<p>No.
The size of the Airy disc, and rings, is defined for a point source and depends on the system magnification, and the system aperture. It does not depend on brightness other than depending on the low-level detectability limit of your sensor.    </p>

<p>It is true that the diffraction pattern of an extended object is roughly the superposition of Airy disks from every (geometric) point in the object's image. There will be a 'halo' around the image of width (approximately) Airy disk radius, and then there will be secondary maxima generated by the (interfering) Airy rings from all the image points.  </p>

CC BY-SA 4.0
2019-12-19T20:23:05.167Z
@ebv no argument - I meant that the physical size of the disc on the sensor focal plane depends on mag, as you noted.
Carl Witthoft
CC BY-SA 4.0
2019-12-19T18:23:35.537Z
2019-12-19T18:23:35.537Z
Carl Witthoft
<blockquote>
  <p>Is the reason an “airy disc”, concentric rings as a diffraction effect, is seen around stars and not Jupiter because Jupiter has an angular size that is thousands of times larger?</p>
</blockquote>

<p>Yes, for a wide range of apertures and wavelengths, from Earth, this is true.</p>

<p>The angle of the <a href="https://en.wikipedia.org/wiki/Airy_disk#Size" rel="nofollow noreferrer">first minimum</a> of the Airy disk is  </p>

<p><span class="math-container">$$\approx 1.22 \frac{\lambda}{d}$$</span> </p>

<p>where <span class="math-container">$\lambda$</span> is the observational wavelength and <span class="math-container">$d$</span> is the diameter of the aperture. Since it's wavelength dependent we actually see colors in the Airy disk fringes if we look hard enough!</p>

<p>The angular radius of Jupiter is the planets radius divided by its distance</p>

<p><span class="math-container">$$\frac{R_J}{r_J} \approx \frac{7.2 \times 10^4 km}{7.8 \times 10^8 km}$$</span> </p>

<p>or about <span class="math-container">$9 \times 10^{-5}$</span> radians +/- 20% as the planets move in their orbits.</p>

<p>If the planets angular diameter is much bigger than the Airy disk you probably won't be able to see fringes but the reason is complicated and I'll explain below. If it's much smaller than the Airy disk, i.e. "star-like", then you will.</p>

<h2>Where is the dividing line?</h2>

<p><strong>Solving <span class="math-container">$1.22 \frac{\lambda}{d} = 9 \times 10^{-5}$</span> gives <span class="math-container">$d = 14,000 \lambda$</span>.</strong></p>

<p>So if you are looking at visible light with a wavelength of <span class="math-container">$5 \times 10^{-7}$</span> meters then the crossover is at about a 7 millimeter aperture.</p>

<p>That means that for a cellphone camera's <em>lens</em>, Jupiter is pretty much star-like. Whether its pixels are are able to discern the Airy disk is a separate question.</p>

<p>At 1 GHz the wavelength is about 0.3 meters, and that makes the "cross-over" aperture about 4 kilometers. A single dish, even <a href="https://en.wikipedia.org/wiki/Five-hundred-meter_Aperture_Spherical_Telescope" rel="nofollow noreferrer">the biggest one on Earth</a> FAST (see <a href="https://astronomy.stackexchange.com/q/16240/7982">also</a>), if equipped with a <a href="https://astronomy.stackexchange.com/q/16237/7982">focal plane array</a> or used to <a href="https://astronomy.stackexchange.com/a/31656/7982">scan across Jupiter</a> (see <a href="https://astronomy.stackexchange.com/q/16238/7982">also</a>) would see its diffraction-limited Airy disk rather than Jupiter's disk. However, a large radiotelescope array tens of kilometers wide could see Jupiter's actual disk.</p>

<hr>

<h2>Does Jupiter have fringes?</h2>

<p>If jupiter were a circular, optically flat mirror, we would propagate each photon from the Sun to illuminate the entire thing, then propagate that reflection back to every point in our telescope. If we did that exercise assuming a large aperture so that Jupiter was well resolved, we would see a knife-edge diffraction-like effect near the edge. The Wikipedia article for this is insufficient so I won't link to it, but hyperphysic's <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/difopa.html" rel="nofollow noreferrer">Diffraction by Opaque Barrier</a> gives an idea what it looks like. </p>

<p><a href="https://i.sstatic.net/oRF4j.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oRF4j.png" alt="enter image description here"></a></p>

<p>Surprisingly to us at first, the fringes are on the <em>inside</em> of the disk, and there's a smooth decay outside of the edge!</p>

<p>But that's for a hard edge, and Jupiter's edge is diffuse for both geometrical and atmospheric reasons, and of course it is not an optical flat.</p>

<p>So on a good clear night, it might be extremely interesting to use a long focal length camera that can be variably stopped down to very small apertures and scan through the transition region from diffraction limited to resolved and see what happens. (long focal length so that sensor pixelation will not be a limiting factor). I think you will see something close to Jupiter's intensity distribution folded with the intensity distribution of your aperture's Airy function as other answers elude to, but I'm not sure.</p>

CC BY-SA 4.0
2019-12-20T00:39:04.340Z
2019-12-20T00:39:04.340Z
uhoh
<p>Jupiter has a minimum brightness that is brighter than stars. Is the reason an “airy disc”, concentric rings as a diffraction effect, is seen around stars and not Jupiter because Jupiter has an angular size that is thousands of times larger? With similar reasoning, if the field of view was scaled up a thousand fold, would the optical effects be similar to that around stars?</p>

