# Lens equivalent depth of field [duplicate]

This question is ported directly from Physics SE, because the guys there believe this is the right place for it.

In digital photography, there's the principle of lens equivalency, according to which, when you shoot with a camera which has a smaller sensor than a full frame sensor (which is the same dimension as classic photography film), you effectively reduce your field of view (relative to shooting with the same optics and a full frame sensor). The resulting crop of the optical image is associated with saying that you are shooting a full frame sensor, but a lens with a greater focal length.

So in terms of the resulting image, one should think of the lens as having a focal length of ff (the real focal length) times the crop factor CC associated with the sensor being used (C≥1). this is all well and good intuitively but then comes the 'weird' part.

It is said that if you're to think of your lens to have a focal length fC, then you must also think of the camera's f-number as being multiplied by C as well. This means that if you increase your f-number (which is proportional to the numerical aperture), then you also increase the depth of field of the camera - and surely enough, for a smaller sensor, you yield a greater depth of field (for the same field of view) - even when your actual physical f-number is the same in both cases.

How can two optical systems with the same f-number\NA, have different depths of field?

I realize that some of you might say it's related to the pixel size of the cameras, but it goes the other way around: the system with smaller pixels (which means less tolerance for the spread of the PSF) has the greater depth of field. This is best demonstrated in modern days smartphones, where for the same f-number as full frame cameras, the depths of field are worlds apart.

What is the physical explanation for this phenomenon? My best guess is that for systems with smaller sensors, one must position the camera further away from the object to capture the same field of view, such that the light rays travel a greater distance and thus become more parallel before being collected by the optics, but I'd like to know for sure.

• You've got the middle section backwards. If you want the DoF to be the same you have to move the f-number the other way. A 100mm FF lens at f/4 is equivalent to a 50mm micro four thirds (2X crop factor) lens at f/2, not at f/8. Aug 10, 2017 at 8:11
• On the other hand, if you want the exposure to be the same, you use 50mm at f/4 and get deeper DoF. Aug 10, 2017 at 8:12

Within the context of creative photography (which is how all questions here should be understood), Depth of Field (DoF) is defined based upon blur perceptible by the viewer.¹ Any blur circle in an image at a specific display size and distance small enough to be perceived as a point by a viewer with a defined visual acuity is deemed to be within the DoF. Any blur circle large enough that it can be perceived as larger than a point by the same viewer at the same display size and distance is deemed to be outside the DoF.

With that context in mind, a word about what depth-of-field is and is not:

In a way, depth-of-field is an illusion. There is only one plane of focus. Everything in front of or behind the point of focus is out of focus to one degree or another. What we call DoF is the area where things look, to our eyes, like they are in focus. This is based on the ability of the human eye to resolve certain minute differences at a particular distance. If the slightly out-of-focus blur is smaller than our eye's capability to resolve the detail then it appears to be in focus. When you magnify a portion of an image by making it larger or moving closer to it you allow your eye to see details that before were too close together to be seen by your eyes as separate pieces of the image.

Since things are gradually blurrier the further they are from the point of focus, as you gradually magnify the image the perceived depth of field gets narrower as the near and far points where your eyes can resolve fine details moves closer to the focus plane.

There is no intrinsic physical property that makes everything within a specific distance 'sharp' and everything outside that distance 'blurry'. There's no magic line inside of which everything is equally sharp and outside of which everything is equally blurry. There is the focus distance that is the 'least blurry' and there is everything else that gets increasingly blurrier the further one moves from the point of focus. Within the context of creative photography (which is how all questions here should be understood), the question to be answered is not, "Is it blurry?" The question to be answered is, "Is it blurry enough that a viewer with a specific visual acuity looking at it at a certain magnification from a certain distance will be able to tell that it is blurry?" Or to put it in the way most people in the photographic field state it, "Is it acceptably sharp?"

In fact, you can take the same exact photo and print it at two different sizes, view them from the same distance, and they will have different depths of field. It is a gradual transition from sharp to blurry. How much we magnify the virtual image cast by the lens onto the sensor as it is recorded by the digital sensor determines the exact size of blur (measured on the sensor) that we can perceive when we view it. There are several variables that affect just how far from the point of focus things start becoming noticeably blurry to our eyes.

