Starship Asterisk*

Chris Peterson wrote: ↑Sun Nov 10, 2024 8:27 pm

johnnydeep wrote: ↑Sun Nov 10, 2024 6:47 pm

Chris Peterson wrote: ↑Sun Nov 10, 2024 2:30 pm
As long as that distant object is a point source (like a star) that lies on the optical axis. Any extended object (regardless of distance) will result in rays entering at different angles, and a point source off of the optical axis will create an image that isn't at the focal point.

But aren't those different angles from a far far away object like so close to parallel that it makes essentially no difference? I guess you're saying not. And, for example, Andromeda's 3° by 1° areal extent is not negligible at all?

The rays from any point are a parallel bundle that enter the entire aperture. But the angle they enter depends on the angle of the source with respect to the optical axis.

Just consider three stars in a line. You aim the telescope at the center star, and all of its light enters the aperture parallel to the optical axis and forms an image at the focal point. A star a little above the center has all of its light enter the aperture tilted slightly downwards, and forms an image at the focal plane just below the focal point. A star a little below the center has all of its light enter the aperture tilted slightly upwards, and forms an image at the focal plane just above the focal point. So you have three little star images at the focal plane. And if you now have an eyepiece with a focal point coincident with the objective's focal point, you'll have three bundles of parallel rays exiting that eyepiece at steeper angles than they entered the objective. When the lens of your eye focuses those three bundles back into an image on your retina, you'll see three stars that appear farther apart than what you see in the sky without the telescope. (With a camera you just place it at the focal plane and record the image directly, as the objective is a focal system, not an afocal one like you have with a true optical telescope... which is not what a "telescope" actually is when used with a camera.)

johnnydeep wrote: ↑Sun Nov 10, 2024 6:47 pm

Chris Peterson wrote: ↑Sun Nov 10, 2024 2:30 pm

johnnydeep wrote: ↑Sun Nov 10, 2024 2:25 pm

Ok, thanks. So...googling... all parallel rays entering at a given angle will converge at a specific point on the focal plane. But since all rays from a very remote object are parallel to the optical axis, they will all converge at the same (focal) point, right?

As I keep saying, I will never understand optics.

As long as that distant object is a point source (like a star) that lies on the optical axis. Any extended object (regardless of distance) will result in rays entering at different angles, and a point source off of the optical axis will create an image that isn't at the focal point.

But aren't those different angles from a far far away object like so close to parallel that it makes essentially no difference? I guess you're saying not. And, for example, Andromeda's 3° by 1° areal extent is not negligible at all?

Chris Peterson wrote: ↑Sun Nov 10, 2024 2:30 pm

johnnydeep wrote: ↑Sun Nov 10, 2024 2:25 pm

Chris Peterson wrote: ↑Sat Nov 09, 2024 10:55 pm

Rays converge at the focal plane, not the focal point. Otherwise you'd never have an image at the focal plane. A bundle of rays around the optical axis will converge at the focal point. Parallel bundles coming at other angles will converge above or below the focal point, but nominally on a plane (ignoring aberrations like field curvature). Google optical ray trace simulator or something like that. There seem to be a bunch of web-based ones out there you could play with.

Ok, thanks. So...googling... all parallel rays entering at a given angle will converge at a specific point on the focal plane. But since all rays from a very remote object are parallel to the optical axis, they will all converge at the same (focal) point, right?

As I keep saying, I will never understand optics.

As long as that distant object is a point source (like a star) that lies on the optical axis. Any extended object (regardless of distance) will result in rays entering at different angles, and a point source off of the optical axis will create an image that isn't at the focal point.

beryllium732 wrote: ↑Sun Nov 10, 2024 3:39 am
Can't thank you enough! You bring joy to this forum and always helpful! It's crazy that Proxima Centauri is such a small star. I wonder if all those red M-stars since the early days have calmed down enough that it gives chance for life to develop on any planets they might contain. And the the night sky would really look bright and spectacular back then with even gas clouds shining bright from all the o-stars but at the same time it also would be deadlier time because of the novas and supernovas going off in the vicinity.

Recently one astronomy team say they found out that stars back then was even hotter than the hottest stars we have found today, so much that they will light up gas clouds which in turn shines brighter than combined stars themselves. You never get tired getting fascinated of the universe!

John Bingamon answered the question, How long does it take for a red dwarf to calm down from being a flare star?

Answer: Nobody knows.

