<p>The resolution of a telescope is the resolution of the image created by the primary mirror at the focal plane.  It provides the minimum separation between two equal brightness stars that appear separate in that image.  Often, astronomers put a photographic plate or a CCD at the focal plane, create long exposure photos and these can then be examined at arbitrary resolution (with a magnifying glass, microscope, or zoom software on a screen).  But, no matter how much you magnify the image, stars separated by less than the resolution will appear merged.  </p>

<p>It is true that if you use, say, a wide-angle lens when observing directly with a telescope the resolution that you get can be limited by the resolution of your eye.  What is happening is the telescope is forming an image on your retina that has the full resolution, but your eye cannot make use of the finest details.  On the other hand, a hawk looking through the same eyepiece may get full resolution benefits since hawks' eyes have better resolution than humans'. </p>

CC BY-SA 3.0
2015-06-10T17:51:45.173Z
So the eye-piece is just for magnification but the resolution was already set when the light entered the telescope? I think it makes sense, thank you.
mmh
CC BY-SA 3.0
2015-06-10T16:40:57.273Z
2015-06-10T16:40:57.273Z
eshaya
<p>The Raleigh criterion is the maximum theoretical limit that ignores the architecture, quality, and state of maintenance of optics. It basically says "assuming the optics in this instrument are PERFECT, this is the resolution you could get out of it". It's a calculation that looks only at the diameter and ignores everything else. In other words, no matter how good the instrument, you cannot beat Raleigh - but you could make things worse.</p>

<p>In practice, of course things are worse. Take a parabolic mirror, as used in many telescope architectures, such as the newtonian. All parabolic mirrors generate perfect images only in the center of the field of view. Anything off-center is subject to coma, an aberration that even "perfect" paraboloids will exhibit. So the real resolving power gets worse as you move towards the edge of the field.</p>

<p>On top of that, you have to add real-world manufacturing imperfections that any mirror will have. Also add distortions caused by temperature differences, etc.</p>

<p>All these contribute to distort the image formed in the focal plane of the primary mirror. The eyepiece's role is to examine and magnify that image, for you to see. That's how telescopes work - primary optics form an image in the primary focal plane, which is then examined with the eyepiece.</p>

<p>Of course the quality of the eyepiece will contribute to further degradation of what you actually see. Even with "magic" primary optics, if you had a "perfect" image in prime focus, a bad eyepiece will blur it. In real life, most eyepieces are at least half-decent in the center of the field (some are not), but quality degrades towards the edge. High quality eyepieces will not introduce visible degradation in the center and over most of the field. Top of the line eyepieces will not degrade the image visibly anywhere.</p>

CC BY-SA 3.0
2015-06-10T18:03:57.710Z
2015-06-10T18:03:57.710Z
Florin Andrei
<p>Suppose a telescope has aperture $D = 20$ cm. The angular resolution of such telescope, according to the Rayleigh criterion (if I have understood it correctly), is given by </p>

<p>$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}$$</p>

<p>where $\lambda$ is the wavelength of light and $D$ is the aperture. Suppose $\lambda = 550$ nm (mid-point of visible light). Then </p>

<p>$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}
= 1.22\cdot\frac{550 \text{ nm}}{20\text{ cm}} \text{ rad}
= 1.22\cdot\frac{5.5\cdot10^{-5}\text{ cm}}{20\text{ cm}}\cdot 206264.8''
\approx 0.69''
$$</p>

<p>In other words, the angular resolution of the telescope is $0.69$ seconds of arc for $\lambda=550$ nm.</p>

<p>Now, is it really so that the eye-piece of the telescope has no play in this at all? <a href="http://en.wikipedia.org/wiki/Angular_resolution" rel="nofollow">Wikipedia</a> says that</p>

<blockquote>
  <p>and D is the diameter of the lens' aperture.</p>
</blockquote>

<p>which I would understand to be taken as the size of the primary mirror alone in a Newtonian telescope.</p>

<p>So if there is for example a double (binary) star, I will see them as one no matter what kind of magnification/eye-piece I am using if the angle between the stars (from my point of view) is less than $0.69''$? And vice versa, if the angle between the stars is greater than $0.69''$ I will see them as separate stars with <em>any</em> eye-piece or magnification?</p>

<p>(In theory, I know there's a lot of limitations from optics, seeing, etc.)</p>

<telescope><binary-star><newtonian-telescope><angular-resolution>
Is the angular resolution of a telescope irrespective of used eye-piece?
mmh
# Is the angular resolution of a telescope irrespective of used eye-piece?

