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Focusing and other musings.

torger

Active member
On the comparison of f/64 8x10" and MFD f/11 I made;

If we assume infinite resolution on the sensor and infinite resolving power of the lens, ie diffraction is the only factor limiting resolution, there's a zero sum game between formats in terms of resolving power and DoF.

If you make the format larger, you need a longer focal length to match field of view (FoV) and then you need a smaller aperture to match DoF, and then you get punished more by diffraction so you end up not gaining any resolving power despite the larger format.

If you go the opposite direction from larger to smaller you need a shorter focal length to match field of view, but to resolve as much as the larger format you need to reduce diffraction and thus open up the aperture, ending up on a zero-sum.

Conveniently enough, the you can use the crop factor both on focal length and f-number to find the corresponding match when going between formats.

8x10" diagonal is about 320mm, 48x36mm MFD is about 60mm, that's a 320/60=5.33 crop factor. 64/5.33=12, that is f/12. For full-frame 645 it becomes ~f/13, so f/11 was a bit of exaggeration on my part, you need to get down to 44x33mm for that...

That is if you want to resolve as much as 8x10" at f/64, you can't shoot at a smaller aperture than f/11 with your 44x33mm sensor (and you need like 600 megapixels on that sensor of course...). The point was showing an example that f/64 is not that deep DoF as it might sound in modern ears.
 

MGrayson

Subscriber and Workshop Member
Sorry to go on about mathematical abstractions when you would all rather consider Art, but I find the corollaries to the above analysis very interesting.

For instance, want to know how big the disks of distant OOF lights will be in your image? Imagine your lens opening sitting at the plane of focus. If you are shooting a 50mm at f/1.0, then the opening is 5 cm, or 2". So imagine a 2" diameter disk however far out you are focused, and that's what the distant lights will look like. Focus closer, disks get bigger. Here's an example:



That's probably a 50 f/1.4 wide open, so those disks should be 1.4" = 50mm/1.4 across at the plane of focus, which is the window grill. That middle section is 10" across, so that looks right. In another shot, where I'm closer to the window, the disks look bigger:



but they still look about 1.4" across at the plane of focus, which is now closer. :cool:

For specular lights not so far away, it's a bit more complicated. I'll spare you.

We now return you to your photography forum... ;)

--Matt
 
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Dogs857

New member
Thanks Matt, Torger and JLM

I get it now, for some reason it was just escaping me but I wasn't taking into account the size of 8x10 and the length of the lenses. It's quite simple once you stop and think.
 

Dogs857

New member
There is a lot of good points being raised here.

I have finally gotten a break in the weather so will get out and try some different techniques and see which I feel works for me best.

I have a GG with my Cambo but only a 3x loupe so was thinking I will use that more for framing than critical focusing. My eyesight is still pretty good so I may be able to get things pretty close.

Time for some testing.
 

Shashin

Well-known member
Sorry to go on about mathematical abstractions when you would all rather consider Art, but I find the corollaries to the above analysis very interesting.

For instance, want to know how big the disks of distant OOF lights will be in your image? Imagine your lens opening sitting at the plane of focus. If you are shooting a 50mm at f/1.0, then the opening is 5 cm, or 2". So imagine a 2" diameter disk however far out you are focused, and that's what the distant lights will look like. Focus closer, disks get bigger. Here's an example:



That's probably a 50 f/1.4 wide open, so those disks should be 1.4" = 50mm/1.4 across at the plane of focus, which is the window grill. That middle section is 10" across, so that looks right. In another shot, where I'm closer to the window, the disks look bigger:



but they still look about 1.4" across at the plane of focus, which is now closer. :cool:

For specular lights not so far away, it's a bit more complicated. I'll spare you.

We now return you to your photography forum... ;)

--Matt
Those disks are an image of the aperture. You notice the disks are becoming oval as they move away from the optical axis (and perpendicular to the radial axis). Those show the mechanical vignetting in your optics. The difference in area between the circular aperture image and the vignetted aperture are proportional to the light falloff.
 