There is considerable evidence that many, probably most, and possibly all, YOUNG red dwarf stars exhibit a LOT of flaring - both much more frequent and more violent than solar flares, despite the smaller size of these stars.
This flaring is driven by magnetic fields, which in turn are driven by the rapid rotation of the stars, and their fully convective interiors. Pretty much all stars have part of their interiors convective, and START their lives with rapid rotation. Thus, young stars have flaring (many readers who are parents will likely shudder and nod their heads at this “terrible twos temper tantrums” behavior). But, they soon slow down, magnetic fields weaken or at least are more well-behaved, and flaring dies down with it.

However, red dwarfs continue to spin rapidly for far longer, and frequent flares and “super-flares” continue.
(...)

More importantly, even if it would still be hard for life to develop then, if the flaring doesn’t blast away the planets’ atmospheres, AND the flaring dies down after a few billion years, then AFTER that life could develop there, and still look forward to hundreds of billions of years of fairly safe conditions.

That could make red dwarfs the most common places for life in the universe, possibly already, and if not, in the future universe.

Wikipedia wrote:

Modern evidence suggests that planets in red dwarf systems are extremely unlikely to be habitable. In spite of their great numbers and long lifespans, there are several factors which may make life difficult on planets around a red dwarf. First, planets in the habitable zone of a red dwarf would be so close to the parent star that they would likely be tidally locked. For a nearly circular orbit, this would mean that one side would be in perpetual daylight and the other in eternal night. This could create enormous temperature variations from one side of the planet to the other. Such conditions would appear to make it difficult for forms of life similar to those on Earth to evolve.

And it appears there is a great problem with the atmosphere of such tidally locked planets: the perpetual night zone would be cold enough to freeze the main gases of their atmospheres, leaving the daylight zone bare and dry. On the other hand, though, a theory proposes that either a thick atmosphere or planetary ocean could potentially circulate heat around such a planet.

Wikipedia wrote:

N6946-BH1 is a disappearing supergiant star and failed supernova candidate formerly seen in the galaxy NGC 6946, on the northern border of the constellation of Cygnus...

In March through to May 2009 its bolometric luminosity increased to at least a million solar luminosities, but by 2015 it had disappeared from optical view...

The brightening was insufficient to be a supernova; the process that created the outburst is still uncertain...

One hypothesis is that of the failed supernova. In this scenario, the core of the star collapsed to form a black hole. The collapsing matter formed a burst of neutrinos that lowered the total mass of the star by a fraction of a percent. This caused a shock wave that blasted out the star's envelope to make it brighter. N6946-BH1 has supplied evidence contrary to the conventional idea that black holes are usually formed after a supernova, suggesting instead that a star may bypass this eventuality and yet collapse into a black hole.

johnnydeep wrote: ↑Sun Nov 10, 2024 2:25 pm

Chris Peterson wrote: ↑Sat Nov 09, 2024 10:55 pm

johnnydeep wrote: ↑Sat Nov 09, 2024 10:28 pm

So any ray entering the objective lens at any angle will converge at the focal point (more or less, depending on "aberration" perhaps?)? I suppose there's a way to calculate the exact path of any given ray...I need an interactive model to play with!


telescope magnification angles.jpg

Rays converge at the focal plane, not the focal point. Otherwise you'd never have an image at the focal plane. A bundle of rays around the optical axis will converge at the focal point. Parallel bundles coming at other angles will converge above or below the focal point, but nominally on a plane (ignoring aberrations like field curvature). Google optical ray trace simulator or something like that. There seem to be a bunch of web-based ones out there you could play with.

Ok, thanks. So...googling... all parallel rays entering at a given angle will converge at a specific point on the focal plane. But since all rays from a very remote object are parallel to the optical axis, they will all converge at the same (focal) point, right?

As I keep saying, I will never understand optics.

Chris Peterson wrote: ↑Sat Nov 09, 2024 10:55 pm

johnnydeep wrote: ↑Sat Nov 09, 2024 10:28 pm

Chris Peterson wrote: ↑Sat Nov 09, 2024 7:42 pm

There's nothing special about them. A ray entering straight will exit straight. No matter what angle a ray comes in, it has to exit at a steeper angle, and that angle is just the input angle times the angular magnification.

So any ray entering the objective lens at any angle will converge at the focal point (more or less, depending on "aberration" perhaps?)? I suppose there's a way to calculate the exact path of any given ray...I need an interactive model to play with!


telescope magnification angles.jpg

Rays converge at the focal plane, not the focal point. Otherwise you'd never have an image at the focal plane. A bundle of rays around the optical axis will converge at the focal point. Parallel bundles coming at other angles will converge above or below the focal point, but nominally on a plane (ignoring aberrations like field curvature). Google optical ray trace simulator or something like that. There seem to be a bunch of web-based ones out there you could play with.

