The lenses we use

Have questions about the equipment used for macro- or micro- photography? Post those questions in this forum.

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ray_parkhurst
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Post by ray_parkhurst »

Justwalking wrote:
ray_parkhurst wrote: The crop you showed appears to be beyond 100%. I see nothing like you are showing at 100% zoom. Please point out exactly where you took the crop so I can see what you're talking about.
Yes, Ray. I did zoom more to draw arrows and show you better what i mean when i see it at 100. Probably i'm not one and only who can see this on monitor. But i can on my Dell U-series 24". The crop was in at "5" left upper corner, but also around in others places.

edited to add:
I shot a new coin using the 6-image 2x configuration. I shot raw, stacked tif, composited tif, and published jpg with light sharpening. Here's the result:
https://easyzoom.com/manage/images/124047
Please let me know if you still see the same issue...
Now i did not see such artefacts at 100%. It is much better.
I think so too. Much more effort but maybe worth it to get rid of the artifacts. Thanks for the feedback.

Justwalking
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Post by Justwalking »

rjlittlefield wrote: With same FOV, DOF only depends on NA and on how you choose to calculate DOF.
I still have hope that maybe the numbers can help you to understand this, since you do not have a large sensor to work with physically.
But what gives you the great DoF is the lens, not the sensor. It is the combination of 7 mm FOV and NA 0.04 . That same combination with a larger sensor would give you the same DOF.
Desagree at all.
First, DOF relates to a print or other reproduction of an image. It's NOT an intrinsic property of a lens. If you put a lens on an optical bench you can measure focal length, you can measure aperture, but you can't measure depth of field.

No! You can't take this combination just of more magnification you must have with large sensor and due to sensor size. DoF is function of magnification. And DoF is inversely proportional to format size.

That's why you need to stitch the stacked frames for coin. But my coin have only 6 shots without stacking at all.

rjlittlefield
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Post by rjlittlefield »

@Justwalking

OK, I give up. Believe what you want to.

But if you give incorrect information to other people, we will correct it then. This forum does not tolerate disseminating incorrect information.

--Rik

Justwalking
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Post by Justwalking »

rjlittlefield wrote:@Justwalking

OK, I give up. Believe what you want to.

But if you give incorrect information to other people, we will correct it then. This forum does not tolerate disseminating incorrect information.

--Rik
Let's do the mathematics that is not about my belief.

And see which information is incorrect

I hope ppl working with macro well known this formula for Dof (simplified but pretty accurate)
DOF ~ 2F'c/M^2, where F' - working number, c - circle of confusion, M - magnfication.

Let's do calculate for the lenses with same working number
for the FF and 5.5 crop

Lets take F'=10
For the FF: 2x10x0.03/4.4/4.4 = 0.031
For the crop: 2x10x0.00545/0.8/0.8 = 0.170
For given another F' you got same difference (diffraction limits is another calc)
So DoF with crop 5.5 is 5.5 larger than on FF in macro.

Want to correct this, Rik?

ray_parkhurst
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Post by ray_parkhurst »

Justwalking wrote:

DOF ~ 2F'c/M^2, where F' - working number, c - circle of confusion, M - magnfication
JW, please just take a close look at the formula you just provided for DOF. Nowhere in that formula is there anything about sensor size, only effective aperture, circle of confusion (which relates to pixel pitch, not sensor size), and magnification. Effective aperture depends on magnification and the nominal lens aperture. So ultimately, as long as you know the lens nominal aperture, and the magnification, and the sensor pitch, you can calculate DOF. Sensor size has no bearing in calculation.

rjlittlefield
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Post by rjlittlefield »

Justwalking wrote:I hope ppl working with macro well known this formula for Dof (simplified but pretty accurate)
DOF ~ 2F'c/M^2, where F' - working number, c - circle of confusion, M - magnfication.

Let's do calculate for the lenses with same working number
for the FF and 5.5 crop

Lets take F'=10
For the FF: 2x10x0.03/4.4/4.4 = 0.031
For the crop: 2x10x0.00545/0.8/0.8 = 0.170
For given another F' you got same difference (diffraction limits is another calc)
So DoF with crop 5.5 is 5.5 larger than on FF in macro.

Want to correct this, Rik?
Sure.

At same FOV and same NA, the working aperture on FF will be 5.5 times larger, so 55, not 10. Just like in my tables above, which you agreed to. Everything scales, including the working f-number.

