# Trig in telescopes

1. Sep 22, 2005

### skiboka33

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. Thanks.

2. Sep 22, 2005

### SpaceTiger

Staff Emeritus
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.

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

3. Sep 22, 2005

### skiboka33

thanks again. Yeah that's what made sense to me, and I assume you can approximate tan theta for theta for distance objects since the angle is so small.

4. Sep 22, 2005

### SpaceTiger

Staff Emeritus
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:

$$\theta=\frac{1.22\lambda}{D}$$

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.

5. Sep 22, 2005

### skiboka33

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.

Last edited: Sep 22, 2005
6. Sep 22, 2005

### SpaceTiger

Staff Emeritus
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:

$$\theta_0=\frac{1.22\lambda}{D}$$

Now let's say there's an object of diameter Dobj and distance dobj. You can calculate its angular diameter with

$$\theta_{obj}=\frac{D_{obj}}{d_{obj}}$$

In order to resolve, this object, one needs

$$\theta_0<\theta_{obj}$$

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:

$$\theta_0'=\frac{1.22\lambda}{D'}$$

then the diameter required for this new telescope is:

$$D'=\frac{1.22\lambda}{\theta_{obj}}=\frac{1.22\lambda d_{obj}}{D_{obj}}$$

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:

$$\theta_0 > \frac{1.22\lambda}{D}$$

7. Sep 22, 2005

### skiboka33

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

8. Sep 22, 2005

### SpaceTiger

Staff Emeritus
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:

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.

9. Sep 22, 2005

### Labguy

From here you can link to almost any telescope parameter and / or glossary of terms you could probably ever use. Hundreds of pages of terminology with explanations. This is just one of the calculators you can reach from the first site above.

Last edited: Sep 22, 2005
10. Sep 23, 2005

### Chronos

HST is the chit when it comes to ST's explanation. Astrophysicists are dang near psychic when predicting this stuff.