MGrayson

Subscriber and Workshop Member
Yep, you can see the aperture vignetting clearly. I just never appreciated that they were images at 1:1 scale at the plane of focus. It's also a credit to the optics that the disks are so clean.

--Matt
 

jlm

Workshop Member
since diffraction was mentioned...it is my understanding that diffraction starts to show when the actual opening size gets relatively small. it is related to the wavelength of the light and the opening size, not the f number directly. meaning f/64 on a 200mm lens will show less diffraction than f64 on a 50mm lens. another way to think about it is that an edge will diffract light and with smaller physical openings, the surrounding "edges" start to affect each other more.
 

Shashin

Well-known member
Diffraction is proportional to the f-number. It is the angular size of the exit pupil that is important--an f/64 light cone has the same angular size regardless of focal length. The impact of diffraction changes with the format size.

The entrance pupil is related to resolution at the object plane, unlike the exit pupil which is resolving power at the image plane. A larger entrance pupil gives greater resolution at the object plane, but the diffraction, the spot size, at the image plane is a product of the f-number. Note, magnification changes with focal length. At a given f-number and FoV, the larger format has greater resolution at the object plane, the same spot size/diffraction at the image plane, less depth of field, and more depth of focus. Crazy world...
 

Shashin

Well-known member
Keep the lens at infinity and stop down for your depth of field, then you can calculate what will and won't be in focus.
Not really. DoF and CoC simply models our perception of sharpness in an image--if you don't like what your DoF scale tells you, find a better CoC and recalculate the values for DoF. Changing the value of the CoC does not change the image, it is just adjusting the model to reflect how you perceive an image. Two people can see the same image and think different values for the CoC are correct and neither would be wrong.

Only the image plane is in focus. But as DoF shows, you can go away from the plane of focus and still get what appears to be a sharp image. "In focus" has nothing to do with sharpness, it is simply an optical property--you can have an in focus image that does not appear sharp.

Because DoF is simply a perceptual model, pixel resolution has no effect on it. Take two images with two 35mm sensors, one with 24MP and another with 37MP and make two 40" prints. The DoF in each print will be identical. The only difference is the 37MP image will have higher frequency detail.
 

Shashin

Well-known member
Jim, do you need help? It is a long entry. From your Wikipedia article:

The ability of an imaging system to resolve detail is ultimately limited by diffraction. This is because a plane wave incident on a circular lens or mirror is diffracted as described above. The light is not focused to a point but forms an Airy disk having a central spot in the focal plane with radius to first null of

d = 1.22λN
where λ is the wavelength of the light and N is the f-number (focal length divided by diameter) of the imaging optics. In object space, the corresponding angular resolution is

sin theta = 1.22(λ/D)
where D is the diameter of the entrance pupil of the imaging lens (e.g., of a telescope's main mirror).
As you can see, diffraction at the image plane is d = 1.22λN where N is the f-number of the system. Entrance pupil is not a factor, but simply f-number. This determines resolving power at the image plane from diffraction.

In object space, the angular resolution is dependent on the diameter of the entrance pupil: sin theta = 1.22(λ/D). That is the resolution in the object space in front of the camera. But this is unrelated to f-number and does not change diffraction at the image plane. BTW, that is based on an object at infinity. When objects are closer to the camera, then the angular size of the entrance pupil defines resolution at the object plane (just like the angular size of the exit pupil in image space). This is also known as the numeric aperture--commonly used with microscopes.

That is what I actually wrote in my post above.
 

MGrayson

Subscriber and Workshop Member
I think Will is right. It's true that diffraction scales inversely as the width of the opening, but the light travels further to the sensor in the larger camera, and that cancels out the scale. Or think of the view from a spot on the sensor looking out. You can't tell how far away the aperture is, i.e., the focal length of the lens. You can only see a disk whose angular size is determined by the f-ratio.

It sounds paradoxical that depth of focus would be bigger on the larger camera, as the optical system doesn't care which is the sensor snd which is the subject. But when we scale up the camera, we don't change the subject distance, while we do change the sensor to exit pupil distance. That's the source of the asymmetry.

Crazy world, indeed.
 
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