Ann wrote: ↑Sat Nov 09, 2024 7:39 am

beryllium732 wrote: ↑Fri Nov 08, 2024 10:14 pm

Ann wrote: ↑Fri Nov 08, 2024 3:43 pm

Oops! Yes, it looks like it! Thanks, Johnny!

Ann

Crazy! So by it's sheer size Andromeda still has a lot of stars compared to our galaxy. That's mind boggling! You can be fooled that seeing that picture you would think it's a ellipctical galaxy and not a spiral galaxy with arms. I wonder what galaxies looked like within the first two billion of years of Bing Bang.

They've must been really bright and visible back then and both surface brightness and have a absolut magnitude well under zero. Really interesting stuff you're posting always great quality!

Thanks, Beryllium!

Yes, the first galaxies must have been bursting with brilliant nebulas, giant star clusters and very, very massive stars.

The first galaxies must also have been small, because they had not had time to grow very large. After all, galaxies grow by absorbing more and more gas and forming more and more stars. Also they grow by merging with other galaxies.

In the nearby universe, starburst galaxies are also typically small. That's because today's large galaxies have typically undergone so many episodes of star formation and several minor mergers that they have, frankly, "used up" most of their gas. And one way that they use up their gas is by forming so many small red M-type dwarf stars! The red dwarf stars are surprisingly massive for their faint light.

Consider the very nearest star after the Sun, Proxima Centauri:

The colors and sizes of nearby stars, including Proxima Centauri. Note that Sirius is not a yellow-white star, as the illustration suggests, and the Sun is not that yellow. Credit: ESO.

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As the Color Commentator, I feel compelled to show you that Sirius is not a yellow-white star:

Spectrum of Sirius. The light output of Sirius culminates in the near ultraviolet part of the spectrum. It produces much more blue than red and yellow light. Credit: SpectraWizard/StellarNet.us

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Back to Proxima Centauri. Take a look at this picture to see how faint it is (Proxima is located in the center of the red circle). The bright star at left is the binary star Alpha Centauri A+B, located at approximately the same distance from us as Proxima.

Alpha Centauri A+B (left), distant Beta Centauri (right) and Proxima Centauri in the center of the red circle. Credit: Skatebiker at English Wikipdia.

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Alpha Centauri A just a tiny bit more massive than the Sun, 1.0788±0.0029 solar, but its light output is about 1.5 times the luminosity of the Sun. Alpha Centauri B is a tiny bit less massive than the Sun, 0.9092±0.0025 the mass of the Sun, but its light output is only about 0.5 times the luminosity of the Sun.

What about Proxima? Its mass is only 0.1221±0.0022 the mass of the Sun, some 12% the mass of the Sun. The bolometric (total) luminosity of Proxima, including the infrared light it emits, is 0.001567±0.000020 solar, or ~ 1.6 parts per mille of the solar energy output. But the visual luminosity is only 0.00005 L_☉. It is frankly extremely faint!

Here is the deal. The tiny red stars like Proxima Centauri (and even the red M-type dwarfs that are considerably more massive than Proxima, some 0.5 times the mass of the Sun) fuse their hydrogen to helium so slowly that their hydrogen supply lasts for many times the current age of the Universe. That means that not a single red dwarf star that has ever been born in our Universe has died, unless it has fallen into a black hole or something. And since the red dwarf stars happily hold on to their gas supply as they slowly, slowly fuse a bit of hydrogen to helium, this in turn means that practically every hydrogen atom that has ever been trapped inside a red dwarf star is trapped there "forever", either to remain unchanged or to be fused into helium. Also, by far most stars that belong to our own galaxy (some 80%) are in fact red dwarfs.

If you ask me, I would guess that most of the hydrogen that was formed in the Big Bang has since been trapped inside red dwarf stars. And thus, this hydrogen can't be used for any fresh star formation again, at least not until the Universe is several times its current age. By contrast, massive stars live short lives and give back much of their gas to the Universe, either because they explode or, far far more common, because they shed their outer layers when they become white dwarfs.

The central star of the planetary nebula NGC 6302 is forcefully giving back much of its gas to the universe as it is turning into a white dwarf star. Credit: NASA, ESA and the Hubble SM4 ERO Team

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Now let's consider large galaxies versus small galaxies. In the large galaxies, many things have already happened, repeated episodes of star formation have taken place, and there have been galactic mergers. The gas in large galaxies gets jostled round, either so that it concentrates in certain places and gives rise to star formation again, or so that it gets so turbulent and hot that it can't form stars any more, certainly not for a long time.

