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What is the faintest "source" that can be spectroscopically analysed
via a telescope for fraunhofer lines and elemental composition? I know bright galaxies and quasars produce ample quantities of light for spectroscopy, but surely the multiple stellar make-up of these objects produces meaningless 'noise' at that level... Ta. Abdul Ahad |
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Ioannis wrote:
? "Abdul Ahad" ?????? ??? ?????? om... What is the faintest "source" that can be spectroscopically analysed via a telescope for fraunhofer lines and elemental composition? I see no reason why it shouldn't be precisely the same magnitude limit that exists for the larger scopes. If it can be photographed, it can be spectrographed. You're going to need a good few photons to be able to remove the uncertainty as to peaks and troughs in the wavelength profile. In this recent story: http://www.astronomynow.com/news/040...t_galaxy.shtml I imagine they had enough photons to calculate an approximate blackbody curve to determine the redshift, but I very much doubt they could tell you what the composition of the galaxy was! -- Andrew Urquhart Reply: www.andrewu.co.uk/about/contact/ |
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In sci.astro.amateur, Abdul Ahad wrote:
What is the faintest "source" that can be spectroscopically analysed via a telescope for fraunhofer lines and elemental composition? How big is the telescope? What is the throughput and resolution of the spectrograph, the QE of the detector, and how long are you willing to integrate on the target? What is the FWHM of the seeing, and how bad is the light pollution or sky background in your wavelength range? It's not a simple question. I know bright galaxies and quasars produce ample quantities of light for spectroscopy, but surely the multiple stellar make-up of these objects produces meaningless 'noise' at that level... In most cases you can't resolve individual stars, so a spectrum of the galaxy consists of the blended contribution from various types of stars. That's not "noise" though. Quasar spectra are quite different from stars, and it is often simple to distinguish the two sources and subtract out the stellar contribution to the spectra. Dave David Whysong DWhysong (at) physics (dot) ucsb (dot) edu |
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In sci.astro.amateur, Ioannis wrote:
"Abdul Ahad" What is the faintest "source" that can be spectroscopically analysed via a telescope for fraunhofer lines and elemental composition? I see no reason why it shouldn't be precisely the same magnitude limit that exists for the larger scopes. Spectrographs usually have much less throughput than cameras; there are losses due to the grating (or grism) and re-imaging optics. Furthermore, the light is spread over many pixels, each of which has read noise and dark current. So you do lose substantial SNR with a spectrograph, compared to imaging. Dave David Whysong DWhysong (at) physics (dot) ucsb (dot) edu |
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In message , Ioannis
writes ? "Abdul Ahad" ?????? ??? μ???μ? . com... What is the faintest "source" that can be spectroscopically analysed via a telescope for fraunhofer lines and elemental composition? A devious answer to this is that we can analyse clouds of hydrogen gas that happen to be in the line of sight between us and a distant quasar no matter how faint the cloud itself may be. The light from the quasar allows us to see the composition and redshift of the intervening gas cloud. I see no reason why it shouldn't be precisely the same magnitude limit that exists for the larger scopes. If it can be photographed, it can be spectrographed. You have it exactly backwards. Anything that a spectrograph will work on will be plenty bright enough to photograph. The main reason the biggest scopes are in demand is to feed enough light into spectrographs. Imaging a star requires obtaining enough signal to noise on a handful of pixels. Imaging a spectrum can require a few thousand times more SNR. Try to make a spectrum with 0.1nm resolution across the visible to look in detail at Fraunhofer lines and that same amount of starlight is spread across 4000 pixels with consequent loss of limiting magnitude (roughly about 9 mags ignoring other secondary factors). Fortunately you don't always need very high resolution spectra to see interesting things - some of the wilder active objects have bright emission lines that stand out even at relatively low dispersion. There are amateur spectra of some brighter quasars and Seyferts about. Regards, -- Martin Brown |
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On Thu, 04 Mar 2004 07:19:13 GMT, David Whysong
wrote: In sci.astro.amateur, Ioannis wrote: "Abdul Ahad" What is the faintest "source" that can be spectroscopically analysed via a telescope for fraunhofer lines and elemental composition? I see no reason why it shouldn't be precisely the same magnitude limit that exists for the larger scopes. Spectrographs usually have much less throughput than cameras; there are losses due to the grating (or grism) and re-imaging optics. Furthermore, the light is spread over many pixels, each of which has read noise and dark current. So you do lose substantial SNR with a spectrograph, compared to imaging. SNR = signal to noise ratio Dave David Whysong DWhysong (at) physics (dot) ucsb (dot) edu |
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On Thu, 04 Mar 2004 07:16:29 GMT, David Whysong
wrote: In sci.astro.amateur, Abdul Ahad wrote: What is the faintest "source" that can be spectroscopically analysed via a telescope for fraunhofer lines and elemental composition? How big is the telescope? What is the throughput and resolution of the spectrograph, the QE of the detector, and how long are you willing to integrate on the target? What is the FWHM of the seeing, and how bad is the light pollution or sky background in your wavelength range? QE= Quantum Efficiency the energy needed to split off measurable electrons. with perfect efficiency, one photon would split off one electron. there's inherent inefficiency in this transfer based on work function and black body properties of the detector material, amongst other reasons. FWHM = Full Width at Half Maximum it applies to a well shaped peak on a graph. its the point halfway up to the maximum of the peak at the theoretical fattest part of that peak. its the value of the weighted average of the peak. It's not a simple question. I know bright galaxies and quasars produce ample quantities of light for spectroscopy, but surely the multiple stellar make-up of these objects produces meaningless 'noise' at that level... In most cases you can't resolve individual stars, so a spectrum of the galaxy consists of the blended contribution from various types of stars. That's not "noise" though. Quasar spectra are quite different from stars, and it is often simple to distinguish the two sources and subtract out the stellar contribution to the spectra. Dave David Whysong DWhysong (at) physics (dot) ucsb (dot) edu |
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Thread | Thread Starter | Forum | Replies | Last Post |
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