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SpaceTiger

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Theta is the angular resolution and represents the approximate minimum angular size of an object that can be distinguished from a point. For example, if I had an angular resolution of an arcminute, I wouldn't be able to distinguish two objects separated by 10 arcseconds. They would just appear as a single source of light. Part of the reason we build big telescopes is to improve the angular resolution in our images.skiboka33 said:Confused to what the theta represents in the theta = wavelength/D formula. Is it the same theta as you can find using trig if you know the distance and size of the object you're trying to see? And does anyone know a good site for astronomy formulas involving telescopes, my textbook isn't cutting it.

And yes, you can determine the angular size of an object by using the trig method you described.

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SpaceTiger

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Hmm, in retrospect, you may have meant that you could derive the equation from a simple trig argument. That's not actually the case, I just meant that you can use trig to find the angular size of an object with known distance and size. In actuality, the equation is more precisely given as:skiboka33 said:

[tex]\theta=\frac{1.22\lambda}{D}[/tex]

for a circular aperture. The result comes from computing the diffraction of electromagnetic waves. There might be some value in thinking of the equation in terms of the angle subtended by a wavelength of light at a distance equal to the aperture size, but I wouldn't recommend it before getting a more thorough understanding of the diffraction effects.

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Well what I was wondering is the relationship between the size and distance from telescope of an object to the size of the telescope (to just resolve the image)... so I kind of guessed that maybe

ang. diameter = diameter of object/distance from telescope

then plugging that into D = wavelength/ang. diameter.

But like I said, that was just a guess.

ang. diameter = diameter of object/distance from telescope

then plugging that into D = wavelength/ang. diameter.

But like I said, that was just a guess.

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SpaceTiger

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This is correct if what you're looking for is the diameter of the telescope required to resolve the object. Let's review. You have a telescope of diameter, D. Its smallest angle it can resolve is:skiboka33 said:ang. diameter = diameter of object/distance from telescope

then plugging that into D = wavelength/ang. diameter.

[tex]\theta_0=\frac{1.22\lambda}{D}[/tex]

Now let's say there's an object of diameter D

[tex]\theta_{obj}=\frac{D_{obj}}{d_{obj}}[/tex]

In order to resolve, this object, one needs

[tex]\theta_0<\theta_{obj}[/tex]

If our old telescope isn't good enough to resolve the object, maybe we want to buy another telescope that will be able to. If its angular resolution is:

[tex]\theta_0'=\frac{1.22\lambda}{D'}[/tex]

then the diameter required for this new telescope is:

[tex]D'=\frac{1.22\lambda}{\theta_{obj}}=\frac{1.22\lambda d_{obj}}{D_{obj}}[/tex]

Be careful when performing this calculation at home, however, because diffraction is not the only thing limiting your resolution. Atmospheric effects will, in general, lead to:

[tex]\theta_0 > \frac{1.22\lambda}{D}[/tex]

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I see, thank you you've been very helpful. And this 1.22 is just a given constant?

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SpaceTiger

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It's the first "zero" of the diffraction pattern, meaning basically that most of the light from a point source is spread out within that angle. There are several caveats:skiboka33 said:I see, thank you you've been very helpful. And this 1.22 is just a given constant?

1. This angle is only a ballpark number for the practical resolution limit. We can sometimes distinguish objects with separation smaller than this. In addition, really bright objects can sometimes obscure their companions at separations larger than the resolution limit.

2. It's only for a circular aperture. If the telescope has, for example, a secondary in the path of the incoming light, the resulting pattern will be more complicated and that equation won't be exactly right.

3. As I said, there are other things that contribute to the "blurring" of an image (like the atmosphere), so even a perfectly designed telescope will not experience exactly this resolution limit.

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Labguy

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From here you can link to almost any telescope parameter and / or glossary of terms you could probably ever use.skiboka33 said:And does anyone know a good site for astronomy formulas involving telescopes, my textbook isn't cutting it. Thanks.

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