<blockquote>
<p>Can we apply <a href="https://en.wikipedia.org/wiki/Angular_resolution#The_Rayleigh_criterion" rel="nofollow noreferrer">Rayleigh's criterion</a> for the resolution of a primary mirror of a reflective telescope?</p>
</blockquote>
<p>Yes, but...</p>
<p>You've asked about the criterion, not the use of the 1.22 equation.</p>
<p>From <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/Raylei.html" rel="nofollow noreferrer">hyperphysics</a> (and pretty much everywhere else):</p>
<blockquote>
<p>The Rayleigh criterion is the generally accepted criterion for the minimum resolvable detail - <strong>the imaging process is said to be diffraction-limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.</strong></p>
</blockquote>
<p>So as long as you can identify a &quot;maximum&quot; in one spot, and a &quot;first diffraction minimum&quot; in the other, you can define what &quot;resolved&quot; means.</p>
<p>However, to distinguish this from the other answers, this does not mean that you can always necessarily use the <span class="math-container">$\theta = 1.22 \lambda/d$</span> equation blindly. The 1.22 comes from the Airy disk which is the diffraction pattern from an optical system with a single, clean, circular aperture.</p>
<p>Some telescopes have pretty strange apertures, central obstructions, secondary mounts or <a href="https://astronomy.stackexchange.com/q/35182/7982">other pathologies</a> in their optics, and so their diffraction patterns differ from what you would calculate for a clear circular aperture's <a href="https://en.wikipedia.org/wiki/Airy_disk" rel="nofollow noreferrer">Airy disk</a>.</p>
<p>Below is a quickie calculation for an aperture the size of the <a href="https://en.wikipedia.org/wiki/Vera_C._Rubin_Observatory" rel="nofollow noreferrer">Vera C. Ruben observatory</a> (8.360 meters). I've done it twice - once without the huge 5.116 meter central obstruction and once with, at 550 nm. We can see that the first theoretical minimum is at 8.25 mas for the unobstructed aperture, but <em>decreases</em> to 6.4 mas with the obstruction!</p>
<p><strong>That's a ~30% <em>improvement</em> in resolution based on Rayleigh's criterion, just by putting a giant hole in the mirror!</strong></p>
<p>That's not a complete surprise, you can always engineer an aperture to boost the <a href="https://en.wikipedia.org/wiki/Optical_transfer_function" rel="nofollow noreferrer">modulation transfer function</a> at one spatial frequency if you sacrifice others - in this case we can see that the additional outer rings (more widely scattered light) are much higher in intensity for the obstructed aperture.  See also</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/a/24182/7982">this answer about aperture interferometry</a></li>
<li><a href="https://astronomy.stackexchange.com/q/29335/7982">Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?</a></li>
</ul>
<hr />
<p><a href="https://i.sstatic.net/1648T.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1648T.png" alt="simulated diffraction pattern of Vera C. Ruben (5.116, 8.360 meters, λ=550nm)" /></a></p>
<pre><code>import numpy as np
import matplotlib.pyplot as plt

r2, r1 = 5.116, 8.360  # meters

lam = 550E-09

n = 1000

ym, xm = np.mgrid[-8.4:8.4:(2*n+1)*1j, -8.4:8.4:(2*n+1)*1j]

rsq = xm**2 + ym**2

ap_clear, ap_obstructed = np.zeros_like(rsq), np.zeros_like(rsq)

ap_clear[(rsq &lt;= r1**2)] = 1.

ap_obstructed[(rsq &gt;= r2**2) * (rsq &lt;= r1**2)] = 1.

theta_max = 5 * lam / (2 * r1)
thetas = np.linspace(0, theta_max, 501)
thetas_arcsec = np.degrees(thetas) * 3600

I_clear, I_obstructed = [], []
for theta in thetas:
    phase = 2 * np.pi * xm * theta / lam # avoid complex by symmetry (xm=0 in center)
    E_clear = np.sqrt(ap_clear) * np.cos(phase) # sqrt for amplitude
    E_obstructed = np.sqrt(ap_obstructed) * np.cos(phase) # sqrt for amplitude
    I_c = np.abs(E_clear.sum())**2
    I_o = np.abs(E_obstructed.sum())**2
    I_clear.append(I_c)
    I_obstructed.append(I_o)

I_clear = np.array(I_clear)
I_obstructed = np.array(I_obstructed)

Imax = max(I_clear.max(), I_obstructed.max())

