In response to "A question af wide angle lenses", I offer the following optical formulae for photographers, which should help clarify much of the confusion over DOF versus focal length...

Aperture or “f stop” = F/a Where F = lens focal length; and a = aperture diameter.

Circle of confusion. c = (v x D)/(1000 x S) Where S = print width in mm; D = negative width in mm; d = viewing distance,; and 1000 = a numeric constant. Thus, viewing a 200mm x 250mm print made from a 35mm negative at 250mm, would give us a maximum acceptable circle of confusion of 0.036mm. (Please note this formula is convention, and may or may not give you the precise result you want. Thus, you need to test this formula for your use, and modify the constant to suit you needs. This is also why many photographers dismiss the DOF scales manufacturers place on lenses.)

Depth of field. d near = U x F x F/[(U x c x f)+(F x F)] and d far = U x F x F/[(U x c x f)–(F x F)] Where U = focus point in mm; F = focal length of lens; c = circle of confusion; f = aperture and d far = farthest point in focus; d near = nearest point in focus. So, using the above circle of confusion with a 50mm lens focused at 10m (10,000mm) and set at f4, would render subjects in focus from 6.3m to 23.6m.

And so, if you shoot a photo from a distance of 10m with a 24mm lens, and enlarge the central portion of the negative by 2X so that it equals an image shot from the same point with a 50mm lens, you have just doubled the circle of confusion on the 24mm negative, thereby eliminating the advantage it had over the 50mm in the first place; and in fact, the resultant DOF’s are now equivalent. We can therefore conclude that relative DOF is strictly a function of aperture selected, subject distance, and final magnification of the subject on the film, and has nothing to do with lens focal length or film format. (!) However format, or negative size, does affect our calculation of an acceptable value for "c".

I hope this helps you all out -- Have fun with your calculators! I will be happy to answer any questions after the 7th. (I will be on vacation until then, and won’t be checking email… But I will be shooting with my M!)

;-)

Jack

-- Jack Flesher (jbflesher@msn.com), March 30, 2001

Answers

Before everybody jumps on me, I want to correct my statements in the second to the last paragraph. My statements in that paragraph are entirely incorrect. Focal length is important in the calculation for DOF, and is only partially cancelled out when multiplied by the new smaller factor for "c" in the denominator. My appologies. Obviously since lens focal length is in the formula, it is relevant. When you perform the math, you find that the shorter focal length image cropped and enlarged does in fact have a DOF advantage over the same image shot with a longer focal length lens. (However, while you are making inverse gains in DOF, you are losing resolution linearly and increasing grain exponentially, so in the end it is probably not a desireable trade-off.)

Also, in the formula for "c", it should read "v = viewing distance in mm".

I should know better than to comment on technical topics when I'm in a hurry.

Embarrassingly,

Jack

-- Jack Flesher (jbflesher@msn.com), March 31, 2001.

Nothing exceeds like excess!

Jack, down under, you would be considered a "tall poppy", !:}

but I do enjoy your test reports, and all the other filings; please keep it up, even if over the top, mate!

Paul Nelson

-- Paul Nelson (clrfarm@comswest.net.au), March 31, 2001.

Jack

As an aside I find virtually all such postings end up being corrected by the author at some stage, or being corrected by someone else - as a result I tend to ignore them! However, don't get me wrong, it is perfectly reasonable to post them! Thanks for all your contributions.

-- Robin Smith (smith_robin@hotmail.com), April 02, 2001.

Jack,

"Aperture or “f stop” = F/a Where F = lens focal length; and a = aperture diameter"

Maybe vice versa, as it is well known from the classic optic:

Aperture or “f-stop” = a/F, where F = lens focal length; and a = aperture diameter, isn't it? We got used to read a value of max aperture of a lens as 1:1.0; 1:1.4; 1:2.0; 1:2.8....and so on. Using your formula we would read this as 1.0:1; 1.4:1; 2.0:1; 2.8:1 .....32:1... So, this doesn't mean the value of the relative aperture.

Though it doesn't matter for your formula of a DOF ;)

best pics from M,

Victor

-- Victor Randin (www.ved@enran.com.ua), April 03, 2001.

Victor:

You may well be correct, but I quote the following two sources FYI: (1)Stroebel's *View Camera Techniques*, page 64: "The f-number (or relative aperture) is determined by dividing the focal length by the effective aperture." (2)The Ansel Adams book *The Camera*, page 46: "The lens aperture is simply the diameter of the lens opening, expressed as a fraction of its focal length. Thus a lens of 4-inch focal length with a diameter of one inch has a relative aperture of 4/1, or 4."

Regards,

-- Jack Flesher (jbflesher@msn.com), April 09, 2001.

Jack,

of course Ansel Adams is a great photografer, and I love him,

however:)

F/a = relative value of Focal length of lens, a/F = relative value of Aperture of lens,

where F is an absolute value of focal length of lens (mm), a - absolute value of aperture of lens (diameter, mm).

So, the DOF depends inversely on a/F, or depends directly on F/a, isn’t it?

Victor

-- Victor Randin (www.ved@enran.com.ua), April 10, 2001.

Victor:

What you say is correct. However, in convention within the optical formulae I've seen published, the F-stop, or relative aperture, is always expressed as the value F/a, not the inverse. Perhaps there are other formulae where a/F is used??? Perhaps when they publish data on a lens, they arbitrarily choose which method they use to express maximum aperture on that lens, as I've seen it done both ways, even within a given manufacturers line. I really don't know why it is done that way. Perhaps it is just a way to designate maximum aperture over relative aperture?

Regards,

Jack

-- Jack Flesher (jbflesher@msn.com), April 10, 2001.