<p>I'm going to make the assumption that the &quot;aperture&quot; of the instrument refers to the width of a fibre that feeds a spectrograph (I can't think of any other plausible scenario).</p>
<p>What you want is to enclose the most power you can within a 2 arcsec diameter in the telescope focal plane <em>at the wavelengths of interest</em>. One confounding effect is the seeing. If 80% of the energy is enclosed within 1 arcsec diameter, then it is a reasonable assumption that you almost get all of it (<span class="math-container">$&gt;95$</span>%) within a 2 arcsec fibre.</p>
<p>The above consideration applies to light of <em>any</em> single, monochromatic wavelength. If you wish to observe a spectrum over a range of wavelengths, then you must take account of atmospheric dispersion. Light of different wavelengths is refracted by the atmosphere and thus light of different wavelengths is brought to a different focal point depending on its wavelength. The amplitude of this differential displacement in the focal plane depends on how much atmosphere is in the way and so increases with zenith distance.</p>
<p>The graph in your question shows you what that displacement is (in arcseconds) with respect to light at 5000A. What this means is that the <em>range</em> of wavelengths you can get successfully into your 2 arcsec fibre will depend on the central wavelength and the zenith distance.</p>
<p>For example, trying to get a spectrum between the atmospheric cut-off at 3200A and 5000A in a single observation will not be possible above a zenith distance if 50 degrees because the centroid of the images at these two wavelengths differ by more than 2 arcsec. And in fact you will start to lose signal before that because the displacement in the graph is just of the image centroid and obviously if that centroid gets near the edge of the aperture then the blurring effects of seeing will lead to light loss.</p>
<p>If instead the &quot;aperture&quot; refers to a slit width on a more traditional spectrograph, then you can get around this dispersion problem by rotating the slit to be at the parallactic angle, so that the slit lies along the dispersed image of the star and so collects light at all wavelengths (though they will appear in different pixels on the camera).</p>
<p>If you are doing imaging the effects of atmospheric dispersion will depend on the bandwidth of your filter and the zenith distance. With a narrow band filter the size of your images will be limited by the seeing. If you are using broadband filters, especially U and B, then your images can become vertically elongated at large zenith distances due to the differential dispersion.</p>
<p>A solution is to introduce an atmospheric dispersion corrector prior to the focal plane, which reverses the calculated effects of atmospheric dispersion (at the expense of some loss of signal and image quality).</p>

CC BY-SA 4.0
2021-01-10T22:15:45.993Z
2021-01-10T22:15:45.993Z
ProfRob
<p>@ProfRob 's answer helped me understand what is going on, but I would like to add something.</p>
<p>If the telescope <em>pointing is optimized for 5000 angstrom</em> the aperture of 2&quot; will completely contain an area of 1&quot; that corresponds to 80% of the energy. If you keep the telescope pointing fixed, but trace the light at 3200 angstrom the &quot;disk&quot; of 1&quot; diameter will shift down due to atmospheric dispersion. As long as this 1&quot; disk stays inside the 2&quot; that you observe with that telescope pointing, you will catch 80% of the 3200 angstrom light.
So what you need to calculate is how far you can maximally displace the center of the 1&quot; disk for it to remain inside the larger disk. This will be when the small circle touches the big circle on the inside, which means shifting the center down by half the diameter of the smallest circle, i.e. 0.5&quot;. I hope this makes sense...</p>
<p>In the plot, -0.5&quot; corresponds to a zenith distance of 20°. If you'd like to determine how many correctors (ADC) you need (in a wheel, not all of them in series) at a certain zenith distance, e.g. 70°, while keeping the loss below 20%, you need to find how many of the 1&quot; disks have shifted outside the 2&quot; disk for that telescope pointing that is optimized at 5000 angstrom. At 70° you would only need to correct for 4 of the wavelengths shown in the plot.</p>
<p><a href="https://i.sstatic.net/1naQY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1naQY.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-01-11T13:47:05.367Z
That's not quite right. You can point the telescope so your seeing disk at 5000 A is at the top of the grey circle you have drawn and its centre would be separated by 1 arcsec from the centre of the blue circle you have drawn. Both interior circles would then contain 80% of the flux. (This is if you wanted to simultaneously image at 3200 and 5000A.) Also, there is usually only one ADC and you either have it in or out.
ProfRob
CC BY-SA 4.0
2021-01-11T14:19:34.687Z
@ProfRob very true, but I wouldn't say that scenario is "optimized" for 5000 A. Anyway, thank you very much!
theWrongAlice
CC BY-SA 4.0
2021-01-11T12:59:18.593Z
2021-01-11T14:15:26.737Z
2021-01-11T14:15:26.737Z
theWrongAlice
<p>I'm trying to understand energy losses due to atmospheric dispersion, given the plot below from the Keck telescope.
For a typical seeing I'm told that there's 80% encircled energy in 1&quot;. The aperture of the instrument is 2&quot;. The plot shows the dispersion as a function of zenith distance for different wavelengths.</p>
<p><em>What is the maximum zenith distance with &lt; 20% loss at 3200 angstrom due to atmospheric dispersion?</em></p>
<p>I'm not used to thinking about aperture in arcseconds, but I suppose it's bigger than the seeing, usually? Also, the fact that 80% of the energy is encircled in 1&quot;, does that imply that you catch 100% with an aperture of 2&quot; or am I overlooking things?
I don't really know how to use this plot. At a certain fixed zenith distance, if for example the y-value is -3, does that mean the seeing increases with 3&quot; (e.g. if seeing was 1&quot; without dispersion, would it be 4&quot; with dispersion)? Or does it not make sense to think about seeing &quot;without&quot; atmospheric dispersion?</p>
<p><a href="https://i.sstatic.net/brCoI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/brCoI.png" alt="Keck" /></a></p>