(Asked by: username_0 on 2015-06-10)

Suppose a telescope has aperture $D = 20$ cm. The angular resolution of such telescope, according to the Rayleigh criterion (if I have understood it correctly), is given by

$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}$$

where $\lambda$ is the wavelength of light and $D$ is the aperture. Suppose $\lambda = 550$ nm (mid-point of visible light). Then

$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}
= 1.22\cdot\frac{550 \text{ nm}}{20\text{ cm}} \text{ rad}
= 1.22\cdot\frac{5.5\cdot10^{-5}\text{ cm}}{20\text{ cm}}\cdot 206264.8''
\approx 0.69''
$$

In other words, the angular resolution of the telescope is $0.69$ seconds of arc for $\lambda=550$ nm.

Now, is it really so that the eye-piece of the telescope has no play in this at all? [Wikipedia](http://en.wikipedia.org/wiki/Angular_resolution) says that

> and D is the diameter of the lens' aperture.

which I would understand to be taken as the size of the primary mirror alone in a Newtonian telescope.

So if there is for example a double (binary) star, I will see them as one no matter what kind of magnification/eye-piece I am using if the angle between the stars (from my point of view) is less than $0.69''$? And vice versa, if the angle between the stars is greater than $0.69''$ I will see them as separate stars with *any* eye-piece or magnification?

(In theory, I know there's a lot of limitations from optics, seeing, etc.)

**Tags:** telescope, binary-star, newtonian-telescope, angular-resolution


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## 2 answers

### Answer by username_1 on 2015-06-10. Score: 3 | Accepted answer

The resolution of a telescope is the resolution of the image created by the primary mirror at the focal plane. It provides the minimum separation between two equal brightness stars that appear separate in that image. Often, astronomers put a photographic plate or a CCD at the focal plane, create long exposure photos and these can then be examined at arbitrary resolution (with a magnifying glass, microscope, or zoom software on a screen). But, no matter how much you magnify the image, stars separated by less than the resolution will appear merged.

It is true that if you use, say, a wide-angle lens when observing directly with a telescope the resolution that you get can be limited by the resolution of your eye. What is happening is the telescope is forming an image on your retina that has the full resolution, but your eye cannot make use of the finest details. On the other hand, a hawk looking through the same eyepiece may get full resolution benefits since hawks' eyes have better resolution than humans'.

> **Comments:**
> **username_0**: So the eye-piece is just for magnification but the resolution was already set when the light entered the telescope? I think it makes sense, thank you.  

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### Answer by username_2 on 2015-06-10. Score: 2

The Raleigh criterion is the maximum theoretical limit that ignores the architecture, quality, and state of maintenance of optics. It basically says "assuming the optics in this instrument are PERFECT, this is the resolution you could get out of it". It's a calculation that looks only at the diameter and ignores everything else. In other words, no matter how good the instrument, you cannot beat Raleigh - but you could make things worse.

In practice, of course things are worse. Take a parabolic mirror, as used in many telescope architectures, such as the newtonian. All parabolic mirrors generate perfect images only in the center of the field of view. Anything off-center is subject to coma, an aberration that even "perfect" paraboloids will exhibit. So the real resolving power gets worse as you move towards the edge of the field.

On top of that, you have to add real-world manufacturing imperfections that any mirror will have. Also add distortions caused by temperature differences, etc.

All these contribute to distort the image formed in the focal plane of the primary mirror. The eyepiece's role is to examine and magnify that image, for you to see. That's how telescopes work - primary optics form an image in the primary focal plane, which is then examined with the eyepiece.

Of course the quality of the eyepiece will contribute to further degradation of what you actually see. Even with "magic" primary optics, if you had a "perfect" image in prime focus, a bad eyepiece will blur it. In real life, most eyepieces are at least half-decent in the center of the field (some are not), but quality degrades towards the edge. High quality eyepieces will not introduce visible degradation in the center and over most of the field. Top of the line eyepieces will not degrade the image visibly anywhere.


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