The blur becomes noticeable at smaller distances from the point of focus (and thus the perceived DoF is smaller) if we:

• Use a longer focal length/narrower angle of view
• Use a shorter subject distance
• Use a wider aperture
• Use a larger display size
• View the displayed image from a closer distance
• Have better vision

The blur increases more gradually at larger distances from the point of focus (and thus the DoF is perceived to be larger) if we:

• Use a shorter focal length/wider angle of view
• Use a longer subject distance
• Use a narrower aperture
• Use a smaller display size
• View the displayed image from a larger distance
• Have weaker vision

Additionally, if the imaging system is either diffraction limited or resolution limited so that everything smaller than a specific size is equally blurry, this will affect the perceived DoF. When nothing is seen as 'sharp' anything with the same amount of blur is seen as within the DoF.

To properly calculate depth of field all of these factors must be taken into account. Many DoF calculators make (often unspoken) assumptions about some of them. Most DoF calculators, such as DOF Master, assume an 8x10 display size viewed from a distance of 10 inches by a person with 20/20 vision.

Since depth-of-field is dependent upon viewing size and distance as well as the visual acuity of the viewer it is hard for a DoF calculation to indicate depth-of-field if it doesn't know what the display size of the photo will be. The same goes for lenses that may be used with different cameras that have different sensor sizes. The DoF scale for the same lens will be different for an APS-C camera than it would be for a Full Frame camera if we plan to display images from both at the same size and distance.

Assuming the standard circle of confusion used for an image produced with a 36x24mm sensor, displayed at 8x10, and viewed at 10 inches by a person with 20/20 vision will accurately predict perceived DoF for most images is too broad in today's environment. The digital photography revolution has pretty much eliminated any idea of a standard display size and viewing distance. For Depth of Field calculations to be accurate they must be based on all of the variables listed above including the display size and viewing distance as well as the focal length, f-number, and angle of view determined by the sensor size (which directly affects the magnification ratio needed to display an image at a specific size). What is 'acceptably sharp' when viewed scaled to fit entirely onto a monitor will be different than the same image file viewed at 100% (1 image pixel equals one screen pixel). Viewing a 24MP image on a 23" HD (1920x1080) monitor at 100% is like looking at a small part of a 60x40" print!

If you want to account for differing display sizes and distances, you can use the Flexible Depth of Field Calculator from Cambridge in Colour and click show advanced to enter those variables.

## Now, let's look at the idea of true equivalency:

There's no such thing.

• When we change the size of the camera's sensor and the focal length of the lens we don't change the wavelength of visible light. The light does not scale at the same proportions as our lenses/cameras do. Specifically with regard to the question, the same f-number on a smartphone with a focal length of 3.3mm that yields the same FoV as a FF camera with 25mm lens is 7.2 times narrower. If we are using f/1.8 the FF lens has an opening 13.9 mm wide. The smartphone has an opening 1.83 mm wide. The size of the light waves trying to squeeze through that 1.83 mm hole are NOT 7.2 times smaller than the light waves passing through that 13.9 mm hole, though. The effects of diffraction will substantially impact the image projected onto the smaller smartphone sensor compared to the image projected onto the FF sensor because a much higher percentage of the light rays going through the smaller aperture of the smartphone will be scattered by interacting with the surface edges of the aperture diaphragm. Light rays travel in waves, not in straight lines as we like to draw them in ray diagrams.
• We can't change the camera position without also changing the perspective which is solely dependent upon the camera position and the relative positions of everything in the field of view. To get the same photo, the optical center of the camera's lens must be in the same location for both cameras. To get an 'equivalent' photo we must use different focal lengths that are in the same proportions as the linear measurements of the sensors.
• When we change the f-number to compensate for the combined changes in magnification and focal length, we also change the exposure. To keep the exposure constant with the same ISO and shutter time, we can't change the f-number at all. If we change the shutter time to compensate for the difference in f-number, any scene that includes objects in motion (or any image taken with the camera in motion) will no longer look the same. Don't forget that an f-number is a dimensionless ratio between the focal length of a lens and the diameter of the entrance pupil. Also keep in mind that any aperture wider than f/13.6 with the 25mm lens for the FF camera will be less affected by diffraction than the 3.3mm lens at f/1.8 in the smartphone. And that is before we even begin to consider the difference in pixel sizes between the two cameras.

Depth of field is all about angles and whether the narrowest angle the viewer can discriminate is larger or smaller than the angle between the smallest discreet details in a photograph as they are viewed.

