0
$\begingroup$

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

F/# $= \frac{f}{D}$ (same reference as above)

where F/# is the f-number, $f$ is the focal length, and $D$ is the lens diameter.

An infinity corrected microscope objective is essentially an infinite conjugate. Therefore:

$\frac{D}{f} = 2 \times tan(\theta)$

Olympus objectives are typically designed such that the power is related to the focal length using the following:

$f = \frac{180mm}{P}$

where $P$ is power.

Taking this to the logical conclusion reaches a paradox. Use a 60X objective with a $NA = 0.95$.

$f = \frac{180mm}{60x}$ = 3.0 mm

$\theta = a sin(0.95)$ = 71.8 deg

$D = 2\times tan(\theta) \times f$ = 18.2 mm!

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"?

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

$\endgroup$
  • 1
    $\begingroup$ Please use MathJax for your formulas and equations. $\endgroup$ – Ellie Aug 27 '14 at 18:48
  • $\begingroup$ @user3533030 I edited the formulas to include MathJax. Are they accurate? $\endgroup$ – HDE 226868 Aug 27 '14 at 19:57
1
$\begingroup$

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 $\sin\theta\not\approx\tan\theta$).

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

Assuming a real objective has no aberrations, can the principal sphere / plane model be used for off-axis imaging?

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

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.

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.

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ An other explanation could be that the F# is D/f... Your link says otherwise. $\endgroup$ – Kyle Kanos Dec 2 '14 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.