Understanding energy loss in a telescope aperture due to atmospheric dispersion 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.
What is the maximum zenith distance with &lt; 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&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?
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).
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 ($&gt;95$%) 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 &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).
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).