I_clear /= Imax
I_obstructed /= Imax

if True:
    extent = [xm.min(), xm.max(), ym.min(), ym.max()]
    fig = plt.figure(figsize=[14, 7.5])
    gs = fig.add_gridspec(nrows=2, ncols=2)
    ax_clear = fig.add_subplot(gs[0, 0])
    ax_obstructed = fig.add_subplot(gs[1, 0])
    ax_plot = fig.add_subplot(gs[:, 1:])
    ax_clear.imshow(ap_clear, cmap='gray', extent=extent)
    ax_clear.set_title('clear')
    ax_clear.set_ylabel('meters')
    ax_obstructed.imshow(ap_obstructed, cmap='gray', extent=extent)
    ax_obstructed.set_title('obstructed')
    ax_obstructed.set_ylabel('meters')
    ax_plot.plot(thetas_arcsec, np.log10(I_clear), '-', label='clear')
    ax_plot.plot(thetas_arcsec, np.log10(I_obstructed), '--', label='obstructed')
    ax_plot.set_xlim(0, 0.03)
    ax_plot.set_xlabel('arcsec')
    ax_plot.set_ylabel('log(I)')
    plt.suptitle('Vera C. Ruben (5.116, 8.360 meters, λ=550nm)')
    plt.legend()
    plt.show()
                 
# first minimum
# clear: .00825 arcsec
# obstructed: .00640 arcsec
</code></pre>

CC BY-SA 4.0
2023-02-21T13:10:07.877Z
2023-02-21T13:10:07.877Z
uhoh
<p>Rayleigh's criterion concerns whether two point-sources will appear to merge by diffraction into a single dot.  It is independent of the method of forming an image. So applies equally to mirror- and lens-based telescopes.</p>

CC BY-SA 4.0
2023-02-20T07:13:21.193Z
So ,in a reflector telescope, should I calculate the angle of resolution of primary mirror or eyepiece
Tapan Gupta
CC BY-SA 4.0
2023-02-20T07:22:15.560Z
What would you do for a refraction telescope, the objective lens or the eyepiece?
James K
CC BY-SA 4.0
2023-02-21T02:01:29.990Z
Objective lens...?
Tapan Gupta
CC BY-SA 4.0
2023-02-20T05:42:50.720Z
2023-02-20T05:42:50.720Z
James K
<p>Rayleigh’s criterion is good for both refractors and reflectors.</p>

CC BY-SA 4.0
2023-02-20T05:41:55.147Z
2023-02-20T05:41:55.147Z
Pierre Paquette
<p>Can we apply <a href="https://en.wikipedia.org/wiki/Angular_resolution#The_Rayleigh_criterion" rel="nofollow noreferrer">Rayleigh's criterion</a> for the resolution of a primary mirror of a reflective telescope?</p>

CC BY-SA 4.0
2023-02-20T04:15:26.927Z
Rayleigh's criterion of angle of resolution
Tapan Gupta
CC BY-SA 4.0
2023-02-20T05:20:09.707Z
You can use edit to amend your question directly or to add clarifications
planetmaker
<telescope>
Is rayleigh's criterion only applicable to refractive telescopes?
Tapan Gupta
# Is rayleigh's criterion only applicable to refractive telescopes?

(Asked by: username_0 on 2023-02-20)

Can we apply [Rayleigh's criterion](https://en.wikipedia.org/wiki/Angular_resolution#The_Rayleigh_criterion) for the resolution of a primary mirror of a reflective telescope?

**Tags:** telescope

> **Comments:**
> **username_0**: Rayleigh's criterion of angle of resolution  
> **username_1**: You can use edit to amend your question directly or to add clarifications  

---

## 3 answers

### Answer by username_2 on 2023-02-21. Score: 3 | Accepted answer

Can we apply Rayleigh's criterion for the resolution of a primary mirror of a reflective telescope?

Yes, but...
You've asked about the criterion, not the use of the 1.22 equation.
From hyperphysics (and pretty much everywhere else):

The Rayleigh criterion is the generally accepted criterion for the minimum resolvable detail - the imaging process is said to be diffraction-limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.