Also remember that every episode of star formation creates a large number of new red dwarfs, so for each episode of star formation, more gas is "trapped" and made unavailable for new star formation. Take a look at this JWST image of cluster NGC 602 in the Small Magellanic Cloud. Look how many small stars are formed for every large star (and then remember that infrared-detecting JWST will make the red dwarfs look brighter):

Cluster NGC 602 in the Small Magellanic Cloud. Credit: ESA/Webb, NASA & CSA, P. Zeidler, E. Sabbi, A. Nota, M. Zamani (ESA/Webb)

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But in the nearby universe, small galaxies are small because they haven't had much activity going on inside them for a long time. They haven't had a new episode of star formation for a long time, and they haven't had any mergers. Many small galaxies have an untapped reservoir of gas that they could use to create new stars, if they just get the right trigger.

Take a look at this diagram of the star formation history of the Large Magellanic Cloud:

Star formation history of the Large Magellanic Cloud Sidney van den Bergh.jpg

The star formation history of the Large Magellanic Cloud.
Credit: Sidney van den Bergh.

The diagram of the star formation history of the Large Magellanic Cloud, which I photographed from a book, goes from right to left, so that the oldest episodes of star formation are at right and the more recent ones at left. As you can see, the Large Magellanic Cloud formed few or no stars for much or most of its history. (Please note that when Sidney van den Bergh created his diagram, the Universe was believed to be 20 billion years old. Now it is considered to be less than 14 billion years old.)

If Sidney van den Bergh is right about the star formation history of the Large Magellanic Cloud, then it is perhaps possible that the LMC used to look like NGC 205 (the largest satellite galaxy of Andromeda) when it was at the end of its longest period of inactivity. Maybe?

Is NGC 205 all red and dead, then? Not totally red and dead, because there are a few small dust lanes near the center of NGC 205, and at the very center, there is a tiny bluish (i.e., relatively young) population.

NGC 205 in visible light. Note the dust lanes and the very small bluish population at the center of the galaxy. Credit: KPNO/NOIRLab/NSF/AURA/Adam Block

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NGC 205, the largest satellite galaxy of Andromeda. Maybe the Large Magellanic Cloud used to look like this when it was at the end of its longest period of inactivity? Credit: Two Micron All Sky Survey.

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But the reason why NGC 205 is "almost red and dead" is that it has undoubtedly orbited Andromeda for quite a long time, and it has been "harassed" into losing most of its gas. The Large Magellanic Cloud, by contrast, has only recently started interacting with the Small Magellanic Cloud, and it has been captured by the Milky Way even more recently.


Anyway, the Large Magellanic Cloud undoubtedly had a considerable gas reservoir of its own even before it started interacting with the Small Magellanic Cloud. But the interaction has rejuvenated the LMC and given it "fresh blood", as it has been proved that gas is streaming from the Small Cloud to the Large Cloud, and the Large Cloud has become a starburst galaxy.

Starbursting Large Magellanic Cloud. Ground-based image from Eckhard Slawik.

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Gas and stars are streaming from the Small Magellanic Cloud to the Large Magellanic Cloud. Credit: V. Belokurov / D. Erkal / A. Mellinger.

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So, yes. The early Universe was chock full of gas, and it must have been full of small starbursting galaxies.

Small starburst galaxies. Credit for Henize 2-10: X-ray (NASA/CXC/Virginia/A.Reines et al); Radio (NRAO/AUI/NSF); Optical (NASA/STScI). Credit for NGC 4214: NASA, ESA and the Hubble Heritage Team (STScI/AURA)-ESA/Hubble Collaboration Acknowledgment: R. O’Connell (University of Virginia) and the WFC3 Scientific Oversight Committee. Credit for NGC 4038/4039: Probably ESA/Hubble, and credit for NGC 1313: VLT/ESO, Henri Boffin.

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So, yes. That is probably what the early universe looked like.

Ann

johnnydeep wrote: ↑Sat Nov 09, 2024 10:28 pm

Chris Peterson wrote: ↑Sat Nov 09, 2024 7:42 pm

johnnydeep wrote: ↑Sat Nov 09, 2024 5:43 pm

Thanks. That would almost makes sense if I only I could understand why those particular light ray lines are the appropriate ones to choose to indicate that this is actually happening! The rays seem suspiciously convenient.

