It seems odd to me that Canon EF 100-400mm f/4.5-5.6L does away with a front element of only about 63 mm, as reported by @jrista - which would be enough for only f/6.3 at 400mm, missing the spec by third of a stop.
It makes me wonder if it's possible to measure what aperture is actually used during taking a photo. It'd be useful both in the described case and exploring how exact stopping down to a smaller aperture actually is.
So my question is - how to measure what aperture is actually used to take a photo? It's okay if the scene has to be specially constructed/measured for performing the test.
Answer
You can probably calculate this by rearranging the DOF formula to solve for c, or circleOfConfusion, as @MattGrum stated. I haven't tried to rearrange a formula as complex as DOF for a while, so I hope my math is correct here:
DOF = (2 Ncƒ²s²)/(ƒ⁴ – N²c²s²)
The terms of that equation are as so:
DOF = depth of field
N = f-number
ƒ = focal length
s = subject distance
c = circle of confusion
For simplicity sake, I'm going to reduce the DOF term to just D.
Now, the term for c appears twice in this equation, one of them to the power of two, so were probably looking at a polynomial of some sort in the end. To rearrange:
D = (2Ncƒ²s²)/(ƒ⁴ – N²c²s²)
D * (ƒ⁴ – N²c²s²) = (2Ncƒ²s²)
Dƒ⁴ – DN²c²s² = 2Ncƒ²s²
0 = 2Ncƒ²s² + DN²c²s² – Dƒ⁴
DN²c²s² + 2Ncƒ²s² – Dƒ⁴ = 0 <-- QUADRATIC!
As Indicated, rearranging terms produces a quadratic polynomial. That makes it pretty strait forward to solve, since quadratics are a common type of polynomial. We can simplify for a moment by substituting some more general terms:
X = DN²s²
Y = 2Nƒ²s²
Z = –Dƒ⁴
That gives us:
Xc² + Yc + Z = 0
Now we can use the quadratic equation to solve for c:
c = (–Y ± √(Y² – 4XZ) ) / (2X)
Replacing the X, Y, and Z terms with their originals and reducing:
c = (–2Nƒ²s² ± √(4N²ƒ⁴s⁴ + 4D²N²ƒ⁴s²) ) / (2DN²s²)
(Whew, thats pretty nasty, and I hope I got all the right terms replaced and typed in correctly. Apologies for discrepancies.)
My brain is a bit too fried right now to figure out exactly what it means for the circleOfConfusion to be quadratic (i.e. having both a positive and negative result.) My first guess would have to be that c grows both when you move towards the camera from the focal plane (negative?), as well as away from the camera and focal plane (positive?), and since quadratic equations grow to infinity pretty quickly, that would indicate the limit on how large or small the circle of confusion could actually become. But again, take that analysis with a grain of salt...I scratched out the solution to the formula and that took the last bit of brainpower I had left today. ;)
If that is the case, then you should be able to determine a maximum CoC for a given aperture and focal length, which would, hopefully, be (or allow deriving) the diameter of the aperture (entrance pupil.) I am willing to bet, however, that this is not actually necessary. My analysis on the linked answer of @Imre's question was rather rough...I don't quite have the ability to observe my 400mm lens' aperture at "infinity", so I am probably seeing the entrance pupil incorrectly. I would be willing to bet that at a sufficient distance that you could call "infinity", the 100–400mm lenses f/5.6 aperture at 400mm would indeed appear to be the same diameter as the front lens element, so at least 63mm in diameter. My measurement of the diameter of that lens was a bit rough too, and it could be off by ±3mm as well. If Canon's patent for a 100–400mm f/4-5.6 lens is telling, the actual focal length of the lens is 390mm, and the actual maximum aperture at "f/5.6" is really f/5.9. That would mean the entrance pupil would only need to appear 66mm in diameter "at infinity", which is within margin of error for my measurements. As such:
I believe the EF 100–400mm f/4.5–5.6 L IS USM lens from Canon is probably spot-on as far as aperture goes, with a 390mm actual focal length and a 66mm entrance pupil diameter, all of which would jive with my own actual measurements of this lens.
No comments:
Post a Comment