Image

Then the calculation is
For the FF: 2x55x0.03/4.4/4.4 = 0.170
For the crop: 2x10x0.00545/0.8/0.8 = 0.170

The calculation that you did would be for same FOV and same working f-number, which would be different NA. In fact your calculation for FF is for NA = 0.22, which would give much less diffraction and much more detail, to go along with the reduced DOF.

--Rik

Justwalking
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Post by Justwalking »

ray_parkhurst wrote:
Justwalking wrote:

DOF ~ 2F'c/M^2, where F' - working number, c - circle of confusion, M - magnfication
JW, please just take a close look at the formula you just provided for DOF. Nowhere in that formula is there anything about sensor size, only effective aperture, circle of confusion (which relates to pixel pitch, not sensor size), and magnification. Effective aperture depends on magnification and the nominal lens aperture. So ultimately, as long as you know the lens nominal aperture, and the magnification, and the sensor pitch, you can calculate DOF. Sensor size has no bearing in calculation.
There two thing about sensor size. It is a DoF CoC and magnification that is correspond to sensor size. And yes, both cameras in this example is with same 16 MP, same aspect ratio and same FoV. Pixel pitch size is exactly 5.5 times less on crop.
In this equation F' is Working F-number that is correspond to magnification also.
You can find this formula even in wiki
https://en.wikipedia.org/wiki/Depth_of_field

Justwalking
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Post by Justwalking »

rjlittlefield wrote:
Justwalking wrote:I hope ppl working with macro well known this formula for Dof (simplified but pretty accurate)
DOF ~ 2F'c/M^2, where F' - working number, c - circle of confusion, M - magnfication.

Let's do calculate for the lenses with same working number
for the FF and 5.5 crop

Lets take F'=10
For the FF: 2x10x0.03/4.4/4.4 = 0.031
For the crop: 2x10x0.00545/0.8/0.8 = 0.170
For given another F' you got same difference (diffraction limits is another calc)
So DoF with crop 5.5 is 5.5 larger than on FF in macro.

Want to correct this, Rik?
Sure.

At same FOV and same NA, the working aperture on FF will be 5.5 times larger, so 55, not 10. Just like in my tables above, which you agreed to. Everything scales, including the working f-number.

Image

Then the calculation is
For the FF: 2x55x0.03/4.4/4.4 = 0.170
For the crop: 2x10x0.00545/0.8/0.8 = 0.170

The calculation that you did would be for same FOV and same working f-number, which would be different NA. In fact your calculation for FF is for NA = 0.22, which would give much less diffraction and much more detail, to go along with the reduced DOF.

--Rik
Is this formula is incorrect o what? Why you try to insert NA everywere?
My calc is for my real lens and we compared different size sensors with lens working with same F-working number. This formula did not consist any NA.
Let's take a lenses with same F-working wich are not in diffraction field for both systems. Then it is F'=4. And we take same 5.5 times with DoF again.
Last edited by Justwalking on Sun Aug 12, 2018 3:02 pm, edited 1 time in total.

ray_parkhurst
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Post by ray_parkhurst »

Justwalking wrote:
There two thing about sensor size. It is a DoF CoC and magnification that is correspond to sensor size. And yes, both cameras in this example is with same 16 MP, same aspect ratio and same FoV. Pixel pitch size is exactly 5.5 times less on crop.
In this equation F' is Working F-number that is correspond to magnification also.
No. CoC does not have anything to do with sensor size. Magnification does not have anything to do with sensor size. The formula has nothing to do with sensor size.

Justwalking
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Post by Justwalking »

ray_parkhurst wrote:
No. CoC does not have anything to do with sensor size. Magnification does not have anything to do with sensor size. The formula has nothing to do with sensor size.
Wow! This formula is about negative era only? This formula is about DoF, nagnification and CoC. Not so long ago you have said that you need more magnification to cover FF and it is proportional to size of the sensor.

ray_parkhurst
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Post by ray_parkhurst »

Justwalking wrote:
ray_parkhurst wrote:
No. CoC does not have anything to do with sensor size. Magnification does not have anything to do with sensor size. The formula has nothing to do with sensor size.
Wow! This formula is about negative era only? This formula is about DoF, nagnification and CoC. Not so long ago you have said that you need more magnification to cover FF and it is proportional to size of the sensor.
I would still say that, of course, assuming you want to cover the same FOV.

rjlittlefield
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Post by rjlittlefield »

Justwalking wrote:Why you try to insert NA everywere?.
Perhaps a diagram will help.