CC BY-SA 4.0
2021-01-10T17:44:29.443Z
What do you mean by the "aperture of the instrument"? Which instrument? Is this a fiber spectrograph?
ProfRob
CC BY-SA 4.0
2021-01-10T17:47:04.233Z
The graph is presumably showing the displacement of the image of a point source as a function of wavelength and zenith distance due to refraction in the atmosphere. It doesn't have anything to do with seeing, other than that both effects can blur an image.
ProfRob
CC BY-SA 4.0
2021-01-10T18:14:36.603Z
@ProfRob the aperture is the primary mirror, I guess? I think what is meant is "telescope" and not so much "instrument".
theWrongAlice
CC BY-SA 4.0
2021-01-10T18:30:01.053Z
The aperture of the Keck telescopes are 10 metres. You need to provide more details with your question
ProfRob
CC BY-SA 4.0
2021-01-10T18:44:16.840Z
@ProfRob the question isn't specifically about Keck. It's a hypothetical and quite generic telescope. It should be possible to answer with the information that was given.
theWrongAlice
<observational-astronomy><telescope><atmospheric-effects><dispersion>
Understanding energy loss in a telescope aperture due to atmospheric dispersion
theWrongAlice
# Understanding energy loss in a telescope aperture due to atmospheric dispersion

(Asked by: username_0 on 2021-01-10)

I'm trying to understand energy losses due to atmospheric dispersion, given the plot below from the Keck telescope.
For a typical seeing I'm told that there's 80% encircled energy in 1". The aperture of the instrument is 2". The plot shows the dispersion as a function of zenith distance for different wavelengths.

*What is the maximum zenith distance with < 20% loss at 3200 angstrom due to atmospheric dispersion?*

I'm not used to thinking about aperture in arcseconds, but I suppose it's bigger than the seeing, usually? Also, the fact that 80% of the energy is encircled in 1", does that imply that you catch 100% with an aperture of 2" or am I overlooking things?
I don't really know how to use this plot. At a certain fixed zenith distance, if for example the y-value is -3, does that mean the seeing increases with 3" (e.g. if seeing was 1" without dispersion, would it be 4" with dispersion)? Or does it not make sense to think about seeing "without" atmospheric dispersion?

[![Keck](https://i.sstatic.net/brCoI.png)](https://i.sstatic.net/brCoI.png)

**Tags:** observational-astronomy, telescope, atmospheric-effects, dispersion
> **Comments:**
> **username_1**: What do you mean by the "aperture of the instrument"? Which instrument? Is this a fiber spectrograph?  
> **username_1**: The graph is presumably showing the displacement of the image of a point source as a function of wavelength and zenith distance due to refraction in the atmosphere. It doesn't have anything to do with seeing, other than that both effects can blur an image.  
> **username_0**: @ProfRob the aperture is the primary mirror, I guess? I think what is meant is "telescope" and not so much "instrument".  
> **username_1**: The aperture of the Keck telescopes are 10 metres. You need to provide more details with your question  
> **username_0**: @ProfRob the question isn't specifically about Keck. It's a hypothetical and quite generic telescope. It should be possible to answer with the information that was given.  