There's nothing special about them. A ray entering straight will exit straight. No matter what angle a ray comes in, it has to exit at a steeper angle, and that angle is just the input angle times the angular magnification.

So any ray entering the objective lens at any angle will converge at the focal point (more or less, depending on "aberration" perhaps?)? I suppose there's a way to calculate the exact path of any given ray...I need an interactive model to play with!


telescope magnification angles.jpg

Chris Peterson wrote: ↑Sat Nov 09, 2024 7:42 pm

johnnydeep wrote: ↑Sat Nov 09, 2024 5:43 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 11:39 pm
Maybe the easiest way to imagine it is by considering the angle of rays. A ray coming into the objective at some angle will exit the eyepiece at a steeper angle (assuming the focal length of the eyepiece is less than that of the objective). A fan of rays spanning a wider angle is the same thing you'd get (without the telescope) if the object were closer or larger.
_
TelescopeMagnification_p1small.gif

Thanks. That would almost makes sense if I only I could understand why those particular light ray lines are the appropriate ones to choose to indicate that this is actually happening! The rays seem suspiciously convenient.

There's nothing special about them. A ray entering straight will exit straight. No matter what angle a ray comes in, it has to exit at a steeper angle, and that angle is just the input angle times the angular magnification.

johnnydeep wrote: ↑Sat Nov 09, 2024 5:43 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 11:39 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 10:32 pm

Al right. My problem is that I keep ignoring the effect that the magnification has. And frankly, now that I realize that, I realize I'm not even clear how the magnification is happening, in apparent opposition to the fact that the objective lens is larger than the eyepiece lens. So, why does the image appear magnified?

Maybe the easiest way to imagine it is by considering the angle of rays. A ray coming into the objective at some angle will exit the eyepiece at a steeper angle (assuming the focal length of the eyepiece is less than that of the objective). A fan of rays spanning a wider angle is the same thing you'd get (without the telescope) if the object were closer or larger.
_
TelescopeMagnification_p1small.gif

Thanks. That would almost makes sense if I only I could understand why those particular light ray lines are the appropriate ones to choose to indicate that this is actually happening! The rays seem suspiciously convenient.

Chris Peterson wrote: ↑Fri Nov 08, 2024 11:39 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 10:32 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 10:01 pm

That is what's happening. Absolutely, more photons are entering the eye. But the photon flux per unit area doesn't change.

Aperture: 50mm
F1 (objective): 500mm
F2 (eyepiece): 50mm
Pupil size: 5mm

The magnification of this system is 10X. The pupil of the eye maps exactly back to the diameter of the objective, so 100% of the light hitting the aperture is transmitted to the eye. The telescope aperture collects 100 times more photons than the eye alone, and all that extra light enters the eye. The image on the retina is 10X larger than without the telescope, so it covers 100X more area. So any spot on the retina sees exactly the same brightness, telescope or no telescope. Can you find some combination of aperture and focal lengths that ever results in a retinal image with a greater flux per unit area? Do you see that if we took the system described here and made the aperture 100mm there would be no difference at the retina?

Al right. My problem is that I keep ignoring the effect that the magnification has. And frankly, now that I realize that, I realize I'm not even clear how the magnification is happening, in apparent opposition to the fact that the objective lens is larger than the eyepiece lens. So, why does the image appear magnified?

Maybe the easiest way to imagine it is by considering the angle of rays. A ray coming into the objective at some angle will exit the eyepiece at a steeper angle (assuming the focal length of the eyepiece is less than that of the objective). A fan of rays spanning a wider angle is the same thing you'd get (without the telescope) if the object were closer or larger.
_
TelescopeMagnification_p1small.gif

beryllium732 wrote: ↑Fri Nov 08, 2024 10:14 pm

Ann wrote: ↑Fri Nov 08, 2024 3:43 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 1:22 pm

You're off by 1000 or so.

Oops! Yes, it looks like it! Thanks, Johnny!

Ann

Crazy! So by it's sheer size Andromeda still has a lot of stars compared to our galaxy. That's mind boggling! You can be fooled that seeing that picture you would think it's a ellipctical galaxy and not a spiral galaxy with arms. I wonder what galaxies looked like within the first two billion of years of Bing Bang.