Unless I'm missing something, you haven't answered my question. Either way, true DOF does exist, and it has a well-defined physical definition, which is very much independent from the observer (I won't go into here). There's no illusion in a smartphone and a full frame DSLR which have the same f# (which supposedly determines your DOF), yet given the same FOV, the two will yield a vastly different DOF.

Within the context of creative photography (which is how all questions here should be understood), the 'well-defined physical definition' would only be applicable if the image is enlarged enough that the viewer can discriminate individual pixels in the image. In the case of current hardware, that would require enlargements and viewing distances that are well beyond what is typical.

At one foot viewing distance a person with 20/15 vision can resolve only about 290 ppi. At 8 inches it is 440 ppi. So the image from a camera such as the Nikon D810 would need to be enlarged to 25x17 inches yet still viewed from only a 12 inch distance before the circle of confusion is reduced to the same size as the camera's pixel pitch. To display that image at full resolution one would need a 30 inch monitor with a resolution of 7360x4912 (that is, a pixel pitch of 290 ppi). Typical 30 inch monitors have resolutions of only around 1920x1080 which is around 75-80 ppi. So even looking at the thing with a magnifying glass we would not be able to discriminate details smaller than about 4 pixels wide (in the original 7360 pixels wide image file before it was scaled down to be displayed at 1920 pixels wide). But who looks at a 30 inch monitor from only one foot away? When we back up to a more typical viewing distance of two to three feet, the circle of confusion grows even larger and the perceived DoF increases.

Depth of field is anything but subjective. It can be quantified using with points spread functions and Reighley's criterion for optical resolution.

I've edited the answer to make clearer that within the context of creative photography, which is what this entire site is about, the perception of blur by the viewer is the defining factor in the definition of DoF. Any definition of DoF that includes differences too small to be perceived by the viewer are moot within the context of creative photography. Even if an optical system and/or recording medium is capable of much higher resolution than the viewer can perceive those details are irrelevant when talking about DoF within the context of creative photography.

Point spread functions and Reighley's criteria are useful for defining the absolute limits within which an optical system can discriminate details. But they must take into account all of the limiting factors in an optical system.

In the context of a human viewing a photograph, the most limiting factor in the entire system is most often the human's ability to perceive the difference between minute amounts of blur and points.

• Unless I'm missing something, you haven't answered my question. Either way, true DOF does exist, and it has a well-defined physical definition, which is very much independent from the observer (I won't go into here). There's no illusion in a smartphone and a full frame DSLR which have the same f# (which supposedly determines your DOF), yet given the same FOV, the two will yield a vastly different DOF. Aug 9, 2017 at 10:38
• @YuvalWeissler Yes he did. You probably didn't understand his detailed answer. I used the technical terms with my brief duplicate of Michael's.
– Stan
Aug 9, 2017 at 14:43
• @ Yuval Weissler -- Sorry to report that depth-of-field is subjective. However the industry usually uses: Near distance P/1+PDf/F^2 Far distance P/1-PDf/F^2 P= point focused upon --- D = diameter of circle of confusion --- f = f-number --- F = focal lenght Aug 9, 2017 at 16:40
• Depth of field is anything but subjective. It can be quantified using with points spread functions and Reighley's criterion for optical resolution. Aug 10, 2017 at 5:03
• @YuvalWeissler I've edited the answer to make clearer that within the context of creative photography, which is what this entire site is about, the perception of blur by the viewer is the defining factor in the definition of DoF. Any definition of DoF that includes differences too small to be perceived by the viewer are moot within the context of creative photography. Even if an optical system and/or recording medium is capable of much higher resolution than the viewer can perceive those details are irrelevant when talking about DoF within the context of creative photography. Aug 10, 2017 at 5:21

They don't.

For any given lens system, nothing changes with the exception of the image size due to the reduced sensor. The Hyperfocal distance remains the same. The circle of confusion remains the same.

Proof: Make two prints of the same scene with the same image size (to compensate for the reduced sensor field size) and the differences will be negligible.

Your perception of the two uncompensated images is what confuses you.

The apparent depth of the plane of focus seems reduced due to the increased size of the print/display. The out of focus points can be observed and discriminated more easily as they are enlarged and become more obvious.

