<blockquote>
  <p>It is well known that diffraction resolution of a telescope depends on the exterior radius of aperture: One can, for instance, use just a narrow outer ring of a parabolic mirror and the resolution will be the same as it were the whole mirror (obviously, intensity will be much much smaller)</p>
</blockquote>

<p>The premise of your question is false. I tried to convince you to separate it out as a separate question, but to no avail. So I'll answer it here instead.</p>

<p>From <a href="https://astronomy.stackexchange.com/questions/34238/number-of-actuators-in-adaptive-optics/34247#comment65497_34238">comment</a>:</p>

<blockquote>
  <p>I personally made computation of the telescope transfer function for annular aperture and found that support of the function in the fourier space is the same as for circular aperture. This is an elementary excercise. </p>
</blockquote>

<p>To help illustrate that a narrow annular aperture has much worse resolution than the full aperture, I'll do a quick comparison. The first plot is the diffraction pattern (scale is +/- 0.5 arcseconds) for a clear aperture of 2 million wavelengths (e.g. 1 meter radius with  1 micron wavelength), and the second is for the same aperture with a 0.9 meter obstruction.</p>

<p>I use <a href="https://en.wikipedia.org/wiki/Airy_disk" rel="nofollow noreferrer">Airy disks</a> and <a href="https://en.wikipedia.org/wiki/Babinet%27s_principle" rel="nofollow noreferrer">Babinet's principle</a> to subtract the E-field of the obstruction before squaring. For this symmetry there's no need to do an explicit Fourier analysis, <a href="https://en.wikipedia.org/wiki/Airy_disk#Mathematical_formulation" rel="nofollow noreferrer">the Airy disk is the Fourier transform of the circular aperture step function</a>. (see <a href="https://physics.stackexchange.com/q/451400/83380">also</a>)</p>

<p>This is for an annulus with outer radius 1.0 and inner 0.9. If it was 1.0 and 0.99 it would look <em>much much worse!</em></p>

<blockquote>
  <p>Now, take such a ring telescope, say, 3cm wide ring which is 100 meters in diameter.</p>
</blockquote>

<p>That would give you a giant splat pattern of roughly the same size as a simple 3 cm diameter telescope would.</p>

<p><a href="https://i.sstatic.net/nfdjN.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nfdjN.png" alt="annular telescopes are bad"></a></p>

<p>Python script:</p>

<pre><code>from scipy.special import j1
import matplotlib.pyplot as plt
import numpy as np

def get_E(a, lam, theta):
    ka  = (twopi/lam) * a
    x   = ka * np.sin(theta)
    return a * (2. * j1(x) / x) 

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

arcsecs = np.arange(0, 0.5, 0.0001)[1:]
theta   = (arcsecs/3600.) * rads

r1, r2  = 1.0, 0.9
lam     = 1E-06

arcsecs   = np.linspace(-1, 1, 1000) # arcsec (skips zero to avoid singularity)
Xas, Yas, = np.meshgrid(arcsecs, arcsecs)
Ras       = np.sqrt(Xas**2 + Yas**2)
theta     = Ras * pi / 3600. / 180.

E1, E2    = [sign*get_E(r, lam, theta) for r, sign in ((r1, +1), (r2, -1))]
asec      = np.linspace(0, 1.0, 500)[1:]
th        = asec * pi / 3600. / 180.
e1, e2    = [sign*get_E(r, lam, th) for r, sign in ((r1, +1), (r2, -1))]

Etot = E1 + E2
etot = e1 + e2

if True:
    plt.figure()
    plt.subplot(2, 2, 1)
    plt.imshow(E1**2, cmap='gray')
    plt.title('r = 1.0, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 3)
    plt.imshow(Etot**2, cmap='gray')
    plt.title ('r1, r2 = 1.0, 0.9, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 2)
    plt.plot(1E+06*th, th*e1**2) # weight by th for azimuthal (phi) integration
    plt.xlabel('urad', fontsize=14)
    plt.title('r = 1.0, lam=1E-06, integ theta')
    plt.subplot(2, 2, 4)
    plt.plot(1E+06*th, th*etot**2) # weight by th for azimuthal (phi) integration
    plt.title('r1, r2 = 1.0, 0.9, lam=1E-06, integ theta')
    plt.xlabel('urad', fontsize=14)
    plt.show()
</code></pre>

