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?
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1$\begingroup$ Not the FOV, but your angular resolution, which is defined by the separation of two neighbouring Airy discs relative to their size. $\endgroup$– AtmosphericPrisonEscapeCommented Dec 19, 2019 at 17:33
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3 Answers
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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.
answered Dec 19, 2019 at 18:23
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$\begingroup$ It depends on the wavelength and diameter. The magnification only makes it more visible to the sensor/eye, up to a degree that from a certain magnification on one only magnifies the Airy disk (or other optical deficiencies of the gear). $\endgroup$– user31179Commented Dec 19, 2019 at 18:44
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1$\begingroup$ @ebv no argument - I meant that the physical size of the disc on the sensor focal plane depends on mag, as you noted. $\endgroup$ Commented Dec 19, 2019 at 20:23
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tl,dr: The Airy disc of a scope has a fixed size and a magnification or choice of fov (sensor) has its limits. Jupiter is too big for a decent amateur telescope to reach its resolution limits.
The Airy disc or diffraction limit is defined by wavelength of the incoming light and the diameter of the telescope's main mirror/frontal lense.
https://en.wikipedia.org/wiki/Airy_disk#Size
A telescope produces an image circle in the focal plane that is brightest on the optical axis and gets ever darker (and distorted and color aberrated) to the edges. The field of view is derived from the diameter of that image divided by the focal length of the scope (https://www.skyandtelescope.com/astronomy-blogs/imaging-foundations-richard-wright/understanding-field-of-view-pixel-scale/).
Depending on observation goal and what amount of distortion and attenuation one is willing to accept, one chooses a sensor size or an eyepiece focal length. Something can be done here with optical tricks like correctors or reducers, but in principle, you cannot amplify the fov beyond the image that the scope produces, which is usually in the range of arcminutes or a few degrees. You can be smaller than that (e.g. for observing planets), by choosing a small sensor, or a large magnification (relation of eyepiece focal length to scope focal length) or even optical converters to multiply the focal length, but the smaller you go, the more you loose of the scope's resolution power, and the higher the influence of the scope's optical errors get.
Remains to say: usually the atmosphere limits the observational experience, not the gear.
With these figures d=115mm, f=805mm, green light wavelength you could calculate the diameter of the Airy disc and compare it to the angular size of Jupiter or its moon Titan to see if they will be properly resolved.
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$\begingroup$ Example: Jupiter in a 25cm amateur telescope. Random youtube video, images processed: youtube.com/watch?v=fR_cAIe7kRY In the short visible wavelengths the resolution is pretty high. $\endgroup$– user31179Commented Dec 20, 2019 at 10:24
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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 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 FAST (see also), if equipped with a focal plane array or used to scan across Jupiter (see also) 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 gives an idea what it looks like.
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.