So as long as you can identify a "maximum" in one spot, and a "first diffraction minimum" in the other, you can define what "resolved" means.
However, to distinguish this from the other answers, this does not mean that you can always necessarily use the $\theta = 1.22 \lambda/d$ equation blindly. The 1.22 comes from the Airy disk which is the diffraction pattern from an optical system with a single, clean, circular aperture.
Some telescopes have pretty strange apertures, central obstructions, secondary mounts or other pathologies in their optics, and so their diffraction patterns differ from what you would calculate for a clear circular aperture's Airy disk.
Below is a quickie calculation for an aperture the size of the Vera C. Ruben observatory (8.360 meters). I've done it twice - once without the huge 5.116 meter central obstruction and once with, at 550 nm. We can see that the first theoretical minimum is at 8.25 mas for the unobstructed aperture, but decreases to 6.4 mas with the obstruction!
That's a ~30% improvement in resolution based on Rayleigh's criterion, just by putting a giant hole in the mirror!
That's not a complete surprise, you can always engineer an aperture to boost the modulation transfer function at one spatial frequency if you sacrifice others - in this case we can see that the additional outer rings (more widely scattered light) are much higher in intensity for the obstructed aperture.  See also

this answer about aperture interferometry
Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?



import numpy as np
import matplotlib.pyplot as plt

r2, r1 = 5.116, 8.360  # meters

lam = 550E-09

n = 1000

ym, xm = np.mgrid[-8.4:8.4:(2*n+1)*1j, -8.4:8.4:(2*n+1)*1j]

rsq = xm**2 + ym**2

ap_clear, ap_obstructed = np.zeros_like(rsq), np.zeros_like(rsq)

ap_clear[(rsq <= r1**2)] = 1.

ap_obstructed[(rsq >= r2**2) * (rsq <= r1**2)] = 1.

theta_max = 5 * lam / (2 * r1)
thetas = np.linspace(0, theta_max, 501)
thetas_arcsec = np.degrees(thetas) * 3600

I_clear, I_obstructed = [], []
for theta in thetas:
    phase = 2 * np.pi * xm * theta / lam # avoid complex by symmetry (xm=0 in center)
    E_clear = np.sqrt(ap_clear) * np.cos(phase) # sqrt for amplitude
    E_obstructed = np.sqrt(ap_obstructed) * np.cos(phase) # sqrt for amplitude
    I_c = np.abs(E_clear.sum())**2
    I_o = np.abs(E_obstructed.sum())**2
    I_clear.append(I_c)
    I_obstructed.append(I_o)

I_clear = np.array(I_clear)
I_obstructed = np.array(I_obstructed)

Imax = max(I_clear.max(), I_obstructed.max())

I_clear /= Imax
I_obstructed /= Imax

if True:
    extent = [xm.min(), xm.max(), ym.min(), ym.max()]
    fig = plt.figure(figsize=[14, 7.5])
    gs = fig.add_gridspec(nrows=2, ncols=2)
    ax_clear = fig.add_subplot(gs[0, 0])
    ax_obstructed = fig.add_subplot(gs[1, 0])
    ax_plot = fig.add_subplot(gs[:, 1:])
    ax_clear.imshow(ap_clear, cmap='gray', extent=extent)
    ax_clear.set_title('clear')
    ax_clear.set_ylabel('meters')
    ax_obstructed.imshow(ap_obstructed, cmap='gray', extent=extent)
    ax_obstructed.set_title('obstructed')
    ax_obstructed.set_ylabel('meters')
    ax_plot.plot(thetas_arcsec, np.log10(I_clear), '-', label='clear')
    ax_plot.plot(thetas_arcsec, np.log10(I_obstructed), '--', label='obstructed')
    ax_plot.set_xlim(0, 0.03)
    ax_plot.set_xlabel('arcsec')
    ax_plot.set_ylabel('log(I)')
    plt.suptitle('Vera C. Ruben (5.116, 8.360 meters, λ=550nm)')
    plt.legend()
    plt.show()
                 
# first minimum
# clear: .00825 arcsec
# obstructed: .00640 arcsec


---

### Answer by username_3 on 2023-02-20. Score: 3

Rayleigh's criterion concerns whether two point-sources will appear to merge by diffraction into a single dot. It is independent of the method of forming an image. So applies equally to mirror- and lens-based telescopes.

> **Comments:**
> **username_0**: So ,in a reflector telescope, should I calculate the angle of resolution of primary mirror or eyepiece  
> **username_3**: What would you do for a refraction telescope, the objective lens or the eyepiece?  
> **username_0**: Objective lens...?  

---

### Answer by username_4 on 2023-02-20. Score: 2

Rayleigh’s criterion is good for both refractors and reflectors.


---

