Mathematical formulas for N/A and f aperture value.

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harisA
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Mathematical formulas for N/A and f aperture value.

Post by harisA »

1)Which is the formula that gives the f number of an objective of known N/A and focal distance.How this formula changes for a combination of an objective and a tube lens of known focal distance and aperture.

2)What is the best way to reduce the n/a of an objective) (a)close the aperture of the condenser,b) put an iris at the back of the objectives.
What is the difference of these methods?

Thank you in advance for your help.

rjlittlefield
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Post by rjlittlefield »

At the camera, the effective f-number of any objective that is used in its correct focus arrangement is:

f_eff = magnification / (2*NA)

For example an NA 0.25 objective at 10X has an effective f-number of f_eff = 10/(2*0.25) = f/20.

If you put a nominal 10X NA 0.25 objective with a short tube lens so as to give 5X, then its effective f-number becomes f_eff = 5/(2*0.25) = f/10.

------

Putting an iris at the back of the objective will block light on all paths, exactly the same as if you used a smaller NA objective.

Closing the aperture of the condenser will block the edges of the objective from seeing direct light, but light diffracted or diffused by the subject can still go through the entire aperture.

For a highly transparent subject they are about the same. For a dense subject that causes the light to spread out a lot, closing the condenser aperture may have little effect except to reduce the light intensity.

--Rik

harisA
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Location: Greece

Post by harisA »

Rik thank you so much for your fast answer.

Kind Regards

Haris

Hokan
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Location: Sacramento, CA, USA

Converting effective f number to NA?

Post by Hokan »

So would the same formula, (solving for NA rather then effective f number), be the generally accepted formula to use?

Hokan
SOM, (Son Of Multiphot), a DIY macro/micro rig.w/120, 65, 35, 19mm Macro-Nikkors, Nikon AZ100 1X and 4X objectives. Nikon Plan Fluor W objectives, 10X, 20X, 40X.
Zeiss Axiomat, incident BF, DF, DIC, Pol, Phase. With Zeiss infinity objectives, LD, Epi, APO, Plan types.

rjlittlefield
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Post by rjlittlefield »

Hokan, I don't understand your question. What problem are you trying to solve? What is the application? If you had a formula, how would you use it?

--Rik

Hokan
Posts: 59
Joined: Fri Apr 06, 2007 11:58 am
Location: Sacramento, CA, USA

Formula for f number

Post by Hokan »

Hi Rik,

Back in my earlier days of photomicrography when I was suffering from NAS, (in remission now), I acquired a Nikon Ultra Micro Nikkor 30mm, effective aperture, according to Nikon, of f/1.2, and magnification 1/25X. Now that I have my DIY macro rig up and running I plan on reversing the lens and seeing how well the lens works.

The lens has a overall working distance of 810mm, object area of 50mm, image area of 2mm when used as designed. So my inspiration is to reverse the lens and do some test step photographs, (with Zerene of course). So I was just curious how the lens's specifications compared to a good 25X Nikon microscope objective.

In 1964 Nikon was claiming a 1,250 lines/mm resolution through out the entire 2 mm image area. Using a 546 nm, green, wavelength.

Hope I answered your question.

Regards,

Hokan
SOM, (Son Of Multiphot), a DIY macro/micro rig.w/120, 65, 35, 19mm Macro-Nikkors, Nikon AZ100 1X and 4X objectives. Nikon Plan Fluor W objectives, 10X, 20X, 40X.
Zeiss Axiomat, incident BF, DF, DIC, Pol, Phase. With Zeiss infinity objectives, LD, Epi, APO, Plan types.

rjlittlefield
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Post by rjlittlefield »

Ah, got it! Thank you for the excellent and detailed description.

For your application, a better formula is the one that relates effective f-number and NA on the same side of the lens:
NA = 1/(2*f_eff)
By this formula, a lens that is effective f/1.2 on the camera side is also NA 0.4167 on the camera side.

When you reverse it, and run at the complementary magnification, the lens will be NA 0.4167 on the subject side, very reasonable for a 25X objective.

Effective f-number on the camera side in that setup will then be 25/(2*0.4167) = f/30.

Effective f-numbers on each side of the lens are always related simply by the magnification, so it is no coincidence that 30 = 25*1.2 .
(f_eff_camera = magnification * f_eff_subject)

The claim of 1,250 lines/mm can be cross-checked against the MTF formulas quoted at http://www.photomacrography.net/forum/v ... 831#124831.

Using those formulas with lambda = 0.000564 mm, I calculate that cutoff frequency nu0 = 1526 lp/mm, and MTF(1250 lp/mm) = 0.09 .

So, the resolution they're quoting is consistent with 9% MTF with perfect optics. That would definitely be visible with a high contrast target, so that looks like a reasonable spec.

--Rik

Hokan
Posts: 59
Joined: Fri Apr 06, 2007 11:58 am
Location: Sacramento, CA, USA

Formula for f number

Post by Hokan »

Rik,
I very much appreciate your quick response and understandable explanations.

I have written your formulas into my studio journal for future use.

Your ability to explain how your formulas can be used while at the same time showing how they are related to the physics of optics and light is most, how shall I say; illuminating! ???? :lol:

With kind regards,
Hokan
SOM, (Son Of Multiphot), a DIY macro/micro rig.w/120, 65, 35, 19mm Macro-Nikkors, Nikon AZ100 1X and 4X objectives. Nikon Plan Fluor W objectives, 10X, 20X, 40X.
Zeiss Axiomat, incident BF, DF, DIC, Pol, Phase. With Zeiss infinity objectives, LD, Epi, APO, Plan types.

rjlittlefield
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Post by rjlittlefield »

And I greatly appreciate your kind words and engaging pun!

--Rik

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