Numerical Aperture and mAcro lenses

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Epidic
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Numerical Aperture and mAcro lenses

Post by Epidic »

I was browsing a book and I found this which can calculate the numerical aperture for a photographic lens:

NA = M / square root (4 f squared (M+1)squared + M squared)

Where M is magnification and f is the effective f-number of the system.

The interesting thing is a 1x microscope objective with an 8x eyepiece and bellows extension of 12.5 has a NA of 0.04. A 25mm f/3.5 photographic lens on a 11 inch bellows at 10x has a NA of 0.128. That is more than tree time greater!

Thought you may like to know.
Will

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Post by rjlittlefield »

Will,
NA = M / square root (4 f squared (M+1)squared + M squared)
This appears to be the exact formula for a thin lens approximation, where f = focal length / aperture diameter, that is, the lens's f-number not corrected for magnification.

Note that for large M and large f, the formula simplifies to the classical approximation that NA ~= 1/(2 f).

For M >= 10 and f >= 2, the maximum error is less than 12%, making the simpler approximation quite useful. In your example, it would yield that NA ~= 1/(2*3.5) = 0.143, about 10% more optimistic than the "exact" formula.

I put quote marks around "exact" in that last statement because we have good experimental evidence that these formulas are rather imperfect predictors of actual lens performance. See photos and discussion in http://www.photomacrography.net/forum/v ... .php?t=424 (two pages, be sure to read both).

However, I agree completely with your point that a good macro lens has wider aperture and therefore more resolution than a small-NA objective.

Where objectives start to shine is more like NA = 0.20, far above the NA = 0.04 of your example.

--Rik

Epidic
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Post by Epidic »

Thanks Rick, I will look at it again. I pulled that out of the focal encyclopedia of photography and assumed they would not make a mistake.
Will

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Post by rjlittlefield »

For what it's worth, here's the derivation I used. (See image below.)

Let me know if you see a mistake in here -- nobody else has checked it.

--Rik

Image

Epidic
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Post by Epidic »

Sorry, I have not replied. I have been other places.

It seems to be fine, but I am not the most reliable person when it comes to this kind of thing. Personally, I cannot see why the effective entrance pupil of the lens cannot be used to calculate numerical aperture? After all, the numerical aperture and the entrance pupil are indicators of angular resolution rather than resolving power.

Why is this not true?

NA = (focal length / f-number) / 2
Will

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Post by rjlittlefield »

Epidic wrote:Personally, I cannot see why the effective entrance pupil of the lens cannot be used to calculate numerical aperture?
It can, and the diagram above shows how.

By definition (ignoring media with refractive index > 1),
NA is the sine of half the angular dimension of the entrance pupil,
as seen by the in-focus subject. [ref]

In the diagram above, this angle is labeled α (alpha).

The angle alpha is related to the entrance pupil diameter D as shown in the diagram (for thin lens model).
Why is this not true?

NA = (focal length / f-number) / 2
I'm not sure how to answer this question.

For starters, it has the wrong form -- it should be the sine of an angle, but what you've written is a linear dimension, because focal length / f-number = D, the aperture diameter.

If I've worked the symbols correctly, an alternate form of the exact equation is this:
NA = sin(atan(1/(2*(M+1)/M)*f_number))
At large magnifications, (M+1)/M ~= 1, and for small angles sin(atan(x)) ~= x, leading to the classical approximation that
NA ~= 1/(2*f_number)
This latter formula is very much like what you have, except that focal length is missing because it's already been factored into the f-number.

Does any of this help?

--Rik

Epidic
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Post by Epidic »

Rick, I was just taking the angular resolution forumula (Rayleigh criteria):

1.22 lambda / D

or

1.22 lambda / 2 NA

From what I understand, NA is really an angular measure at the object plane. f-numbers refer to the image plane. So I assume taking the the linear measure of the entrance pupil would be the easiest way to calculate NA. Most likely I have made an error - don't worry, it is not the first time, and I promise it won't be the last.
Will

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Post by rjlittlefield »

Ah, I see!

Wikipedia has a good explanation about these formulas, see http://en.wikipedia.org/wiki/Angular_resolution .

The Rayleigh criterion for angles is

Equation 1: sin(delta_theta) = 1.22 * lambda / D

where delta_theta is the angular resolution. This maps very nicely to telescopes, where effective resolution is best measured as angles.

For microscopes and macro photography, the effective resolution is best measured as length. In that case, the (approximate) Rayleigh criterion is

Equation 2: delta_length = 1.22 * focal_length * lambda / D

or more precisely that

Equation 3: delta_length = 1.22 * lambda / (2 * NA)

I think the confusion results from equating sin(delta_theta) (Equation 1) with delta_length (Equations 2 & 3). That error would produce NA = D/2 = (focal_length / f_number) / 2, as you proposed.

---Rik

Epidic
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Post by Epidic »

Thanks Rick, I knew I was wrong, but until now I did not know why.
Will

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Post by rjlittlefield »

You're very welcome. I have learned much from the discussion. (It usually works that way! :D )

--Rik

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