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Thus the 18.2 mm calculated here could correspond to a plane inside the objective. In this case 18.2mm is a plausible lenght since most of the objective I know have an external diameter superior to 20mm.

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Sep 12, 2024 — In conclusion, microscope objective lenses are an essential part of a microscope and are used to magnify the specimen being observed. They ...

To operate the tutorial, use the Reference Focal Length and Objective Focal Length sliders to alter the specifications of the virtual infinity optical system.

By the way, I've modeled some microscope objectives in Zemax using US Patents. Some obey these equations above, some do not.

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The focal length of the lens determines your field of view (composition) and the magnification of the scene. The graphic shows increasing f-stop values with ...

The lens diameter appears too large. There are no microscope objectives this large. Although I can accept that the optical prescription can increase (decrease) the marginal ray, the exiting beam from a point source at the object plane is collimated. Under what conditions can this beam be smaller than the "lens diameter"?

I agree. Since then, I have tried to build a model (Zemax) of the objective. It seems that microscope objectives have a principal sphere instead of a principal plane. This addresses the first order ray tracing. I'm not really sure how far this model can be used.

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For a lens obeying the Abbe sine condition, e.g., any well corrected microscope objective, $\frac{D}{f} = 2\times\sin\theta$. The explanation for this is in the Wikipedia article you refer to. It is probably the cause of the discrepancy in your results for high NA objectives (where $\sin\theta\not\approx\tan\theta$).

Numerical aperture (NA) $= sin(\theta)$ where $\theta$ is the half-angle (see: http://en.wikipedia.org/wiki/Numerical_aperture)

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An other explanation could be that the F# is D/f where D is the diameter of the entrance pupil of the optical system (cf http://en.wikipedia.org/wiki/F-number), here an microscope objective.

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erratum : my bad F# = f/D thanks Kyle. But my main poin remain valid : D is the pupil diameter therefore while being a proper physical dimension, it may not be related to any physical object (i.e lenses), it's a product of the overall optical assembly.