CC BY-SA 4.0
2019-12-17T20:56:37.363Z
Thanks for computation. I know that the difference between full circular aperture and ring aperture gives pattern like this: In the full aperture case it is (J_1(x)/x)^2 while for the thin ring aperture it is J_0(x)^2 i.e. envelope of intensity of pattern falls approximately as 1/x^3 in the full aperture case, while in the ring case it falls as 1/x. This is the result you got.  I was talking  not about direct observation, but about a possibility of restoring the resolution using post-processing (de-convolution).
Truffaldino
CC BY-SA 4.0
2019-12-17T20:58:23.433Z
In the case of the ring aperture such restoration is possible in theory, Even if the intensity falls in differnt manner, the sizes of cenrtal lobe of the patterns are the same in the both cases.
Truffaldino
CC BY-SA 4.0
2019-12-09T15:54:35.910Z
2019-12-09T17:30:07.657Z
2019-12-09T17:30:07.657Z
uhoh
<p>It is well known that diffraction resolution of a telescope depends on the exterior radius of aperture: One can, for instance, use just a narrow outer ring of a parabolic mirror and the resolution will be the same as it were the whole mirror (obviously, intensity will be much much smaller).</p>

<p>Now, take such a ring telescope, say, 3cm wide ring which is 100 meters in diameter.</p>

<p>Suppose we are using adaptive optics. How many actuators are needed to reach the maximal possible resolution?</p>

<p>I assume that spacing between the actuators should be of an order of Fried radius r0 (i.e. one actuator about every 10cm along the circumference of the ring). I.e. number or actuators should be of order D/r0, where D is the aperture diameter.</p>

<p>Is it correct?</p>

CC BY-SA 4.0
2019-12-09T01:45:32.910Z
The first part about a narrow annular aperture having the same resolution as a full aperture is not right. I think that would make a great *separate question* but as a premise to your question it causes some problems. I recommend that you edit your question and just ask for some guidelines for how many actuators are necessary for a given aperture; that would be a great question by itself! Ask separately about your idea about an annular telescope because that answer will be totally different (and probably a bit disappointing). See https://en.wikipedia.org/wiki/Babinet%27s_principle
uhoh
CC BY-SA 4.0
2019-12-09T14:34:28.343Z
I personally made computation of the telescope transfer function for annular aperture and found that support of the function in the fourier space is the same as for circular aperture. This is an elementary excercise. So the resolution is the same. Moreover I read this statement in several papers.
Truffaldino
CC BY-SA 4.0
2019-12-09T14:58:11.670Z
I am also adding NASA report as an example of a paper that states " A ring-shaped primary mirror loses only collecting area, not resolution" (see page 2, last sentence before figure 1.) https://www.nasa.gov/sites/default/files/atoms/files/2014_phase_i_webster_cash_the_aragoscope.pdf
Truffaldino
CC BY-SA 4.0
2019-12-09T15:11:57.550Z
Ah, I see, the first sentence is correct "Consequentially, it is a well-known trick of sophomore physics that *an obscuration of a* ***small part*** *of a telescope’s aperture* dims the image but does not reduce its resolution." However, the next sentence drops the "small part" and overgeneralizes, and is in your case wrong: "A ring-shaped primary mirror loses only collecting area, not resolution." It's wrong in your case because they dropped those two words, and for your case the obscuration is not small, it's almost the same size as the main aperture. Their 2nd sentence is quite misleading.
uhoh
CC BY-SA 4.0
2019-12-09T16:05:31.197Z
Since I've now answered the premise part of the question, it might be a good idea to ask a more general question about number of actuators as a separate question.
uhoh
<telescope>
Number of actuators in adaptive optics
Truffaldino
# Number of actuators in adaptive optics

(Asked by: username_0 on 2019-12-09)

It is well known that diffraction resolution of a telescope depends on the exterior radius of aperture: One can, for instance, use just a narrow outer ring of a parabolic mirror and the resolution will be the same as it were the whole mirror (obviously, intensity will be much much smaller).

Now, take such a ring telescope, say, 3cm wide ring which is 100 meters in diameter.

Suppose we are using adaptive optics. How many actuators are needed to reach the maximal possible resolution?

I assume that spacing between the actuators should be of an order of Fried radius r0 (i.e. one actuator about every 10cm along the circumference of the ring). I.e. number or actuators should be of order D/r0, where D is the aperture diameter.

Is it correct?

**Tags:** telescope
> **Comments:**
> **username_1**: The first part about a narrow annular aperture having the same resolution as a full aperture is not right. I think that would make a great *separate question* but as a premise to your question it causes some problems. I recommend that you edit your question and just ask for some guidelines for how many actuators are necessary for a given aperture; that would be a great question by itself! Ask separately about your idea about an annular telescope because that answer will be totally different (and probably a bit disappointing). See https://en.wikipedia.org/wiki/Babinet%27s_principle  
> **username_0**: I personally made computation of the telescope transfer function for annular aperture and found that support of the function in the fourier space is the same as for circular aperture. This is an elementary excercise. So the resolution is the same. Moreover I read this statement in several papers.  
> **username_0**: I am also adding NASA report as an example of a paper that states " A ring-shaped primary mirror loses only collecting area, not resolution" (see page 2, last sentence before figure 1.) https://www.nasa.gov/sites/default/files/atoms/files/2014_phase_i_webster_cash_the_aragoscope.pdf  
> **username_1**: Ah, I see, the first sentence is correct "Consequentially, it is a well-known trick of sophomore physics that *an obscuration of a* ***small part*** *of a telescope’s aperture* dims the image but does not reduce its resolution." However, the next sentence drops the "small part" and overgeneralizes, and is in your case wrong: "A ring-shaped primary mirror loses only collecting area, not resolution." It's wrong in your case because they dropped those two words, and for your case the obscuration is not small, it's almost the same size as the main aperture. Their 2nd sentence is quite misleading.  
> **username_1**: Since I've now answered the premise part of the question, it might be a good idea to ask a more general question about number of actuators as a separate question.  