They've must been really bright and visible back then and both surface brightness and have a absolut magnitude well under zero. Really interesting stuff you're posting always great quality!

johnnydeep wrote: ↑Fri Nov 08, 2024 10:32 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 10:01 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 9:52 pm

Yes, exactly as shown in the image: all those photons in the entire larger area of the objective lens are compressed down to the smaller area of the eyepiece lens. Why isn't that what's actually happening?


telescope and eye.jpg

That is what's happening. Absolutely, more photons are entering the eye. But the photon flux per unit area doesn't change.

Aperture: 50mm
F1 (objective): 500mm
F2 (eyepiece): 50mm
Pupil size: 5mm

The magnification of this system is 10X. The pupil of the eye maps exactly back to the diameter of the objective, so 100% of the light hitting the aperture is transmitted to the eye. The telescope aperture collects 100 times more photons than the eye alone, and all that extra light enters the eye. The image on the retina is 10X larger than without the telescope, so it covers 100X more area. So any spot on the retina sees exactly the same brightness, telescope or no telescope. Can you find some combination of aperture and focal lengths that ever results in a retinal image with a greater flux per unit area? Do you see that if we took the system described here and made the aperture 100mm there would be no difference at the retina?

Al right. My problem is that I keep ignoring the effect that the magnification has. And frankly, now that I realize that, I realize I'm not even clear how the magnification is happening, in apparent opposition to the fact that the objective lens is larger than the eyepiece lens. So, why does the image appear magnified?

Chris Peterson wrote: ↑Fri Nov 08, 2024 10:01 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 9:52 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 9:44 pm
An objective lens by itself is not a telescope. A single objective is a focal optical system. A telescope is an afocal optical system. The only reason a telescope forms an image is because the lens of your eye is a focal optical system.

Again, work backwards from your retina. Can you figure out any way that you can get more photons per unit area on your retina with a telescope?

Yes, exactly as shown in the image: all those photons in the entire larger area of the objective lens are compressed down to the smaller area of the eyepiece lens. Why isn't that what's actually happening?


telescope and eye.jpg

That is what's happening. Absolutely, more photons are entering the eye. But the photon flux per unit area doesn't change.

Aperture: 50mm
F1 (objective): 500mm
F2 (eyepiece): 50mm
Pupil size: 5mm

The magnification of this system is 10X. The pupil of the eye maps exactly back to the diameter of the objective, so 100% of the light hitting the aperture is transmitted to the eye. The telescope aperture collects 100 times more photons than the eye alone, and all that extra light enters the eye. The image on the retina is 10X larger than without the telescope, so it covers 100X more area. So any spot on the retina sees exactly the same brightness, telescope or no telescope. Can you find some combination of aperture and focal lengths that ever results in a retinal image with a greater flux per unit area? Do you see that if we took the system described here and made the aperture 100mm there would be no difference at the retina?

Ann wrote: ↑Fri Nov 08, 2024 3:43 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 1:22 pm

Ann wrote:So, how many stars are there in M32? According to Hubblesite, there are roughly 400 million stars within a diameter of only 1,000 light-years in M32! That 's 400,000,000,000 stars. Therefore, if only one out of 400,000 stars is a red giant, there will still be 100,000 1000 red giants within a volume with a diameter of only 1,000 light-years in M32. And a thousand light-years is roughly the distance from here to the bright stars in Orion. Imagine having 100,000 1000 red giant stars between us and Orion. Imagine having 100,000 1000 stars as bright as Arcturus (some 100 times brighter than the Sun) between us and Orion.

You're off by 1000 or so.

Oops! Yes, it looks like it! Thanks, Johnny!

Ann

johnnydeep wrote: ↑Fri Nov 08, 2024 9:52 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 9:44 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 9:30 pm

Still making no sense to me. If I put my pupil at the focal point of the objective lens in the picture, and we're pointed at the Sun, my eye would be severely burned (as would a piece of paper placed there). Right? So why isn't that intensity merely a bit reduced by the expansion of the rays on the way from that focal point to the eyepiece any my eye?

An objective lens by itself is not a telescope. A single objective is a focal optical system. A telescope is an afocal optical system. The only reason a telescope forms an image is because the lens of your eye is a focal optical system.

Again, work backwards from your retina. Can you figure out any way that you can get more photons per unit area on your retina with a telescope?

Yes, exactly as shown in the image: all those photons in the entire larger area of the objective lens are compressed down to the smaller area of the eyepiece lens. Why isn't that what's actually happening?


telescope and eye.jpg

Chris Peterson wrote: ↑Fri Nov 08, 2024 9:44 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 9:30 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 7:49 pm
You certainly collect more photons. But you distribute them over a larger area. That's why there's no change in brightness.