• The circle size may remains the same at the image plane however the smaller format requires more magnification to obtain the same size displayed image. Thus the circle size for the smaller format is more stringent. If the crop (magnification) factor is 1.5, the smaller format requires a smaller circle at the image plane. This will be 1/1.5 X 100 = 66%. In other words the smaller format's requirement is 66% of the larger as to circle size. Aug 9, 2017 at 15:52
• Right, a smaller sensor is NOT the same. . The blur circle may remain the same in a cropped sensor (with same lens and same distances), but CoC as a limit is computed from sensor diagonal, so CoC as a limit does NOT remain the same in a cropped sensor. Therefore, Hyperfocal and DOF do NOT remain the same in a cropped sensor (because of course, the smaller sensor requires more magnification to be compared similarly). Aug 9, 2017 at 15:57
• When the diagonal of a FF camera is compared to the diagonal of a smartphone with a 1/3"-1/3.6" sensor, the CoC changes by a factor of 7-8 or so. Aug 10, 2017 at 5:54

A simple explanation: the blur you get outside the focus plane is directly linked to the absolute size of the entrance pupil of your lens. So, when you reduce the size of each element of your camera (sensor and lens), you also reduce the size of the pupil and hence increase the DoF. If you want a constant DoF, you need to keep the size of the pupil constant. But since the size of the pupil is usually expressed as f-number, i.e. proportionally to the focal length (f/2.8 means "pupil size is equal to the focal length divided by 2.8). So, to keep a constant pupil size while reducing the focal length, you need to increase the f-number.

Obviously, you can't always reduce the f-number. To get the equivalent of an f/1.4 lens on a full frame with an iPhone for example, you'd need an f/0.2 lens, and I don't think anyone has ever built such lens). So, basically, you can't get the equivalent of a full-frame f/1.4 lens on a smartphone.

In short, if you need to augment/reduce the f-number by as much as you augment/reduce the focal length to keep the DoF constant.

How can two optical systems with the same f-number\NA, have different depths of field?

I realize that some of you might say it's related to the pixel size of the cameras, but it goes the other way around: the system with smaller pixels (which means less tolerance for the spread of the PSF) has the greater depth of field. This is best demonstrated in modern days smartphones, where for the same f-number as full frame cameras, the depths of field are worlds apart.

What is the physical explanation for this phenomenon?

TL;DR: The size of the circle of confusion on the image sensor scales down quadratically with crop factor, as opposed to the lens's focal length and sensor's diagonal linear scaling down with crop factor.

Since this question came by way of Physics.SE, I'm going to give the equation-oriented answer to your question.

Wikipedia's Depth of field article provides us with handy equations for the depth of field near and far limits, $$\D_\text{n}\$$ and $$\D_\text{f}\$$, respectively. For the purposes of our work, either equation will work (we will get to the same answer); I'm going to use the equation for the near limit of field depth:

$$D_\text{n} = \frac{sf^2}{f^2 + Nc(s-f)}$$

where

• $$\f\$$ is the focal length of the lens;
• $$\N\$$ is the f-number (relative aperture number) of the lens;
• $$\s\$$ is the subject distance (the distance from the entrance pupil of the lens to the plane/subject of focus);
• $$\c\$$ is the diameter of the circle of confusion on the image plane (i.e., sensor), which is a function of the desired minimum acceptable sharpness of an image viewed at a certain size from a chosen distance. For a full-frame 35 mm format, this number is typically around $$\d/1500\$$ to $$\d/1730\$$ (where $$\d\$$ is the diagonal of the sensor), which correspond to resolving about 5 lines per mm on a print 30 cm diagonal.

Comparing two different camera systems related by the crop factor $$\k\$$ ($$\k\$$ is chosen because $$\c\$$ is already being used), $$\f_k\ = f/k\$$; $$\f\$$ is the referent focal length (e.g., full frame 35mm format), and $$\f_k\$$ is the scaled format (e.g., APS-C, in which case $$\k = 1.5\$$ (Nikon) or $$\k = 1.6\$$ (Canon)).

Now, your question expects that two camera systems with equal f-numbers $$\N_k = N\$$, logically should produce equal depths of field, and wonders why they aren't equal. Also, the subject distance for each camera is equal ($$\s_k = s\$$), meaning we're shooting the same scene with the same subject focus distance — otherwise we aren't comparing apples to apples.