---

## 1 answers

### Answer by username_1 on 2019-12-09. Score: 1

It is well known that diffraction resolution of a telescope depends on the exterior radius of aperture: One can, for instance, use just a narrow outer ring of a parabolic mirror and the resolution will be the same as it were the whole mirror (obviously, intensity will be much much smaller)

The premise of your question is false. I tried to convince you to separate it out as a separate question, but to no avail. So I'll answer it here instead.
From comment:

I personally made computation of the telescope transfer function for annular aperture and found that support of the function in the fourier space is the same as for circular aperture. This is an elementary excercise. 

To help illustrate that a narrow annular aperture has much worse resolution than the full aperture, I'll do a quick comparison. The first plot is the diffraction pattern (scale is +/- 0.5 arcseconds) for a clear aperture of 2 million wavelengths (e.g. 1 meter radius with  1 micron wavelength), and the second is for the same aperture with a 0.9 meter obstruction.
I use Airy disks and Babinet's principle to subtract the E-field of the obstruction before squaring. For this symmetry there's no need to do an explicit Fourier analysis, the Airy disk is the Fourier transform of the circular aperture step function. (see also)
This is for an annulus with outer radius 1.0 and inner 0.9. If it was 1.0 and 0.99 it would look much much worse!

Now, take such a ring telescope, say, 3cm wide ring which is 100 meters in diameter.

That would give you a giant splat pattern of roughly the same size as a simple 3 cm diameter telescope would.

Python script:
from scipy.special import j1
import matplotlib.pyplot as plt
import numpy as np

def get_E(a, lam, theta):
    ka  = (twopi/lam) * a
    x   = ka * np.sin(theta)
    return a * (2. * j1(x) / x) 

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

arcsecs = np.arange(0, 0.5, 0.0001)[1:]
theta   = (arcsecs/3600.) * rads

r1, r2  = 1.0, 0.9
lam     = 1E-06

arcsecs   = np.linspace(-1, 1, 1000) # arcsec (skips zero to avoid singularity)
Xas, Yas, = np.meshgrid(arcsecs, arcsecs)
Ras       = np.sqrt(Xas**2 + Yas**2)
theta     = Ras * pi / 3600. / 180.

E1, E2    = [sign*get_E(r, lam, theta) for r, sign in ((r1, +1), (r2, -1))]
asec      = np.linspace(0, 1.0, 500)[1:]
th        = asec * pi / 3600. / 180.
e1, e2    = [sign*get_E(r, lam, th) for r, sign in ((r1, +1), (r2, -1))]

Etot = E1 + E2
etot = e1 + e2

if True:
    plt.figure()
    plt.subplot(2, 2, 1)
    plt.imshow(E1**2, cmap='gray')
    plt.title('r = 1.0, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 3)
    plt.imshow(Etot**2, cmap='gray')
    plt.title ('r1, r2 = 1.0, 0.9, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 2)
    plt.plot(1E+06*th, th*e1**2) # weight by th for azimuthal (phi) integration
    plt.xlabel('urad', fontsize=14)
    plt.title('r = 1.0, lam=1E-06, integ theta')
    plt.subplot(2, 2, 4)
    plt.plot(1E+06*th, th*etot**2) # weight by th for azimuthal (phi) integration
    plt.title('r1, r2 = 1.0, 0.9, lam=1E-06, integ theta')
    plt.xlabel('urad', fontsize=14)
    plt.show()

> **Comments:**
> **username_0**: Thanks for computation. I know that the difference between full circular aperture and ring aperture gives pattern like this: In the full aperture case it is (J_1(x)/x)^2 while for the thin ring aperture it is J_0(x)^2 i.e. envelope of intensity of pattern falls approximately as 1/x^3 in the full aperture case, while in the ring case it falls as 1/x. This is the result you got.  I was talking  not about direct observation, but about a possibility of restoring the resolution using post-processing (de-convolution).  
> **username_0**: In the case of the ring aperture such restoration is possible in theory, Even if the intensity falls in differnt manner, the sizes of cenrtal lobe of the patterns are the same in the both cases.  

---