Image

My calculation corresponds to just moving the sensor farther away from the lens so as to increase magnification from 0.8 to 4.4 . Same lens diameter at same distance from subject gives same DOF and same available MP in optical image.

Your calculation corresponds to swapping out the lens for one with a lot wider hole. Wider lens at same distance from subject gives less DOF and much more available MP in optical image.

--Rik

Justwalking
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Post by Justwalking »

rjlittlefield wrote:
Justwalking wrote:Why you try to insert NA everywere?.
Perhaps a diagram will help.

Image

My calculation corresponds to just moving the sensor farther away from the lens so as to increase magnification from 0.8 to 4.4 . Same lens diameter at same distance from subject gives same DOF and same available MP in optical image.

Your calculation corresponds to swapping out the lens for one with a lot wider hole. Wider lens at same distance from subject gives less DOF and much more available MP in optical image.
--Rik
Rik, your diagramm and calc have nothing with DoF.
It is all about system resolution lens+sensor and diffraction.
As we see from your previous table the wilder lens can't resolve more on 16 MP sensor on FF compared crop just due to another F-effective.
It seems my poor english is not enough to resolve misunderstanding.
I just leave this sever articles for your reading

http://www.savazzi.net/photography/macrodof.htm

http://www.bobatkins.com/photography/te ... aldof.html

https://www.dpreview.com/articles/30649 ... hotography

rjlittlefield
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Post by rjlittlefield »

Hhmm... Words don't work, formulas don't work, diagrams don't work. Let me try one last technique.
Justwalking wrote:It is all about system resolution lens+sensor and diffraction.
I agree.

So tonight I have shot two images for you.

They are very ugly -- a dead bee head sitting on a resolution chart.

And they are very blurred, because I have deliberately stopped down so far that the diffraction limit is obvious.

But I hope they are informative.

Here are the two images, resized to fit side by side.

ImageImage

I would say that these two images have at least very close to the same DOF, judging for example from the amount of blur of the upper antenna.

It may not be apparent from the above, but both images also have the same resolution on subject. You can see from these crops that both images resolve the same lines in the resolution chart.

ImageImage

Now here's the catch: these two images were shot with a factor of over 9.5X difference in magnification. One of the images occupied 26.3 mm wide on a full-frame sensor. The other image occupied 2.745 mm wide on a 1/2.5" sensor.

So...

We have here two images, one shot at over 9.5 times higher magnification, but both having the same resolution on subject and the same DOF.

In my model, this is no problem. Both images are shot with the same FOV and the same NA, so of course they have the same diffraction and the same DOF.

But as I understand your approach, one of the images should have about 9.5 times more DOF than the other.

So I am curious.

How do you explain these two images?

--Rik

Justwalking
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Post by Justwalking »

rjlittlefield wrote:Hhmm... Words don't work, formulas don't work, diagrams don't work. Let me try one last technique.
Justwalking wrote:It is all about system resolution lens+sensor and diffraction.
I agree.
So tonight I have shot two images for you.
They are very ugly -- a dead bee head sitting on a resolution chart.
And they are very blurred, because I have deliberately stopped down so far that the diffraction limit is obvious.

But I hope they are informative.
Here are the two images, resized to fit side by side.

In my model, this is no problem. Both images are shot with the same FOV and the same NA, so of course they have the same diffraction and the same DOF.

But as I understand your approach, one of the images should have about 9.5 times more DOF than the other.

So I am curious.

How do you explain these two images?

--Rik
Both of your images deep in diffraction field and too small and resized by differnt scale to talk about comparision of DoF.
Rik, if wiki formulas for macro did not work for you and words (not only mine) also all you can do is do clear experiment.
Do the single shot of safety match with same FoV as mine at 0.8X so the nearest point of it's head will be in focus and see were will be better Dof or may be the same.
But first you need the camera FF sensor with 16MP, not 36 or 42.

I'm not at home now and can't show my shot with it.

P.s. if yoy want insert NA in your calculatoin you must use another formula for microscopy
https://www.microscopyu.com/tutorials/depthoffield

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