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Say I have an object at 40 cm from a convex 20 cm focal length lens, then a concave -30 cm focal length lens 10 cm behind that. This would give an image at infinity, which one could take a picture of, but around what magnification should they expect (there would be some leeway in the result if since these measurements wouldn't be exact)? Saying infinite magnification like from normal lens equations wouldn't make sense.

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    \$\begingroup\$ I’m voting to close this question because it has almost nothing to do with the practice of actual photography, as outlined in the group's description. This reads like someone's homework assignment for an optics class. \$\endgroup\$
    – Michael C
    Commented Jan 26, 2023 at 5:50
  • \$\begingroup\$ It's for photographing at infinity in a closed space, no homework \$\endgroup\$
    – user109880
    Commented Jan 26, 2023 at 15:12
  • \$\begingroup\$ Again, is what you are doing the kind of photography defined in our description, or something else? Machine vision (not really on subject here)? Security video (definitely not on subject here)? Something else? \$\endgroup\$
    – Michael C
    Commented Jan 26, 2023 at 15:16
  • \$\begingroup\$ I want to check the focus of a camera at infinity in a small space. I would think adjusting the focus is part of photography. \$\endgroup\$
    – user109880
    Commented Jan 26, 2023 at 21:13
  • \$\begingroup\$ where is the description? \$\endgroup\$
    – user109880
    Commented Jan 26, 2023 at 21:47

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If you mount this lens set like a filter, with the "infinity" side toward your camera's lens, you'd get to photograph an object 20 cm away with infinity focus on the camera's own lens.

This is a Newtonian refracting telescope, in essence, so you'd use diopter arithmetic and then convert diopters to magnification: you have a +5 (= 2.25x and a - 3.33 diopter (= ~1.8x) (diopter strength is the inverse of focal length in meters, while magnification is diopter strength divided by 4, then add 1) -- but with the negative lens acting like the eyepiece of the Newtonian telescope, its power would be the negative of its strength. That means you'd have about 4x for the combination.

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  • \$\begingroup\$ Thanks, do you have a reference to go with your explanation? \$\endgroup\$
    – user109880
    Commented Jan 24, 2023 at 20:05
  • \$\begingroup\$ endmemo.com/bio/microscope.php and bolioptics.com/blog/how-to-convert-diopters-to-magnification -- which latter suggests I misremembered. Editing... \$\endgroup\$
    – Zeiss Ikon
    Commented Jan 24, 2023 at 20:14
  • \$\begingroup\$ They aren't too close to be considered a telescope? \$\endgroup\$
    – user109880
    Commented Jan 25, 2023 at 14:17
  • \$\begingroup\$ @user14094230 Notice that one of those links is for microscopes? Calculating magnification works the same for both. \$\endgroup\$
    – Zeiss Ikon
    Commented Jan 25, 2023 at 14:28
  • \$\begingroup\$ Ok this is not close enough to just add their diopters together, before getting the magnification bbastrodesigns.com/diopterToFocalLengthConverter.html \$\endgroup\$
    – user109880
    Commented Jan 25, 2023 at 17:26

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