My question is: Where in the optical train does the image invert? Is it at the last piece of glass before heading to the plane of precise focus and if so, is the image size minuscule and then does it grow in size from that point to the in focus image circle? My reason for asking is that I have a 2 inch filter wheel mounted to the ASI6200MM Pro, but have a 60mm image circle produced by the scope. Trying to wrap my brain around what is going on.
Image inversion in the optical train
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It’s really hard to answer your question at face value. It just doesn’t make any sense. The chief ray which passes through the stop and the edge of the sensor is what determines the field angle and ultimately the image orientation; but, it doesn’t sound like that’s what you are really asking. It sounds more like you are trying to understand vignetting but again, you are using terms that are confusing your question. The image circle of the scope defines the well-corrected field that the scope is capable of producing. It has nothing to do with the sensor size, which for your camera (which I believe has an IMX455 sensor) has a diagonal of about 42 mm. As long as your filters are kept fairly close to the camera sensor, you should not get excessive vignetting (or any at all) with 2” filters.
I suggest that you watch the presentation that I did on first order optics for the Astro Imaging Channel. It’s on YouTube and you can just Google search for it. It will to a long way toward helping you to better understand the optics in your system and make it easier to ask clear questions.
Good luck with it!
John
I suggest that you watch the presentation that I did on first order optics for the Astro Imaging Channel. It’s on YouTube and you can just Google search for it. It will to a long way toward helping you to better understand the optics in your system and make it easier to ask clear questions.
Good luck with it!
John
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Still looking for the Youtube piece you described. I thought it made perfect sense. Is the image a 60mm circle all the way down the tubes and components once it gets beyond the last correcting lens, or is it a cone that is getting ever larger until it becomes the 60mm image circle at the sensor?
Saw your piece and it's quite a bit above my understanding, not being a physicist. There has to be a far easier answer than that. Yes, my concern is vignetting.
DavesView:
Still looking for the Youtube piece you described. I thought it made perfect sense. Is the image a 60mm circle all the way down the tubes and components once it gets beyond the last correcting lens, or is it a cone that is getting ever larger until it becomes the 60mm image circle at the sensor?
*The corrected image circle is at the focus location (focal plane/surface) and nowhere else, as it doesn't make much sense what happens elsewhere as the image won't be in focus. To know whether a given filter vignettes you need to know the filter clear diameter, the distance between it and the sensor and the focal ratio of the optical system.
If you ray race a Newtonian or a standard refactor you'll see the the light after the lens or mirror forms a cone who's angle of convergence is determined by the focal length. Compound reflectors such as Cassegrains will have two cones, one from the primary and one from the secondary. The only exception might be a petzval refractor, I'm not sure.
Two cones? Doesn't make sense. And why a petzal refractor should be any different from a standard one?
DavesView:
Still looking for the Youtube piece you described. I thought it made perfect sense. Is the image a 60mm circle all the way down the tubes and components once it gets beyond the last correcting lens, or is it a cone that is getting ever larger until it becomes the 60mm image circle at the sensor?
The radius that you need is the sum of the chief ray height and the margial ray height through the system. It will be a conical shape. You can find the presentation here:
https://www.youtube.com/watch?v=M7JvBG_HQuY..
The 60 mm image circle is only relevant to the field correction in the design. It has nothing to do with determining field size. The sensor determines the field size. Watch that presentation and I hope that it will help to clear up some of this stuff.
John
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andrea tasselli:
Two cones? Doesn't make sense. And why a petzal refractor should be any different from a standard one?
I was referring to a compound system, aka Cassegrain. There's a light cone from the primary and another cone from the secondary with a different angle of convergence.
A compound system isn't necessarily having a Cassegrain form, could be a Gregory form as well, for a 2 mirror system with both having optical power. Speaking of two cones makes no sense and it is just confusing as there is only one converging bundle of light rays forming a real image at the system's focus, such that the final focal length is given by:
f = f1*f2/(f1-f2-d), for both forms, where f is the final focal length of the system, d is the separation of the mirrors and f1 and f2 are the focal lengths of primary and secondary.
f = f1*f2/(f1-f2-d), for both forms, where f is the final focal length of the system, d is the separation of the mirrors and f1 and f2 are the focal lengths of primary and secondary.
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andrea tasselli:
A compound system isn't necessarily having a Cassegrain form, could be a Gregory form as well, for a 2 mirror system with both having optical power. Speaking of two cones makes no sense and it is just confusing as there is only one converging bundle of light rays forming a real image at the system's focus, such that the final focal length is given by:
f = f1*f2/(f1-f2-d), for both forms, where f is the final focal length of the system, d is the separation of the mirrors and f1 and f2 are the focal lengths of primary and secondary.
When I say two cones I don't think it's confusing at all. There's the cone from the primary and the cone from the secondary that have different angles of convergence. It seems perfectly logical to take about them that way so I don't know why you find it confusing but, each to his own.
At any rate, I think the OP's original question has been answered well enough.