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Depth of field and focal length

Shashin

Well-known member
The best photographer I know is unaware of any optics theory.
The great thing about photography is seeing is believing. Hard to argue something does not look the way it does.

This is what makes photography as a science extremely difficult. It isn't just describing a physical process, but a perceptual/cognitive one as well. DoF is a perceptual quality--it actually does not exist in any absolute sense--if you have a picture and no one sees it, does it have DoF? Photographic systems ultimately are trying to imitate the human visual system. The fact they work so well is pretty amazing, but the limitations are obvious too.
 

MGrayson

Subscriber and Workshop Member
Depth of field is a function of image magnification and aperture. If both are the same, the DoF will be the same regardless of the focal length, film/sensor format, etc.
This is a consequence of a deeper statement than I first thought. (Or equivalent to it, if by DoF one really means "knowing the size of every OOF spot in the final image"). DoF is a choice we make from that information.

Theorem: Adjusted for magnification, bokeh disks will be identical in two images of the same f-number and same focus plane. Magnification will vary from point to point, but the relationship holds point-wise.

Here we go... We take two random lenses of the same f-stop, physical apertures a1 and a2, and place them anywhere, but focus them on the same plane:



b1, the bokeh spot as seen by lens 1 on the focal plane, has size b1/x=a1/(x+y1), so b1 = a1*x/(x+y1). Similarly (pun intended) b2 = a2*x/(x+y2). Now a1 and a2 are apertures at the same f-stop, so the focal lengths f1 and f2 are f*a1 and f*a2. So magnification m1 = f*a1/(y1+x) and m2 = f*a2/(y2+x).

We calculate: b1 adjusted for magnification is ...

b1*m2/m1 = a1*x/(x+y1)*f*a2/(y2+x)/(f*a1/(y1+x)) = a2*x/(y2+x) = b2.

Hot damn, it's true EVERYWHERE.

Matt
 
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Shashin

Well-known member
But are we confusing depth of focus, the distance on each side of the image plane that will result in a perception of sharpness, which is only dependent on f-number (and the permissible circle of confusion), or depth of field, the distance on each side on the object plane, which is dependent on focal length and f-number (and permissible circle of confusion)? The description of object space and image space is not the same. Any lens at the same f-number will have the same depth of focus, but not the same depth of field.

Or to put it another way: what is the relationship of the Numeric Aperture (the angular size of the entrance pupil from the object plane) to the f-number (the angular size of the exit pupil from the image plane)?
 

MGrayson

Subscriber and Workshop Member
But are we confusing depth of focus, the distance on each side of the image plane that will result in a perception of sharpness, which is only dependent on f-number (and the permissible circle of confusion), or depth of field, the distance on each side on the object plane, which is dependent on focal length and f-number (and permissible circle of confusion)? The description of object space and image space is not the same. Any lens at the same f-number will have the same depth of focus, but not the same depth of field.

Or to put it another way: what is the relationship of the Numeric Aperture (the angular size of the entrance pupil from the object plane) to the f-number (the angular size of the exit pupil from the image plane)?
Isn't that ratio magnification?

My difficulty here is that I never learned the proper names for things. I learned (and taught) optics as a problem in linear algebra - lenses and air gaps are just different 2x2 matrices. A telescope, periscope, or microscope is a matrix with a desired form. How do you combine lens and spacing matrices to get you your telescope? Real optics design uses different names - not always, but often enough to leave me confused.
 

Shashin

Well-known member
Isn't that ratio magnification?

My difficulty here is that I never learned the proper names for things. I learned (and taught) optics as a problem in linear algebra - lenses and air gaps are just different 2x2 matrices. A telescope, periscope, or microscope is a matrix with a desired form. How do you combine lens and spacing matrices to get you your telescope? Real optics design uses different names - not always, but often enough to leave me confused.
Sorry, ratio magnification is a new term for me.

And I have appreciated your approach on this topic. I have never seen it defined in the terms you have shown and I find it interesting. I have a bachelors of science from RIT in imaging and photographic technology and so look at this issue in applied photo science terms. DoF is one of those tricky subjects as you are trying to predict human perception of an image displayed under certain criteria using optical choices of a photographer (format, focal length, f-number, distance) with a subjective constant to join them--permissible circle of confusion. And while we can only perceive the resulting image in an image space, DoF is trying to make predictions in the object space when the image is taken--the photographer wants to be able to predict what will appear sharp in the scene in front of him/her. And we have wandered into a more complex idea of how the OFF area will be percieved.

Personally, this is what I find interesting, how do these factors influence what we perceive in an image. (Beyond an upward pointing arrow in object space will point down in image space. ;) )
 

MGrayson

Subscriber and Workshop Member
Sorry - I meant "Isn't the ratio of angles you just mentioned the same thing as magnification?"

That's why I like to talk about bokeh disks as things like "imagine that your subject has a disk on her shoulder the size of the physical aperture (or entrance pupil, or whatever). That's what a streetlight will look like. That is concrete to the photographer, and they can draw whatever conclusions they want about what is or is not in focus. If the subject is a mountain climber on a distant slope, the disk is insignificant. If they are 20 feet from you and you have a 400mm f/2.8 monster, then it will be a very big disk indeed. To get the sensor size and final print size in the mix is unnecessary confusion, although very important if you're making that print!

I'm also fond of taking the convolution of Bob with a dime to express an image of a person with a dime sized point blur.
 

MGrayson

Subscriber and Workshop Member
But are we confusing depth of focus, the distance on each side of the image plane that will result in a perception of sharpness, which is only dependent on f-number (and the permissible circle of confusion), or depth of field, the distance on each side on the object plane, which is dependent on focal length and f-number (and permissible circle of confusion)? The description of object space and image space is not the same. Any lens at the same f-number will have the same depth of focus, but not the same depth of field.