Say you have a dark adapted pupil size of 6mm. With a scope operating at 100X, you'd need an objective of 600mm to achieve the same surface brightness on your retina that your eye would have without the scope. 10,000 times more photons, but spread out over 10,000 time more area. Same photon flux per unit area on your retina. Anything less than that aperture and the brightness will be less bright. Anything more won't change the brightness beyond that of the 600mm objective because the light at the edges will be outside your pupil.

Still making no sense to me. If I put my pupil at the focal point of the objective lens in the picture, and we're pointed at the Sun, my eye would be severely burned (as would a piece of paper placed there). Right? So why isn't that intensity merely a bit reduced by the expansion of the rays on the way from that focal point to the eyepiece any my eye?

An objective lens by itself is not a telescope. A single objective is a focal optical system. A telescope is an afocal optical system. The only reason a telescope forms an image is because the lens of your eye is a focal optical system.

Again, work backwards from your retina. Can you figure out any way that you can get more photons per unit area on your retina with a telescope?

johnnydeep wrote: ↑Fri Nov 08, 2024 9:30 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 7:49 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 7:40 pm

Well, what if the eyepiece diameter is the same size as the pupil? Isn't it receiving more photons per area than it would without the optics in place?


telescope and eye.jpg

You certainly collect more photons. But you distribute them over a larger area. That's why there's no change in brightness.

Say you have a dark adapted pupil size of 6mm. With a scope operating at 100X, you'd need an objective of 600mm to achieve the same surface brightness on your retina that your eye would have without the scope. 10,000 times more photons, but spread out over 10,000 time more area. Same photon flux per unit area on your retina. Anything less than that aperture and the brightness will be less bright. Anything more won't change the brightness beyond that of the 600mm objective because the light at the edges will be outside your pupil.

Still making no sense to me. If I put my pupil at the focal point of the objective lens in the picture, and we're pointed at the Sun, my eye would be severely burned (as would a piece of paper placed there). Right? So why isn't that intensity merely a bit reduced by the expansion of the rays on the way from that focal point to the eyepiece any my eye?

Chris Peterson wrote: ↑Fri Nov 08, 2024 7:49 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 7:40 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 7:12 pm
Yes... but how much of that light is making it into the eye? All of the light that is outside the diameter of your pupil is lost. That's why you have to project the size of your pupil backwards through the system. Doing that you'll see that the effective aperture is not the physical aperture, but the magnification times your pupil size. Which can be much smaller. And more light spread out over more area doesn't mean anything got brighter.

There's a reason that you can look at the Moon through a giant telescope and not incur damage. It's because that telescope doesn't increase the brightness of the Moon at all. It can't.

Well, what if the eyepiece diameter is the same size as the pupil? Isn't it receiving more photons per area than it would without the optics in place?


telescope and eye.jpg

You certainly collect more photons. But you distribute them over a larger area. That's why there's no change in brightness.

Say you have a dark adapted pupil size of 6mm. With a scope operating at 100X, you'd need an objective of 600mm to achieve the same surface brightness on your retina that your eye would have without the scope. 10,000 times more photons, but spread out over 10,000 time more area. Same photon flux per unit area on your retina. Anything less than that aperture and the brightness will be less bright. Anything more won't change the brightness beyond that of the 600mm objective because the light at the edges will be outside your pupil.

johnnydeep wrote: ↑Fri Nov 08, 2024 7:40 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 7:12 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 7:05 pm

How does that pertain to this image, which sure looks like it is concentrating more photons onto a lesser area:

Click to view full size image

Yes... but how much of that light is making it into the eye? All of the light that is outside the diameter of your pupil is lost. That's why you have to project the size of your pupil backwards through the system. Doing that you'll see that the effective aperture is not the physical aperture, but the magnification times your pupil size. Which can be much smaller. And more light spread out over more area doesn't mean anything got brighter.

There's a reason that you can look at the Moon through a giant telescope and not incur damage. It's because that telescope doesn't increase the brightness of the Moon at all. It can't.

Well, what if the eyepiece diameter is the same size as the pupil? Isn't it receiving more photons per area than it would without the optics in place?


telescope and eye.jpg

Chris Peterson wrote: ↑Fri Nov 08, 2024 7:12 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 7:05 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 5:52 pm

Ok, now make your 1-m telescope have a magnification of two. Propagate your eye's pupil backwards through that system, and you'll see that the only light entering your eye is from a central disk on the objective that is twice the diameter of your pupil. Does that make sense? So you are collecting four times more light (woo hoo!) and distributing it across four times the area on your retina (oops!). The size is doubled, but the surface brightness is exactly the same. The photon flux encountered by a rod cell in your retina is exactly the same.

