<p>Your question is a bit unclear. Compare in what sense?</p>

<p>The problem boils down to that you have a fixed number of photons in a wavelength band and you can choose, with the appropriate filters or dispersion elements, how many "bins" to divide these photons into. </p>

<p>To accumulate the same total number of photons with the same telescope will take the same amount of time, except putting more optical elements in the way will reduce the efficiency of the instrument. Filters are more efficient than spectrographs by factors of a few to an order of magnitude.</p>

<p>However a further factor to consider, is that although in principle we could divide the light up as finely as we want, there may be static sources of noise that are independent of the signal. In the case of a CCD detector, this means the read noise.</p>

<p>If you want read noise <em>not</em> to limit the observations, then the product of the bin size and exposure time needs to be large enough that the square root of the accumulated counts (including any sky background) in a bin exceeds the read noise.</p>

<p>An example might help. Suppose I have a star that I observe for 10s through a V-band filter with a width of 100 nm and it produces 1000 detected photons on a CCD, where the star image effectively illuminates 10 pixels, each of which has a read noise of 10 counts. The total read noise is <span class="math-container">$10\sqrt{10} = \sqrt{1000}$</span> counts (the square root of the total variance) and equals the Poisson noise from the source plus sky. Ignoring the sky, the signal to noise ratio is <span class="math-container">$1000/\sqrt{1000+1000}=22$</span>.</p>

<p>Now suppose I do slitless spectroscopy, passing the starlight through a diffraction grating, splitting the light into 1 nm bins. My signal strength in each bin is divided by 100, but the read noise is approximately the same. To obtain a spectrum with a similar signal to noise ratio you would need to expose for 100 times longer, i.e.1000 s (probably longer because a diffraction grating is not as efficient as a filter).</p>

<p>So there is no general answer to your question, it depends on the spectrograph efficiency, the "resolution" of the spectrograph, the relative contributions of source, sky and read noise and the signal to noise ratio desired.</p>

<p><strong>A simple rule of thumb to get an exposure that is just limited by read noise for an imaging filter of wavelength width <span class="math-container">$w$</span> and a spectrograph with a resolution element <span class="math-container">$\Delta \lambda$</span>, would be to observe for a factor of <span class="math-container">$w/\Delta \lambda$</span> longer, with an extra factor of a few to account for poorer spectrograph efficiency.</strong></p>

<p>Edit:
To test my ideas I go to the Isaac Newton group of telescopes' signal to noise calculator.<a href="http://catserver.ing.iac.es/signal/" rel="nofollow noreferrer">http://catserver.ing.iac.es/signal/</a> </p>

<p>I try observing a 20th magnitude star for 10s with a V filter on the WHT prime focus, assuming 1 arcsec seeing and no moon. This gives me a SNR of 46.</p>

<p>To get the same SNR in a low resolution spectrum withe ISIS spectrograph and 158R grating (lowest resolution available) takes 15,000 s.</p>

<p>Does this stack up with what I said above? Each pixel in the spectrum is 0.16 nm. If the V band filter is about 100nm wide, my crude formula above suggested an increase in exposure time of 100/0.16 = 625. But the instrument throughput is only 33% (compared with essentially 100% for the filter) so we have to add another factor 3, to get a required exposure that is 1875 times as long as for the photometry. Pretty close!</p>

CC BY-SA 4.0
2019-06-12T21:11:16.643Z
Fair comment. I'm just interested in highlighting the long exposure times necessary to get a spectrum of a dim object, compared to getting a photometric magnitude (say, in V band) of the same object. I appreciate that the situation is not that simple, but are there any resources out there that do a good job of comparing how long it takes to get a spectrum with how long it takes to get photo data?
Jim421616
CC BY-SA 4.0
2019-06-13T06:26:49.993Z
@Jim421616 why can't you work out what you want from the last paragraph of my answer? "Get a spectrum" is incredibly vague. At what resolution?
ProfRob
CC BY-SA 4.0
2019-06-13T09:16:53.903Z
I wrote that comment before you added your edits, I think (unless I just missed them). Your edits do help me answer my question, thank you. I'll mark as answered.
Jim421616
CC BY-SA 4.0
2019-06-12T07:01:39.020Z
2019-06-13T06:45:39.850Z
2019-06-13T06:45:39.850Z
ProfRob
<p>For a report I'm writing, I'm interested in comparing the exposure time required for spectroscopy versus photometry for a given magnitude. I found the figures on pages 7 and 8 of <a href="http://www.astro.wisc.edu/~sheinis/ast500/AY500_lect5.ppt.pdf" rel="nofollow noreferrer">this lecture</a> particularly informative. I tried emailing the author, but his email address is inactive. Where can I get similar information? </p>

