Potential dead horse: how bad is FF's deep DoF disadvantage?

Started Mar 5, 2014 | Discussions thread
jackdan Contributing Member • Posts: 896
Re: Cha-ching!

Great Bustard wrote:

joejack951 wrote:

My guess is that since the aperture of a lens isn't really the physical size of the aperture (as if one used some calipers and measured the opening of the blades) but the size it appears to be as viewed from the front of the lens, the floating lens groups must change this apparent size.


Understanding the fundamental concepts of Equivalence requires making important distinctions between various terms which people often take to mean the same thing. It is very much akin to making the distinction between "mass" and "weight", two terms which most people take to mean the same thing, when, in fact, they measure two different (but related) quantities. While there are circumstances where making the distinction is unnecessary, there are other times when it is critical.

The first of these distinctions that needs to be made is between aperture and f-ratio. The term "aperture", by itself, is vague -- we need a qualifying adjective to be clear. There are three different terms using "aperture":

  1. The physical aperture (iris) is the smallest opening within a lens.
  2. The virtual aperture (entrance pupil) is the image of the physical aperture when looking through the FE (front element).
  3. The relative aperture (f-ratio) is the quotient of the focal length and the virtual aperture.

For example f/2 on a 50mm lens means the diameter of the virtual aperture (entrance pupil) is 50mm / 2 = 25mm, since the "f" in the f-ratio stands for "focal length". Likewise, a 50mm lens with a 25mm virtual aperture has an f-ratio of 50mm / 25mm = 2. The same relative aperture (f-ratio) will result in the same density of light falling on the sensor (exposure) for a given scene luminance and shutter speed for all systems, whereas the same virtual aperture (entrance pupil) will result in the same total amount of light falling on the sensor for a given shutter speed (as well as same DOF for a given perspective, framing, and display size). Thus, equivalent lenses are lenses that have the same AOV (angle of view) and virtual aperture (entrance pupil).

Interestingly, a "constant aperture" zoom is a zoom lens where physical aperture (iris) remains constant, but the virtual aperture (entrance pupil) scales with the focal length, thus keeping the relative aperture (f-ratio) constant as well. Let's consider a 70-200 / 2.8 zoom. Assuming the diagram in the link is accurate, then the proportions of the diagram results in the diameter of the physical aperture (iris) to be 38.5mm, which is based on the assumption that the diameter of the FE (front element) is 77mm, and relative size of the FE and physical aperture (iris) in the diagram.

The virtual aperture (entrance pupil) is the image of the physical aperture (iris), that is, it is how large the physical aperture appears when viewed through the front element. The diameter of virtual aperture is 25mm at 70mm f/2.8 (70mm / 2.8 = 25mm) and 71mm at 200mm f/2.8 (200m / 2.8 = 71mm). So, as the lens zooms, neither the physical aperture (iris) nor the relative aperture (f-ratio) change, but the virtual aperture (entrance pupil) changes in direct proportion to the magnification. Since the diameter of the physical aperture (iris) is 38.5mm in the above design, then we can basically think of this particular 70-200 / 2.8 lens as a 108 / 2.8 prime (assuming that the virtual aperture and physical aperture have the same diameter with the prime, which is probably not a good assumption) with additional elements that magnify or reduce the image as it zooms.

You need to rewrite the Wikipedia entry for the definition of aperture as it pertains to photography. Surely you have considered it.

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