So taking another copy of the near limit of depth of field equation, $$\D_{\text{n}k}\$$, substituting $$\f_k = f/k\$$, $$\N_k = N\$$, $$\s_k = s\$$ and setting them equal to each other,

\begin{align} D_{\text{n}k} &= D_\text{n} \\ \frac{s_kf_k^2}{f_k^2 + N_kc_k\left(s_k - f_k\right)} &= \frac{sf^2}{f^2 + Nc\left(s - f\right)} \\ \frac{s\left(f/k\right)^2}{\left(f/k\right)^2 + Nc_k\left(s - f/k\right)} &= \frac{sf^2}{f^2 + Nc\left(s - f\right)} \\ \end{align}

Algebra yields

$$c_k = \frac{c}{k^2}\cdot\frac{s - f}{s - f/k}$$

Looking at the 2nd rational term on the right, for non-macro distances, $$\s\gg f\$$. For instance, with a 50 mm lens, a rather close portrait at 1m is 20× larger than the lens's focal length. And for a crop sensor camera (looking at the denominator), to get the same field of view, the difference is even greater. Thus, for non-macro distances, this entire 2nd term approaches unity from above for both increasing crop factor $$\k\$$, and increasing subject distance–to–focal length ratio $$\s/f\$$.

Therefore, $$\c_k \approx \displaystyle\frac{c}{k^2}\$$.

So, for fixed f-numbers between two camera systems taking the same picture with the same framing of the same subject and focus, in order for the depths of field to be identical, as you decrease the focal length and sensor diagonal by the crop factor $$\k\$$, you must decrease the diameter of the circle of confusion by $$\k^2\$$.

As you do this, you will quickly bump into limits:

1. Another measure that didn't scale by the crop factor in the compared systems is the wavelength of light. A hard lower limit to the diameter of the circle of confusion is the diameter of the Airy disk produced by the diffraction of light through a circular aperture. The size of the Airy disk is a function of the wavelength of light, and the f-number. The wavelength of light is of course independent of the camera system, and the f-number has been fixed by the constraints of this scenario.

For example, a CoC of d/1730 for a 35 mm full frame camera is 0.025mm. The iPhone 7 has a crop factor of about 7.21 for its small sensor. So to meet the constraints of this scenario, its CoC would have to be 7.212 = 52 times smaller, at 0.48 µm. Choosing an aperture of f/1.8 (the iPhone 7's aperture), the Airy disk produced by violet light is around 922 nm, or twice the CoC diameter constraint of this scenario.

2. Of course, engineering constraints prevent easy scaling as well. Sensor pixel sizes can't feasibly scale down according to crop factor. For example, the previously mentioned Nikon D810's sensor's pixel width is 4.88 µm, about 5× smaller than the 35mm "typical" CoC size of d/1730.

The iPhone's 1.12 µm sensor pixel width is 2.3× larger than the "required" CoC diameter for this scenario. And also note that the comparison isn't otherwise identical: the D810 has 36 megapixels, vs. the iPhone 7's 12 MP.

So even sacrificing resolution to compensate for the $$\k^2\$$ CoC scale factor, the iPhone 7 is both diffraction limited (by the Airy disk), and still under-resolves the comparison scenario's requirements.

It's these lower limits, both fundamental physics and engineering, that keep the depth of field high as you scale down camera sensors.

• "The size of the Airy disk is a function of the wavelength of light, and the f-number." Actually, it is a function of the wavelength of light and the actual diameter of the physical diaphragm (as well as the shape of the aperture opening when it is not perfectly circular), rather than the diameter of the entrance pupil used along with focal length to determine f-number. As light is refracted by a lens, the wavelengths don't change, only the density of the light waves change. Aug 10, 2017 at 4:50
• This answer goes the way I was looking for, so thanks right off the bat :) So, from what I gather, your answer points to my suspicion that the DOF is different for the two systems (with the same f#) because of the different distances from the subject, correct? Aug 10, 2017 at 4:58
• @YuvalWeissler If I understand your question correctly, then no. Two cameras with equal field of view (say, 35mm full frame with 75mm lens, and 35mm crop 1.5 (Nikon) with 50mm lens), both taking the same image of the same subject, framed identically, will be the same distance from the subject. It's the distance between the lens and sensor that gets shorter when you scale down the system but maintain the same subject distance.
– scottbb
Aug 10, 2017 at 5:04
• If the iPhone's max aperture is f/1.8, the equivalent f-number for the FF camera is around f/13. That's where the largest difference lies: lenses for the FF camera used at apertures much wider than f/13 (such as f/8 or even f/1.8) means the systems are much less diffraction limited than the iPhone is, even when the differences in pixel pitch are taken into account. There's no such thing as true equivalency in photography. Aug 10, 2017 at 5:58
• @MichaelClark indeed. The crop-squared in my answer yields ridiculously small CoC only because of the strict equal f-number of the "experiment". Practically, you can "peel off" one of those crop factors and apply it to get equivalent f-number, and get much more sane comparisons. Basically, the contrived constraints of the scenario just swamps the system in the only way it can manifest. But even when accounting for equivalent f-number, everybody's point (such as in your answer) still applies: "equivalence principle" is not so equivalent when you peel back the curtain.
– scottbb
Aug 10, 2017 at 6:05