CC BY-SA 4.0
2019-12-19T17:33:40.807Z
Not the FOV, but your angular resolution, which is defined by the separation of two neighbouring Airy discs relative to their size.
AtmosphericPrisonEscape
<amateur-observing><optics>
Optics: "Airy disc" and size of objects
Stark
# Optics: "Airy disc" and size of objects

(Asked by: username_0 on 2019-12-19)

Jupiter has a minimum brightness that is brighter than stars. Is the reason an “airy disc”, concentric rings as a diffraction effect, is seen around stars and not Jupiter because Jupiter has an angular size that is thousands of times larger? With similar reasoning, if the field of view was scaled up a thousand fold, would the optical effects be similar to that around stars?

**Tags:** amateur-observing, optics
> **Comments:**
> **username_1**: Not the FOV, but your angular resolution, which is defined by the separation of two neighbouring Airy discs relative to their size.  

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## 2 answers

### Answer by username_2 on 2019-12-19. Score: 2

No.
The size of the Airy disc, and rings, is defined for a point source and depends on the system magnification, and the system aperture. It does not depend on brightness other than depending on the low-level detectability limit of your sensor.

It is true that the diffraction pattern of an extended object is roughly the superposition of Airy disks from every (geometric) point in the object's image. There will be a 'halo' around the image of width (approximately) Airy disk radius, and then there will be secondary maxima generated by the (interfering) Airy rings from all the image points.

> **Comments:**
> **username_2**: @ebv no argument - I meant that the physical size of the disc on the sensor focal plane depends on mag, as you noted.  

---

### Answer by username_3 on 2019-12-20. Score: 1

> Is the reason an “airy disc”, concentric rings as a diffraction effect, is seen around stars and not Jupiter because Jupiter has an angular size that is thousands of times larger?

Yes, for a wide range of apertures and wavelengths, from Earth, this is true.

The angle of the [first minimum](https://en.wikipedia.org/wiki/Airy_disk#Size) of the Airy disk is

$$\approx 1.22 \frac{\lambda}{d}$$

where $\lambda$ is the observational wavelength and $d$ is the diameter of the aperture. Since it's wavelength dependent we actually see colors in the Airy disk fringes if we look hard enough!

The angular radius of Jupiter is the planets radius divided by its distance

$$\frac{R\_J}{r\_J} \approx \frac{7.2 \times 10^4 km}{7.8 \times 10^8 km}$$

or about $9 \times 10^{-5}$ radians +/- 20% as the planets move in their orbits.

If the planets angular diameter is much bigger than the Airy disk you probably won't be able to see fringes but the reason is complicated and I'll explain below. If it's much smaller than the Airy disk, i.e. "star-like", then you will.

## Where is the dividing line?

**Solving $1.22 \frac{\lambda}{d} = 9 \times 10^{-5}$ gives $d = 14,000 \lambda$.**

So if you are looking at visible light with a wavelength of $5 \times 10^{-7}$ meters then the crossover is at about a 7 millimeter aperture.

That means that for a cellphone camera's *lens*, Jupiter is pretty much star-like. Whether its pixels are are able to discern the Airy disk is a separate question.

At 1 GHz the wavelength is about 0.3 meters, and that makes the "cross-over" aperture about 4 kilometers. A single dish, even [the biggest one on Earth](https://en.wikipedia.org/wiki/Five-hundred-meter_Aperture_Spherical_Telescope) FAST (see [also](https://astronomy.stackexchange.com/q/16240/7982)), if equipped with a [focal plane array](https://astronomy.stackexchange.com/q/16237/7982) or used to [scan across Jupiter](https://astronomy.stackexchange.com/a/31656/7982) (see [also](https://astronomy.stackexchange.com/q/16238/7982)) would see its diffraction-limited Airy disk rather than Jupiter's disk. However, a large radiotelescope array tens of kilometers wide could see Jupiter's actual disk.

---

## Does Jupiter have fringes?

If jupiter were a circular, optically flat mirror, we would propagate each photon from the Sun to illuminate the entire thing, then propagate that reflection back to every point in our telescope. If we did that exercise assuming a large aperture so that Jupiter was well resolved, we would see a knife-edge diffraction-like effect near the edge. The Wikipedia article for this is insufficient so I won't link to it, but hyperphysic's [Diffraction by Opaque Barrier](http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/difopa.html) gives an idea what it looks like.

[![enter image description here](https://i.sstatic.net/oRF4j.png)](https://i.sstatic.net/oRF4j.png)

Surprisingly to us at first, the fringes are on the *inside* of the disk, and there's a smooth decay outside of the edge!

But that's for a hard edge, and Jupiter's edge is diffuse for both geometrical and atmospheric reasons, and of course it is not an optical flat.

So on a good clear night, it might be extremely interesting to use a long focal length camera that can be variably stopped down to very small apertures and scan through the transition region from diffraction limited to resolved and see what happens. (long focal length so that sensor pixelation will not be a limiting factor). I think you will see something close to Jupiter's intensity distribution folded with the intensity distribution of your aperture's Airy function as other answers elude to, but I'm not sure.


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