<observational-astronomy><spectroscopy><photometry>
Exposure time for spectroscopy compared to photometry
Jim421616
# Exposure time for spectroscopy compared to photometry

(Asked by: username_0 on 2019-06-12)

For a report I'm writing, I'm interested in comparing the exposure time required for spectroscopy versus photometry for a given magnitude. I found the figures on pages 7 and 8 of [this lecture](http://www.astro.wisc.edu/~sheinis/ast500/AY500_lect5.ppt.pdf) particularly informative. I tried emailing the author, but his email address is inactive. Where can I get similar information?

**Tags:** observational-astronomy, spectroscopy, photometry


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## 1 answers

### Answer by username_1 on 2019-06-12. Score: 3 | Accepted answer

Your question is a bit unclear. Compare in what sense?

The problem boils down to that you have a fixed number of photons in a wavelength band and you can choose, with the appropriate filters or dispersion elements, how many "bins" to divide these photons into.

To accumulate the same total number of photons with the same telescope will take the same amount of time, except putting more optical elements in the way will reduce the efficiency of the instrument. Filters are more efficient than spectrographs by factors of a few to an order of magnitude.

However a further factor to consider, is that although in principle we could divide the light up as finely as we want, there may be static sources of noise that are independent of the signal. In the case of a CCD detector, this means the read noise.

If you want read noise *not* to limit the observations, then the product of the bin size and exposure time needs to be large enough that the square root of the accumulated counts (including any sky background) in a bin exceeds the read noise.

An example might help. Suppose I have a star that I observe for 10s through a V-band filter with a width of 100 nm and it produces 1000 detected photons on a CCD, where the star image effectively illuminates 10 pixels, each of which has a read noise of 10 counts. The total read noise is $10\sqrt{10} = \sqrt{1000}$ counts (the square root of the total variance) and equals the Poisson noise from the source plus sky. Ignoring the sky, the signal to noise ratio is $1000/\sqrt{1000+1000}=22$.

Now suppose I do slitless spectroscopy, passing the starlight through a diffraction grating, splitting the light into 1 nm bins. My signal strength in each bin is divided by 100, but the read noise is approximately the same. To obtain a spectrum with a similar signal to noise ratio you would need to expose for 100 times longer, i.e.1000 s (probably longer because a diffraction grating is not as efficient as a filter).

So there is no general answer to your question, it depends on the spectrograph efficiency, the "resolution" of the spectrograph, the relative contributions of source, sky and read noise and the signal to noise ratio desired.

**A simple rule of thumb to get an exposure that is just limited by read noise for an imaging filter of wavelength width $w$ and a spectrograph with a resolution element $\Delta \lambda$, would be to observe for a factor of $w/\Delta \lambda$ longer, with an extra factor of a few to account for poorer spectrograph efficiency.**

Edit:
To test my ideas I go to the Isaac Newton group of telescopes' signal to noise calculator.<http://catserver.ing.iac.es/signal/>

I try observing a 20th magnitude star for 10s with a V filter on the WHT prime focus, assuming 1 arcsec seeing and no moon. This gives me a SNR of 46.

To get the same SNR in a low resolution spectrum withe ISIS spectrograph and 158R grating (lowest resolution available) takes 15,000 s.

Does this stack up with what I said above? Each pixel in the spectrum is 0.16 nm. If the V band filter is about 100nm wide, my crude formula above suggested an increase in exposure time of 100/0.16 = 625. But the instrument throughput is only 33% (compared with essentially 100% for the filter) so we have to add another factor 3, to get a required exposure that is 1875 times as long as for the photometry. Pretty close!

> **Comments:**
> **username_0**: Fair comment. I'm just interested in highlighting the long exposure times necessary to get a spectrum of a dim object, compared to getting a photometric magnitude (say, in V band) of the same object. I appreciate that the situation is not that simple, but are there any resources out there that do a good job of comparing how long it takes to get a spectrum with how long it takes to get photo data?  
> **username_1**: @username_0 why can't you work out what you want from the last paragraph of my answer? "Get a spectrum" is incredibly vague. At what resolution?  
> **username_0**: I wrote that comment before you added your edits, I think (unless I just missed them). Your edits do help me answer my question, thank you. I'll mark as answered.  

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