Holy Depth of Field, Batman (aka, what's your fastest lens?)

[snap58]

Would be my 135 2.0 L

 

[0098386]

My best attempt not a fast lens.

 

[Clix Pix]

[
My best attempt not a fast lens.

A neat idea! In this instance a tripod would have helped significantly so that you could have then had an aperture which would allow for appropriate depth-of-field....

 

[jared_kipe]

Very-wide aperture lenses often exhibit more obvious vignetting (edge darkening) because of the optical effect of light rays having to travel further to the focal plane the further away they are from the optical axis. Also, their minimum aperture is usually less wide (typically only stopping down to f16) since smaller apertures often introduce diffraction (seen as a softening and drop in image contrast). That's maybe what you meant about wide-barrel lenses having lower optical qualities.

...
f1.2 is half a stop wider than f1.4. f1.0 is a stop wider than f1.4. FYI f-numbers increase/decrease by a factor of the square-root of 2 (with a little rounding-up to keep things sane).

Since the f-stop number is the ratio between the focal length and the diameter of the front lens element (the effective diameter after the aperture blades), logic would dictate that f16 will loose exactly the same amount of detail to diffraction as any lens of equal focal length.

As per the increase or decrease by square root of 2 thing, the square root of 2 is 1.414, so if we add to it 1.414 we get 2.828, not 2.0. And if we subtract it we get zero. The square root of two thing is an approximation, unless I don't understand what you're talking about. 5.6+1.414=~7 and 8+1.414 =9.414, not 8 and 11 respectively.

I will now atempt to derive a relationship to give you the correct f stop relationship...

Since the aperture ratio goes down one stop when the area of the widest point

ratio=f/(2r)

area of circle=a=Pi(r^2)

Now if we solve for r and assume we will be taking 1/2 over and over to find each successive aperture number then we'll say

r=squareroot(a/(pi 2^n)) where n is an arbitrary number I will give constraints in a bit

ratio=f/(2squareroot(a/(pi 2^n)))

now we know ratio=1 is a value, and we'll let that value occur when n=0

1=f/(2squareroot(a/(pi 2^n)))
or
f=2squareroot(a/pi 2^0) with 2^0=1 for those who don't know

that lets us simplify everything nicely to
ratio=1/(1/squareroot(2^n)) which cleaning up using rules of exponents to

ratio=2^(n/2) there is where you get your square root of 2 nonsense, its when n=1 ... let us verify

n=0 ratio=1
n=1 ratio=1.414
n=2 ratio=2 (we picked up the missing 2)
n=3 ratio=2.83
n=4 ratio=4
n=5 ratio=5.66
n=6 ratio=8

And so on , note that for even n this ratio will always be a counting number, and for all odd n the ratio will be irrational.

I would assume this would work with n being negative too, to find stops faster than 1.0.
EDIT: Also note that this formulae works for 1/2 or 1/3 stop too, just make it 2^(n/4) or 2^(n/6) respectively.

 

[snap58]

Since the f-stop number is the ratio between the focal length and the diameter of the front lens element (the effective diameter after the aperture blades), logic would dictate that f16 will loose exactly the same amount of detail to diffraction as any lens of equal focal length.

As per the increase or decrease by square root of 2 thing, the square root of 2 is 1.414, so if we add to it 1.414 we get 2.828, not 2.0. And if we subtract it we get zero. The square root of two thing is an approximation, unless I don't understand what you're talking about. 5.6+1.414=~7 and 8+1.414 =9.414, not 8 and 11 respectively.

I will now atempt to derive a relationship to give you the correct f stop relationship...

Since the aperture ratio goes down one stop when the area of the widest point

ratio=f/(2r)

area of circle=a=Pi(r^2)

Now if we solve for r and assume we will be taking 1/2 over and over to find each successive aperture number then we'll say

r=squareroot(a/(pi 2^n)) where n is an arbitrary number I will give constraints in a bit

ratio=f/(2squareroot(a/(pi 2^n)))

now we know ratio=1 is a value, and we'll let that value occur when n=0

1=f/(2squareroot(a/(pi 2^n)))
or
f=2squareroot(a/pi 2^0) with 2^0=1 for those who don't know

that lets us simplify everything nicely to
ratio=1/(1/squareroot(2^n)) which cleaning up using rules of exponents to

ratio=2^(n/2) there is where you get your square root of 2 nonsense, its when n=1 ... let us verify

n=0 ratio=1
n=1 ratio=1.414
n=2 ratio=2 (we picked up the missing 2)
n=3 ratio=2.83
n=4 ratio=4
n=5 ratio=5.66
n=6 ratio=8

And so on , note that for even n this ratio will always be a counting number, and for all odd n the ratio will be irrational.

I would assume this would work with n being negative too, to find stops faster than 1.0.
EDIT: Also note that this formulae works for 1/2 or 1/3 stop too, just make it 2^(n/4) or 2^(n/6) respectively.

Maybe if you multiply by 1.414?