Or to put it another way: what is the relationship of the Numeric Aperture (the angular size of the entrance pupil from the object plane) to the f-number (the angular size of the exit pupil from the image plane)?
I'm confused again. Something in my definitions must be wrong. I'm trying to put everything in the object plane, which is easier to visualize. I think I don't understand something, and I think I'm confusing two meanings of magnification.

I blame it on the pandemic.

M
 

MGrayson

Subscriber and Workshop Member
I'm confused again. Something in my definitions must be wrong. I'm trying to put everything in the object plane, which is easier to visualize. I think I don't understand something, and I think I'm confusing two meanings of magnification.

I blame it on the pandemic.

M
Right. So everything I said in the last post with diagrams was wrong. Not that the math was wrong, but what it MEANT was not what I meant it to mean :loco::facesmack:. I'm one of those "don't look it up, work it out from scratch" kind of annoying mathematicians, but that way you can find new ways of looking at things. More often, you just rediscover the wheel.

I'm going to go back to the simpler case of uniform magnification on the focus plane and see what we get from there.

M
 

Shashin

Well-known member
Right. So everything I said in the last post with diagrams was wrong. Not that the math was wrong, but what it MEANT was not what I meant it to mean :loco::facesmack:. I'm one of those "don't look it up, work it out from scratch" kind of annoying mathematicians, but that way you can find new ways of looking at things. More often, you just rediscover the wheel.

I'm going to go back to the simpler case of uniform magnification on the focus plane and see what we get from there.

M
In imaging, magnification is usually linear magnification, the difference between object size and image size. This should not be confused with visual instruments like telescopes and microscopes, with describes angular magnification (also note, that definition is different between telescopes and microscopes (a telescope working at 1x is not the same as a microscope working at 1x). But the difference is practical. Linear magnification allows us to predict the image size of an object in an imaging system, where angular magnification allow us to predict how something will appear when we look at it.

And this is where linear magnification can give two perceptually different results. If you shoot at 1:1 (or at 1x) on a medium format camera and APS-C camera, the resulting image will appear that the APS-C has more "magnification" simply because it has a smaller sensor. A 1cm object will appear larger on a smaller sensor simply because 1cm is proportionally larger on the small sensor.

I am not sure this will be helpful, but although image and object planes are conjugate, they are not the same appearance and are not directly related to point spread. So an f/2 lens will have the same depth of focus, the focused light cone will intersect the image plane at the same angle regardless of focal length, but the displacement from the image plane will represent different object distances because of focal length--think how far long focal length on a view camera has to be moved to focus from infinity to 1m compared to wide angle illustrating the image-side point spread is not the same as the object-side point spread (although those can coincide as you are doing). DoF is describing the appearance of the object-side point spread.

So how does this impact the visual appearance of the OFF image? So as you know, the difficultly with equivalence is it is like Home Depot--it almost has what you want. So in equivalence you can get many, but not all variables to coincide. And the same thing seems to be happening here. So if you look at the specular highlight on the ruler two feet behind the lens, the out of focus image does not appear the same, even though the DoF is equal. I think it is related to relative size in relation to perspective, as I posted above--the difference in the sizes of two object in an image is proportional to their distances from the camera. So in this case, the 250 is further from the lens, but the ruler end is fixed at 2ft for both. So while the DoF is identical in each image, the linear perspective (at least in terms of identifiable points in the image) is different.

In short, you can make DoF coincide, but not linear perspective and hence magnification, at least for objects within the frame. This may be the reason we can perceive the same DoF, but not Bokeh.

Just a thought.
 

Shashin

Well-known member
Actually, scratch that. I think point spread at the image plane can still work, but the distance from the image plane along the cone does not represent the same distance in object space--it might also be proportional to the linear perspective thing above.

Or something...
 

MGrayson

Subscriber and Workshop Member
My goal is to move everything to the object plane or, even better, to the entire 3D scene. The latter means "what do I do to reality so that a pinhole camera image agrees with the 50 f/2 image taken from here." I understand the truth of what you're saying, but I'm after an "interpretation" that will help a photographer visualize the effect of focal length, aperture, focus distance and OOF objects OUTSIDE the camera.

For instance, a distant streetlight looks like my lens's entrance pupil placed next to the subject in focus. This is universally true.

Or "If I focus on the mountains with a 2cm entrance pupil, then EVERY object in the scene is smeared by a 2cm disk." 2cm at a distance will be smaller than the CoC, and so that part of the image will be "in focus". Every enlargement will have objects that appear smeared by "real" 2cm. Enlarge the the object, enlarge the 2cm disk. If I focus at infinity, then even the mountains have a 2cm blur, but 2cm looks very tiny at that distance.

I'm hoping for other simple universal relationship between these variables. I think there may still be a few.

Matt
 

Shashin

Well-known member
Matt, you absolutely can describe an image from object space--conjugate planes after all. I am more thinking out loud. Most of the formula used for things like DoF are very simple and are only really interested in what looks sharp for one particular condition. The OOF question is interesting. There is a strange idea the sharp and OOF are just binary conditions and so sharp is sharp and OOF and OOF, yet looking at images, it is not that simple.

Looking forward to seeing what you find.
 

Shashin

Well-known member
Just a little thinking...

Images are simply a single plane sample from a projected volume. Using a simple lens, the volume is the same regardless of focal length, just different sizes--magnification. But you should be able to intersect any volume with any size plane (format) and get the same image of the object plane--the in focus image. So, how does the entrance/exit pupil and size of the image plane effect the appearance of the resulting image? And how does the point spread change with those variables to create the OOF area?

Boy, am I glad you are the mathematician. I will stick with pub quizes on 70's pop music...
 
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