I've studied this in a few optics classes before, but I'm drawing a blank at the moment. If I know the aperture size and focal length of a telescope, as well as how far away the object I want to view is, can I calculate how large/wide/big that object must be in order to be seen?

  • $\begingroup$ I would assume that albedo matters just as much as size. $\endgroup$ – HDE 226868 Jul 28 '15 at 19:42
  • $\begingroup$ @HDE226868 What's the albedo of the sun? $\endgroup$ – Michael Sep 16 '17 at 20:45
  • $\begingroup$ I was attempting to suggest it in the context of objects that don't emit their own light, @Michael. I should have provided that context. $\endgroup$ – HDE 226868 Sep 17 '17 at 18:17

A few clarifications.

Telescopes in general operate at very large distances - "at infinity" is the term used in optics parlance.

A bright enough object can be seen from any distance, no matter what its size is. All that matters is that:

  1. It's bright enough to produce an impression on whatever sensor you're using (or your eye)

  2. The background is dark enough to produce sufficient contrast

But then it would be just a bright but tiny spot.

I believe what you're really asking for is: what is the combination of factors that shows the object as bigger than a simple dot? In that case, it's two factors: aperture of the telescope, and angular size of the object.

Assuming a flawless telescope, its aperture is what determines its resolving power. The resolving power is the angle at which two dots can be separated by the telescope. The formula is:

resolving power = 1 / (10 * aperture)

where resolving power is in arcseconds, and aperture is in meters. Examples:

aperture         resolving power
10 cm            1 arcsec
20 cm            0.5 arcsec
1 m              0.1 arcsec

As long as the object's angular size is bigger than the resolving power, it will appear bigger than a dot.

That's all. In astronomy, we don't speak of an object's absolute size, we only speak of the angular size. But that should be enough. As soon as you have the angular size, and say the distance, then you could deduce the absolute size, it's a simple matter of trigonometry.

absolute size = distance * tangent(angular size)

E.g., this is the size of an object of 1 arcsec angular size, situated at 384,000 km (the orbit of the Moon):


It's 1.8 km (in case the link above is unavailable).

In other words, that's the minimum absolute distance resolved by a telescope 10 cm in aperture, for objects on the Moon. Any two dots closer together than 1.8 km, placed on the Moon, are seen as one dot in a 10 cm telescope.

  • $\begingroup$ "at infinity" is a funny expression, since stuff look larger the further away they are, at z++. In astronomy, where else. $\endgroup$ – LocalFluff Jul 28 '15 at 1:42
  • $\begingroup$ Right. Well, as someone who makes telescope optics for fun, I can only say that, unless you're doing something quite special, anything beyond a few kilometers away might as well be at infinity, it makes no difference. $\endgroup$ – Florin Andrei Jul 28 '15 at 3:57
  • $\begingroup$ This is exactly what I was looking for. Thank you for all the detail, and clear explanation! $\endgroup$ – detroitwilly Jul 28 '15 at 5:15
  • $\begingroup$ And, since angular size is quite small, tangent(angular size) can is approximately angular size, so absolute size = distance * angular size. $\endgroup$ – Pete Becker Jul 28 '15 at 12:45

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