# Optimal telescope size?

Consider a diffraction-limited telescope with unobstructed aperture $D$. Such a scope is capable of yielding an angular resolution $\alpha$ that scales as $\lambda/D$, with $\lambda$ denoting the wavelength of the light. However, in reality such a telescope will need to look through a turbulent atmosphere, the refractive index variations of which will cause wavefront errors with a spatial correlation length (Fried parameter) $r_0$. As a result, no stable diffraction pattern but rather a speckle pattern will form. I am interested in the angular resolution corresponding to the average size of this speckle pattern.

What is this angular resolution $\alpha(D/\lambda,r_0/\lambda)$?

More specifically: how does $\alpha$ scale for $D\gg r_0$, and for fixed $r_0$: is there an optimal (highest resolution) aperture $D$? Or more down-to-earth: If I built a 200 inch Hale telescope in my backyard*, would it beat a 10 inch amateur telescope in terms of optical resolution?

*You may assume my backyard to have a typical seeing characterized by a Fried parameter $r_0$ of about 4 inch.

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Including adaptive optics would render the exercise futile. The whole purpose of adaptive optics is to obtain a diffraction-limited resolution $~\lambda/D$ regardless the value of $D/r_0$. –  Johannes May 1 '13 at 15:08
Okay! If $D > r_0$, the resolution will be worse than the theoretical best($D/\lambda$). So, what's the catch? Why the 150 bounty? You are asking what if $D >> r_0$. The answer is simply, a super worse resolution limited by seeing. And to the telescope thing...No, you would not. I am not posting this as an answer since I am making sure whether I get the question or not. –  Cheeku Jul 15 '13 at 23:29
@Cheeku - Your answer doesn't address the specific behavior for $D >> r_0$. Sure, the resolution will be less than the theoretical limit, but will it still be a monotonical function of $D/r_0$? Will it level off? Or will it go thru an optimal resolution for a specific finite value of $D/r_0$? –  Johannes Jul 16 '13 at 2:52
I'm not convinced that there's a universal optimum regardless of how one carries out the imaging and post-processes the images. For fixed values of $D$ and $r_0$, doesn't $\alpha$ depend on the length of the exposure? In a longer exposure, I'd think there would be at least some averaging out of the turbulence effect, whereas averaging doesn't help at all with diffraction. In reality, I think people take a long series of images, pick out the ones that were taken at moments of unusually good seeing, and then "stack" (i.e., average) those. –  Ben Crowell Jul 16 '13 at 15:58
@Ben Crowell -- to arrive at an unambiguous angular resolution relationship, you may take the limit of exposure times going to zero. The typical time scale for 'star twinkle' is about 0.1 s. so any exposure well below that value will do. (Btw. For the same reason the exposures in the 'lucky imaging' you describe are also typically short.) –  Johannes Jul 16 '13 at 16:45

The seeing is a much more natural way of thinking about the effect of the atmosphere. A seeing of 1" (an arcsecond) is good.

If your telescope is diffraction limited, then your angular resolution is $1.22 \lambda / D$. A reasonable limit to the size of your telescope would be to set the seeing disk FWHM equal to your angular resolution, which for the best case (blue light, $\lambda \approx 400 ~\text{nm}$) would give you a diameter of 4 inches (10 cm), and in the worst case (red light, $\lambda \approx 700 ~\text{nm}$) a diameter of 7 inches (17.6 cm). So, if you build a backyard telescope without any adaptive optics, you only need to build a 7 inch diameter telescope to achieve maximum angular resolution!

Why then did people build bigger ground telescopes before adaptive optics were invented? Because larger telescopes can collect more light! Building a bigger backyard telescope won't get you higher resolution, but it will help you to see dimmer sources.

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From the information provided it seems that for fixed seeing (fixed $r_0$) optimal resolution is achieved for apertures $D$ such that $D/r_0$ reaches values close to 2. In other words, under any realistic atmospheric conditions and in terms of angular resolution, a 20" telescope would beat a 200" telescope by a large margin.