How does that pertain to this image, which sure looks like it is concentrating more photons onto a lesser area:

Click to view full size image

Yes... but how much of that light is making it into the eye? All of the light that is outside the diameter of your pupil is lost. That's why you have to project the size of your pupil backwards through the system. Doing that you'll see that the effective aperture is not the physical aperture, but the magnification times your pupil size. Which can be much smaller. And more light spread out over more area doesn't mean anything got brighter.

There's a reason that you can look at the Moon through a giant telescope and not incur damage. It's because that telescope doesn't increase the brightness of the Moon at all. It can't.

johnnydeep wrote: ↑Fri Nov 08, 2024 7:05 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 5:52 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 5:46 pm

Well, yes, I suppose if you put your face in the path of the eyepiece your eyes would get the same number of photons per unit area as if there were no telescope at all. But I think of telescopes with magnification > 1 as concentrating the photons into a smaller area, thus increasing brightness. Is that not correct?

Ok, now make your 1-m telescope have a magnification of two. Propagate your eye's pupil backwards through that system, and you'll see that the only light entering your eye is from a central disk on the objective that is twice the diameter of your pupil. Does that make sense? So you are collecting four times more light (woo hoo!) and distributing it across four times the area on your retina (oops!). The size is doubled, but the surface brightness is exactly the same. The photon flux encountered by a rod cell in your retina is exactly the same.

How does that pertain to this image, which sure looks like it is concentrating more photons onto a lesser area:

Click to view full size image

Chris Peterson wrote: ↑Fri Nov 08, 2024 5:52 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 5:46 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 5:28 pm

Maybe it would help if you considered a telescope with a one meter aperture, and an objective and an eyepiece with the same focal length. That is, a one meter aperture and a magnification of one. Can you see why that won't produce an image brighter than your eye alone?

Well, yes, I suppose if you put your face in the path of the eyepiece your eyes would get the same number of photons per unit area as if there were no telescope at all. But I think of telescopes with magnification > 1 as concentrating the photons into a smaller area, thus increasing brightness. Is that not correct?

Ok, now make your 1-m telescope have a magnification of two. Propagate your eye's pupil backwards through that system, and you'll see that the only light entering your eye is from a central disk on the objective that is twice the diameter of your pupil. Does that make sense? So you are collecting four times more light (woo hoo!) and distributing it across four times the area on your retina (oops!). The size is doubled, but the surface brightness is exactly the same. The photon flux encountered by a rod cell in your retina is exactly the same.

johnnydeep wrote: ↑Fri Nov 08, 2024 5:46 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 5:28 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 5:26 pm

I fear I will never understand this point that you're always making.

Maybe it would help if you considered a telescope with a one meter aperture, and an objective and an eyepiece with the same focal length. That is, a one meter aperture and a magnification of one. Can you see why that won't produce an image brighter than your eye alone?

Well, yes, I suppose if you put your face in the path of the eyepiece your eyes would get the same number of photons per unit area as if there were no telescope at all. But I think of telescopes with magnification > 1 as concentrating the photons into a smaller area, thus increasing brightness. Is that not correct?

Chris Peterson wrote: ↑Fri Nov 08, 2024 5:28 pm

johnnydeep wrote: ↑Fri Nov 08, 2024 5:26 pm

Chris Peterson wrote: ↑Fri Nov 08, 2024 4:20 pm

It's worth mentioning (again) that telescopes don't make anything brighter, only bigger. A perfect telescope produces an image on the back of the eye with the same brightness as the eye alone receives. In actual life, telescopes aren't perfect, and the image is always somewhat less bright. But we don't necessarily need brighter, what we need is more information, and if you can make an image bigger on the retina, our brains have more to work with, and we may see more, even without anything being brighter. But you can only increase the magnification of a telescope to a certain point before the image becomes dimmer, and that point is determined by the aperture of the telescope. Large apertures demanded technological changes that took time after telescopes were first invented. Technology in making large accurate mirrors, and technology in grinding lenses bigger than just a few centimeters. That's what allowed for high magnifications without a reduction in brightness, and the first visual observations of subtle detail in astronomical objects.

I fear I will never understand this point that you're always making.

Maybe it would help if you considered a telescope with a one meter aperture, and an objective and an eyepiece with the same focal length. That is, a one meter aperture and a magnification of one. Can you see why that won't produce an image brighter than your eye alone?