(2NA)?

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Nov 21, 2023 at 22:06	comment	added	Michael C		The most recognizable photograph in all of human history was captured in a vacuum. It's a photo of Buzz Aldrin standing on the surface of the Moon.
Nov 18, 2023 at 0:20	comment	added	scottbb♦		Michael, the entire 1st paragraph is essentially a waste of time. If OP, or any other author, is talking about relative refractive indices with respect to air (the obvious basis medium for photographic lenses), then everything washes, and \$1/2\textit{NA}\$ is the numerical aperture. No need to get pedantic on a 0.028% error of refractive index with respect to vacuum, when nobody is taking pictures in a vacuum.
Sep 20, 2016 at 14:20	comment	added	JEJV		@MichaelClark: the reason I mentioned "What is the maximum aperture consistent with the Nikon F-mount?" was that in that question, there was some mention of whether the second principal surface should be treated as flat or spherical. A spherical second principal surface seems to be illustrated in Fig. 01 here. "Actually, on spherically corrected lenses this surface is not a plane, it’s a sphere centered on the second focal point (F’)"
Sep 20, 2016 at 4:49	comment	added	scottbb♦		@MichaelClark I'm not talking about FoV. I'm referring to the fact that the fastest possible lens you can get will have a f-number equal to N = 1/(2×sin(α/2)), where α is the angle of the cone that focuses collimated light from this "ideal" lens. sin(α/2) has a maximum value of 1 at α=180°, thus the maximum relative aperture is 0.5. See this answer for derivation of this limit based on etendue: In what way does the lens mount limit the maximum possible aperture of a lens?
Sep 20, 2016 at 4:24	comment	added	scottbb♦		@MichaelClark correct on the radians, good call. However, inferring tanΘ ≈ sinΘ is still problematic, because either one (tan or sin) only holds for small Θ. The error ƒ(x) – x is positive for ƒ(x) = tan(x), and negative for sin(x) (again, as you point out, for small x in radians). So setting one approximately equal to the other compounds the error. And again, we're not talking about small angles. We're talking about the largest possible angle that can achieve the mathematically possible largest NA (180°). That is not a valid value for the small-angle approximation.
Sep 20, 2016 at 4:08	comment	added	Michael C		@scottbb shouldn't your comment above say that "... for small Θ, sinΘ ≈ Θ when Θ is expressed in radians? Also that since tanΘ ≈ Θ is also true (when Θ is expressed in radians) then tanΘ ≈ sinΘ can be deduced.
Sep 20, 2016 at 3:57	history	edited	Michael C	CC BY-SA 3.0	added 12 characters in body
Sep 20, 2016 at 3:37	comment	added	Michael C		@JEJV The issue with Nikon, or any other mount system for that matter, has nothing to do with the angles as they enter a lens. Rather, that issue has to do with the angle of light as projected by the rear of the lens being restricted by a narrower opening between the lens and camera. If the cone of light as projected by the rear of the lens is too wide to fit through the hole in the front of the camera, then some of that light will be lost. If the sensor is so large that it is shaded by the hole in the front of the camera, then hard vignetting will also occur.
Sep 19, 2016 at 17:42	comment	added	JEJV		@scotbb: I think this question should be linked, but I don't know how to do it: What is the maximum aperture consistent with the Nikon F-mount?
Sep 19, 2016 at 13:43	comment	added	scottbb♦		@JEJV When Nakamura says "when the value 𝜃' is very small...", he's referring to the small-angle approximation, which is simply that for small 𝜃, sin𝜃 ≈ 𝜃. Even at 𝜃 = 10°, the error is < 1%. This approximation is used all the time in physics and engineering, often to the point that we forget about it being used. That can be a pitfall, because we then extend the use of simple formulas derived from using the approximation, to areas where the simple formulas no longer work because the precondition of small 𝜃 no longer holds
Sep 19, 2016 at 12:25	comment	added	JEJV		So f-number = f/D is an accurate guide to the brightness of a lens focussed at infinity... and ... f-number = n / (2*NA). Part of the Nakamura book that confused me was "when the value of Θ' is very small it can be approximated..." [as] F=f/D. Is that wrong? There seems to be a problem/inconsistency with the Wikipedia Numerical Aperture page where a formula involving arctan should be using arcsin, according to the comment below the formula which refers to the "Abbe sine condition", and then says "[...] the traditional thin-lens definition and illustration of f-number is misleading"
Sep 19, 2016 at 12:13	comment	added	JEJV		So f-number = f/D is an accurate guide to the brightness of a lens focussed at infinity... and ... f-number = n / (2*NA).
Sep 19, 2016 at 11:34	history	edited	Michael C	CC BY-SA 3.0	added 189 characters in body
Sep 19, 2016 at 11:28	history	answered	Michael C	CC BY-SA 3.0