigger aperture and more light and the resolution will be poorer, but it's hard to make more general statements without additional details.  
> **username_0**: A conventional one using a diffraction grating; the noise would come from the background  like heat, noise, stray light, electronic noise etc.  
> **username_2**: @Jay D I don't know how these things are actually constructed, but I'd be very surprised if a spectrograph for a telescope didn't already utilize all the available light from a star.  So I'm not sure where 'slot-width' would come in.  
> **username_3**: What do you mean by an improvement to the FT?  
> **username_3**: @RogerWood almost all spectrographs have slits defining the illumination of the dispersive element (and as a result, do not use all the available light from a star).  
> **username_2**: @ProfRob I'm very very dubious about your statement. Do you have a reference to support that?  Spectrographs of low magnitude objects are invariably starved of photons. No-one would intentionally throw any away. I don't know the technology at all, but I can imagine there might be a physical slit involved in trimming an optically shaped beam. In that case a few photons might not make it through - rather like illuminating a parabolic antenna/mirror and some photons fall off the edge..  
> **username_3**: @RogerWood you're dubious?  Here are two links for 8-m telescopes, though why you couldn't find this I don't know. https://www.gemini.edu/instrumentation/gmos https://www.eso.org/sci/facilities/paranal/instruments/uves/inst.html  
> **username_3**: @RogerWood if you type "spectrograph slit" into Google you'll find plenty of other stuff to read. Slitless spectrographs do exist. They either have fibers defining the entrance (and light is lost when light enters the fiber) or they are used in a slitless mode to do spectrophotometry.  
> **username_2**: @ProfRob Thanks, That's helpful. I've read up some of the references now. But my understanding remains the same.  The slit is placed in the image plane to remove objects above and below (perpendicular to) the slit whose spectra might overlap with the target object. Ideally the slit will admit ALL the light from the target and NONE from the unwanted neighbors. I'm sure there are many circumstances when this ideal cannot be achieved. http://slittlefair.staff.shef.ac.uk/teaching/phy217/lectures/instruments/L16/index.html  
> **username_3**: "The second function of the slit is to provide stable spectral resolution. This can be understood by noting that a spectrum is essentially an infinite number of images of the telescope focal plane, each shifted slightly in wavelength. Hence, without a slit present, the spectral resolution of an object would be defined by the width of the object. For a point source like a star the would be defined by the seeing. However, since seeing varies with time, the spectral resolution would then vary with time." at least read the notes you are quoting at me.  
> **username_3**: .It isn't "ideal" to admit all the light because you make your spectral resolution worse and time variable. It is positively disastrous for radial velocity measurements. That's why almost **all** astronomical spectrographs use a slit or a fiber (and I've observed with plenty of them).  @RogerWood. ps Contrary to what Stuart Littlefair says, it isn't just the seeing width, the longer term differential image motion and tracking errors would destroy the spectral resolution even more  
> **username_2**: @ProfRob The spectral resolution of a point source will highest when all the light is admitted and the resolution is determined by the primary.  Theoretically, passing the light through an unneccesarily narrow slot will both lose photons and make the spectral resolution worse!  But I can understand that if the image is dancing around, any motion perpendicular to the slot would also appear as a loss of resolution. Then, as your say, I can well believe it's better to compromise and include a slot.   BTW, I appreciate your taking the time to educate me!  
> **username_3**: @RogerWood that isn't correct.  
> **username_2**: @ProfRob. Yes, thanks.  Stopping down the slot will not degrade spectral resolution, since the slot is in the image plane.  

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## 0 answers

