<p>1 A.U. is same as infinity. The difference in terms of eyepiece position is infinitesimal, you can't measure it. Anything beyond a few kilometers away is pretty much "at infinity".</p>

<p>Regardless of that - from the practice of designing and building telescopes, calculations only offer you a starting point. You do the math, and the distance is 105 cm. But in practice lenses will deviate from the ideal focal length. Even if they didn't deviate at some temperature, put them in a cold environment, and the focal length will change a fraction of mm.</p>

<p>So take the calculations as a starting point, and build the instrument in such a way as to allow fine adjustments of the position of the eyepiece. There's a device called focuser that allows such fine adjustments. Or simply rely on friction to move the eyepiece back and forth until the image looks best, and hold it there.</p>

<p>When using the instrument in practice, you'll forget the ideal distance. What you will do is adjust the position of the eyepiece until the image looks best. You will do that every time you observe, and often multiple times during the same observation.</p>

<hr>

<p>If you want some math, take a look at the thin lens equation, and apply it to the objective lens.</p>

<p>f = focal length of the lens</p>

<p>o = distance from lens to object</p>

<p>i = distance from lens to image</p>

<p>Then the thin lens equation is:</p>

<pre><code>1/f = 1/i + 1/o

i = 1/(1/f - 1/o)
</code></pre>

<p>If o = infinity, then i = f.</p>

<p>But what happens if o = 1 A.U.?</p>

<p>i = 1/(1/1 - 1/(1.5 * 10^11)) = 1.0000000000067 meters</p>

<p>The difference is something like 6.7 * 10^-12 meters. It's smaller than an atom.</p>

CC BY-SA 3.0
2015-11-17T01:35:37.853Z
When l looked a star using the telescope,I got a circle glowing image,with distance between eyepiece and objective close to 105 cm. As I increased distance between eyepiece and objective so much more than 105 cm I got a inverted image of star just like we see in the sky. So which of this picture is the magnified image of star.
user213892
CC BY-SA 3.0
2015-11-17T05:23:31.633Z
Stars looks like super-tiny dots even in telescopes. That's how you focus a telescope: try and make the star as small as possible. That's also how you collimate the scope too: try and make the star images as small as possible; no flares, no tails, no spikes, just a tiny dot - the smaller, the better. Forget the 105 cm, just move the eyepiece back and forth until the stars look the smallest possible.
Florin Andrei
CC BY-SA 3.0
2015-11-17T05:31:03.613Z
You cannot magnify stars with a telescope, they're just too far away. Even the biggest telescopes, with mirrors over 10 meters in diameter, still can't resolve stellar disks. There are special devices called interferometers that can resolve details on a stellar disk, but the equivalent aperture of such a thing is on the order of hundreds of meters or up to 1 kilometer.
Florin Andrei
CC BY-SA 3.0
2015-11-16T19:06:07.450Z
2015-11-16T19:06:07.450Z
Florin Andrei
<p>Astronomical objects are so far away that they focus the same point as an object at infinite distance. In fact any object more than a hundred metres away or so can be treated as being an infinite distance. </p>

<p>To quantify this consider the <a href="http://www.physicsclassroom.com/class/refln/Lesson-3/The-Mirror-Equation" rel="nofollow">focal distance equation</a>:</p>

<p>$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$</p>

<p>If the focal distance = 1m, and the object distance $d_o$ = 1AU = $150\times 10^9$m</p>

<p>The the image distance ($d_i$) will be 100.000000001cm (less than one atom different)</p>

CC BY-SA 3.0
2015-11-16T19:01:14.493Z
2015-11-16T19:01:14.493Z
James K
<p>I have a home made astronomical telescope with 100 cm objective lens and 5 cm eyepiece.when viewing an object in infinity the distance between eyepiece and objective is 105 cm. So how could I calculate distance between objective and eyepiece for and far away object.For example a planet 1 AU from earth how much should be the distance between objective and eyepiece ?</p>

<telescope><telescope-making>
Equation to find distance between objective and eyepiece
user213892
# Equation to find distance between objective and eyepiece

(Asked by: username_0 on 2015-11-16)

I have a home made astronomical telescope with 100 cm objective lens and 5 cm eyepiece.when viewing an object in infinity the distance between eyepiece and objective is 105 cm. So how could I calculate distance between objective and eyepiece for and far away object.For example a planet 1 AU from earth how much should be the distance between objective and eyepiece ?

**Tags:** telescope, telescope-making

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## 2 answers

### Answer by username_1 on 2015-11-16. Score: 3

1 A.U. is same as infinity. The difference in terms of eyepiece position is infinitesimal, you can't measure it. Anything beyond a few kilometers away is pretty much "at infinity".
Regardless of that - from the practice of designing and building telescopes, calculations only offer you a starting point. You do the math, and the distance is 105 cm. But in practice lenses will deviate from the ideal focal length. Even if they didn't deviate at some temperature, put them in a cold environment, and the focal length will change a fraction of mm.
So take the calculations as a starting point, and build the instrument in such a way as to allow fine adjustments of the position of the eyepiece. There's a device called focuser that allows such fine adjustments. Or simply rely on friction to move the eyepiece back and forth until the image looks best, and hold it there.
When using the instrument in practice, you'll forget the ideal distance. What you will do is adjust the position of the eyepiece until the image looks best. You will do that every time you observe, and often multiple times during the same observation.

If you want some math, take a look at the thin lens equation, and apply it to the objective lens.
f = focal length of the lens
o = distance from lens to object
i = distance from lens to image
Then the thin lens equation is:
1/f = 1/i + 1/o

i = 1/(1/f - 1/o)

If o = infinity, then i = f.
But what happens if o = 1 A.U.?
i = 1/(1/1 - 1/(1.5 * 10^11)) = 1.0000000000067 meters
The difference is something like 6.7 * 10^-12 meters. It's smaller than an atom.

> **Comments:**
> **username_0**: When l looked a star using the telescope,I got a circle glowing image,with distance between eyepiece and objective close to 105 cm. As I increased distance between eyepiece and objective so much more than 105 cm I got a inverted image of star just like we see in the sky. So which of this picture is the magnified image of star.  
> **username_1**: Stars looks like super-tiny dots even in telescopes. That's how you focus a telescope: try and make the star as small as possible. That's also how you collimate the scope too: try and make the star images as small as possible; no flares, no tails, no spikes, just a tiny dot - the smaller, the better. Forget the 105 cm, just move the eyepiece back and forth until the stars look the smallest possible.  
> **username_1**: You cannot magnify stars with a telescope, they're just too far away. Even the biggest telescopes, with mirrors over 10 meters in diameter, still can't resolve stellar disks. There are special devices called interferometers that can resolve details on a stellar disk, but the equivalent aperture of such a thing is on the order of hundreds of meters or up to 1 kilometer.  

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### Answer by username_2 on 2015-11-16. Score: 1

Astronomical objects are so far away that they focus the same point as an object at infinite distance. In fact any object more than a hundred metres away or so can be treated as being an infinite distance.

To quantify this consider the [focal distance equation](http://www.physicsclassroom.com/class/refln/Lesson-3/The-Mirror-Equation):

$\frac{1}{f}=\frac{1}{d\_o}+\frac{1}{d\_i}$

If the focal distance = 1m, and the object distance $d\_o$ = 1AU = $150\times 10^9$m

The the image distance ($d\_i$) will be 100.000000001cm (less than one atom different)


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