## NA to f/#

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Harald
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### NA to f/#

Hi there good people,
I was wondering about posting an question about converting NA to f/#. when I came across this one:

http://www.calctool.org/CALC/phys/optics/NA_to_f

Is it so simple?
Kind Regards
Harald

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ChrisR
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NO because the calculator seems to only work sometimes!

f# = 1/(2*NA)
usually works but it depends exactly what you apply it to.
Chris R

mtuell
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I think I can clarify a couple of points regarding f/# vs NA.

First of all, I will disregard details such as entrance pupil location and principal planes, as well as the index of refraction of the medium.

The f/# is defined as the focal length / diameter. The numerical aperture is n*sin(A), where A is the half-angle of the cone. These are described in the figure below.

So, since f/D is equal to f/(2*(D/2)), and the tangent of A is (D/2)/f, we can see that, with a little algebra, f/# = 1/(2*tan(A)), and NA = sin(A).

Now, in the small angle approximation, (in radians) A = sin(A) = tan(A), so the formula that f/# = 1/(2*NA) is based on the small angle approximation. The graphs of A, sin(A) and tan(A) are shown below.

As seen in the figure, this approximation starts to fall apart at A about 20°, or a cone angle of 40°. The values at 20° are sin(20)=0.342 and tan(20)=0.364. So, the NA here is about 0.35 in air. Faster cones than this do not follow the small angle approximation very well, so f/# = 1/(2*NA) starts to become a worse and worse approximation to reality.

I hope this helps
Mike

mtuell
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I should also point out that the derivation I gave is only strictly true for an image plane at infinity (collimated beam, as drawn). For a finite system, the f/# is still based on the focal length of the lens, but the NA is the cone angle, which is computed from the Gaussian form of the thin lens equation

1/f = 1/i + 1/o, where i and o are the image and object distances.

Mike

rjlittlefield
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mtuell wrote:I hope this helps
Unfortunately, it does not.

You have fallen prey to the common myth that f# and NA are only approximately related by f# = 1/(2*NA) because sin() is only approximately related to tan().

I used to think that also, and then I got enlightened. It turns out that the simple relationship is actually exact, for quite an interesting reason.

Quoting what I wrote back in 2009:
Another interesting thing is that despite all the words written about principal "planes", the fact is that those virtual surfaces aren't actually flat, but rather they are pieces of spheres! As Rudolf Kingslake explains, "...the complete theory of the Abbe sine condition shows that if a lens is corrected for coma and spherical aberration, as all good photographic objectives must be, the second principal plane becomes a portion of a sphere of radius f centered about the focal point."[1]

This actually does matter in practice, as Kingslake explains two sentences earlier: "It is a common error to suppose that the ratio of Y/f is actually equal to tan(theta) and not sin(theta)... The tangent would, of course, be correct if the principal planes were really plane. However...".

In other words, every time you read something like "the effective f-number of a lens is approximately equal to 1/(2*NA), or more precisely 1/(2*tan(asin(NA)))", you should mentally delete that stuff about "approximately" and "more precisely..." The 1/(2*NA) formula is correct, and it's because the principal plane is really a piece of a sphere. I notice this because I've written that stuff more than once, it's wrong, and now I have to figure out whether it's worth going back to flag the errors.

--Rik

[1] "Optics in Photography", Rudolf Kingslake, 1992, page 107 footnote, Google Books extract HERE.
Likewise there's a simple change that means you don't need to worry about entrance pupil location or focus relationships.

Just refine the equation to say that it's the effective aperture that we're talking about, after all those other things are factored in.

Then the equation becomes

f_effective = 1/(2*NA)

which is simple, tidy, and I believe exact -- at least in the limiting case of refractive index equal 1.

A rare combination.

--Rik

P.S. I see that the link to Google Books is still valid today, but just to be sure that this does not get lost for future readers, here is what the page from Kingslake's book looks like:

mtuell
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Thanks, Rik,

I'll have to look into the details, but the essence of what you are saying is that if you look at it approximately, it is an approximation, but if you look at it exactly, it is exact.... interesting phenomenon!

In my (admittedly lame) defense, I did preface it by saying that I was ignoring details such as the principle planes and entrance pupil. :P

Consider this retracted!

Mike

rjlittlefield
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mtuell wrote: if you look at it approximately, it is an approximation, but if you look at it exactly, it is exact....
I never thought of it that way, but I like that formulation!

--Rik

mtuell
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Further discussion of this topic comes from Warren J. Smith, "Modern Optical Engineering". I have the first edition, and on page 132, it says

"Numerical aperture and f-number are obviously two methods of defining the same characteristic of a system. Numerical aperture is more conveniently used for systems that work at finite conjugates (such as microscope objectives) and the f-number is applied to systems for use with distant objects (such as camera lenses and telescope objectives). For aplanatic systems (i.e. systems corrected for coma and spherical aberration) with infinite object distances, the two quantities are related by: f-number = 1/2N.A."

He also shows on p. 18, the principal "planes" as curved (quotes his).

So, overall, 1/2NA is only exact for certain situations, as Rik mentioned. You do have to account for the "effective focal length (EFL)", not just the catalog value.

Mike

rjlittlefield