---

## 2 answers

### Answer by username_1 on 2021-01-10. Score: 2 | Accepted answer

I'm going to make the assumption that the "aperture" of the instrument refers to the width of a fibre that feeds a spectrograph (I can't think of any other plausible scenario).

What you want is to enclose the most power you can within a 2 arcsec diameter in the telescope focal plane *at the wavelengths of interest*. One confounding effect is the seeing. If 80% of the energy is enclosed within 1 arcsec diameter, then it is a reasonable assumption that you almost get all of it ($>95$%) within a 2 arcsec fibre.

The above consideration applies to light of *any* single, monochromatic wavelength. If you wish to observe a spectrum over a range of wavelengths, then you must take account of atmospheric dispersion. Light of different wavelengths is refracted by the atmosphere and thus light of different wavelengths is brought to a different focal point depending on its wavelength. The amplitude of this differential displacement in the focal plane depends on how much atmosphere is in the way and so increases with zenith distance.

The graph in your question shows you what that displacement is (in arcseconds) with respect to light at 5000A. What this means is that the *range* of wavelengths you can get successfully into your 2 arcsec fibre will depend on the central wavelength and the zenith distance.

For example, trying to get a spectrum between the atmospheric cut-off at 3200A and 5000A in a single observation will not be possible above a zenith distance if 50 degrees because the centroid of the images at these two wavelengths differ by more than 2 arcsec. And in fact you will start to lose signal before that because the displacement in the graph is just of the image centroid and obviously if that centroid gets near the edge of the aperture then the blurring effects of seeing will lead to light loss.

If instead the "aperture" refers to a slit width on a more traditional spectrograph, then you can get around this dispersion problem by rotating the slit to be at the parallactic angle, so that the slit lies along the dispersed image of the star and so collects light at all wavelengths (though they will appear in different pixels on the camera).

If you are doing imaging the effects of atmospheric dispersion will depend on the bandwidth of your filter and the zenith distance. With a narrow band filter the size of your images will be limited by the seeing. If you are using broadband filters, especially U and B, then your images can become vertically elongated at large zenith distances due to the differential dispersion.

A solution is to introduce an atmospheric dispersion corrector prior to the focal plane, which reverses the calculated effects of atmospheric dispersion (at the expense of some loss of signal and image quality).


---

### Answer by username_0 on 2021-01-11. Score: 1

@ProfRob 's answer helped me understand what is going on, but I would like to add something.

If the telescope *pointing is optimized for 5000 angstrom* the aperture of 2" will completely contain an area of 1" that corresponds to 80% of the energy. If you keep the telescope pointing fixed, but trace the light at 3200 angstrom the "disk" of 1" diameter will shift down due to atmospheric dispersion. As long as this 1" disk stays inside the 2" that you observe with that telescope pointing, you will catch 80% of the 3200 angstrom light.
So what you need to calculate is how far you can maximally displace the center of the 1" disk for it to remain inside the larger disk. This will be when the small circle touches the big circle on the inside, which means shifting the center down by half the diameter of the smallest circle, i.e. 0.5". I hope this makes sense...

In the plot, -0.5" corresponds to a zenith distance of 20°. If you'd like to determine how many correctors (ADC) you need (in a wheel, not all of them in series) at a certain zenith distance, e.g. 70°, while keeping the loss below 20%, you need to find how many of the 1" disks have shifted outside the 2" disk for that telescope pointing that is optimized at 5000 angstrom. At 70° you would only need to correct for 4 of the wavelengths shown in the plot.

[![enter image description here](https://i.sstatic.net/1naQY.png)](https://i.sstatic.net/1naQY.png)

> **Comments:**
> **username_1**: That's not quite right. You can point the telescope so your seeing disk at 5000 A is at the top of the grey circle you have drawn and its centre would be separated by 1 arcsec from the centre of the blue circle you have drawn. Both interior circles would then contain 80% of the flux. (This is if you wanted to simultaneously image at 3200 and 5000A.) Also, there is usually only one ADC and you either have it in or out.  
> **username_0**: @ProfRob very true, but I wouldn't say that scenario is "optimized" for 5000 A. Anyway, thank you very much!  

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