Depth-of-field is intertwined with the degree of magnification required to enlarge the actual image (imaging chip dimensions) to some suitable size for viewing. With this in mind, most depth-of-field tables assume an 8X10 inch print will be the end result. Now the revered 35mm film size is 24mm height by 36mm length. To make an 8X10 from this format requires a magnification (enlargement) of 8.5X. The compact digital format is also based on a film size popular in the 1990’s. This APS-C format (Advanced Photo System – Classic) measures 16mm height by 24mm length. To compare, we divide the two diagonal measures. Thus 43.3 ÷ 28.8 = 1.5. This value is the crop or magnification factor. To make the same 8X10 final image, the magnification (degree of enlargement is 8.5 X 1.5 = 12.75X. In other words, more magnification is required if the format is smaller, to achieve an image of comparable size.

So now we tackle how depth-of-field is calculated:

The camera lens views a vista as if it consists of countless points of light. A point being something so tiny it has no perceivable dimension. The lens handles each point and projects them as a tiny circle. These are the countless circles of confusion that make up an optical image. Under a microscope, these circles appear as indistinct and overlapping, hence the name “circles of confusion”. We perceive an image to be sharp if these circles are too small to be perceived as discs. As an example, newspaper photos appear un-sharp because the dots of ink that they are comprised of are too large.

So what does it take to render these circles dimensionless? The photo industry generally accepts that a circle of confusion will be perceived as a point when its diameter is ½ mm or less viewed from 500mm. That works out to 3.4 minutes of arc. The key argument is that the circles must be so small that they appear dimensionless. This is rather complex because this circumstance intertwines viewing distance. Additionally it interweaves image contrast, viewing illumination level and the acuity of the observer. In other words, image sharpness is subjective “in the eye of the beholder”.

Now suppose you are making an 8X10 from a full frame 35mm format 24mm by 36mm. As stated, the minimum magnification is 8.5X. How small must the circles of confusion be at the image plane? Answer 0.5mm ÷ 8.5 = 0.06mm. To make a comparable image from a compact digital 16mm by 24mm, the circle size must be 0.5mm ÷ 12.75 = 0.04mm. What we have learned is, the circle size for the compact digital, as projected by the lens, must be 66% of the size of full frame (1/1.5 X 100 = 66%).

How does the industry manage when calculating depth-of-field tables? The industry uses some fraction of the focal length. Most depth-of-field tables are computed using 1/1000 of the focal length. This method, as a rule of thumb roughly takes into account the magnification that will likely be used to make the final print and a guesstimate as to the viewing distance. Now for critical work, the Leica standard is 1/1500 of the focal length, the Kodak standard is 1/1750 of the focal length.

Using the 1/1000 standard, mount a 50mm lens and depth-of-field will be calculated based on a circle size of 50 ÷ 1000 = 0.05mm. What will the circle size be for an 8X10? Answer 0.05 X 8.5 = 0.43mm (image appears sharp. For the compact digital we mount a 30mm. Using 1/1000 as the circle size, 30 ÷ 1000 = 0.03mm. What will the circle size be for an 8X10? Answer 0.03 X 12.75 = 0.38mm (image appears sharp. Remember 0.5mm is limit for a viewing distance of 500mm.

I don’t know about you but I view this stuff as gobbledygook!

Lens equivalency is not a principle. It is a marketing term. It sort of equates to field of view, but that's not a principle either. It's just the name of a particular feature of an optical configuration.

The confusion behind the question is that it intermingles the psychometric terms with terms from Newtonian physics, e.g. "focal length" from Newtonian Physics and the psychometric term "Depth of Field". In a particular context with a particular purpose it may be possible to map depth of field characteristics to information and then apply information theory in a way consistent with Newtonian physics. But only after establishing criteria for judging the quality of photons captured from a particular scene as signal to noise.

The practice of photography is strongly empirical. That does not elevate the The Rule of Thirds to the level of scientific principle...it's not the Third Law of Thermodynamics.