DOF Calculations for Telephotos

[neil poulsen]

If I divide the focal length by the f-stop for a telephoto lens, do I obtain the actual aperture diameter? This works for standard lens designs like a plasmat, but does it work for telephotos?

I've been making dof tables, and I'm not sure whether the same forumlas work for telephotos.

[Dan Fromm]

Why shouldn't they work?

[neil poulsen]

I only know enough to be dangerous. But, telephoto lenses are closer to the image plane then regular lenses, so for the same f-stop, their aperture diameters must be smaller. But, I'm not sure how f-stops are calculated for telephotos, hence my question.

[Dan Fromm]

I think you're confusing flange to film distance with rear nodal point to film distance.

Cheers,

Dan

[sanking]

As far as DOF is concerned lenses are lenses.

There are some pretty neat programs on the net. For example, see http://dfleming.ameranet.com/dofjs.html

Some of the programs, including this one, allow you to set the size of your circle of confusion, a feature that I particulary like.

You can also download the program Palm Pilot OS or Windows.

[Graeme Hird]

"....do I obtain the actual aperture diameter?"

My understanding is that you will get the effective aperture rather than the physical size of the opening.

Anyone is welcome to dispute this.

Cheers, Graeme

[Bob Salomon]

If you do not know the actual focal length of the lens then your answer would be close but not exact.

[Chris Gittins]

Graeme is correct. Dividing focal length by f-number gives you an effective aperture diameter which may or may not correspond to physical size of the opening you're using to stop the lens.

Chris

[Struan Gray]

The f-number is defined as the focal length divided by the diameter of the entrance pupil, not the aperture as such. The entrance pupil is the aperture as seen through the glass of the front lens cell, so if that cell has a high power the size of the entrance pupil can be fairly different from that of the hole formed by the aperture blades.

Depth of field is determined by the position and size of the *exit* pupil. This is the aperture as seen through the *back* of the lens. The exit pupil's apparent size and position fixes a base and angle for the cone of light converging onto the focal plane from the lens, which in turn determines the DOF.

If you look at a telephoto lens from both ends you'll see that the aperture looks smaller when seen from the back of the lens than from the front: i.e. the entrance and exit pupils have different sizes. The asymmetry reflects the asymmetry of the telephoto design, which in its simplest form has a converging element in front of the aperture and a diverging one behind - these necessarily give different appearances to the aperture sandwiched between them.

Crucially, the same asymmetry in the optical design that makes the aperture look different is also responsible for moving the rear nodal point forward, allowing you to use less bellows extension. For things like bellows factor the asymmetry does appear in correct formulea as a 'pupillary magnification ratio', but for DOF calculations two effects cancel out and you don't have to worry. The aperture is indeed closer to the film, but it looks smaller, so you end up with the same cone of converging light, and DOF is the same as for a regular, non-telephoto lens.

[Jeff Conrad]

Because the DoF is determined by the exit pupil, and the lens <var>f</var>-number is
related to the entrance pupil, don’t you need to consider the pupillary
magnification in any DoF equation that involves the lens <var>f</var>-number?

It’s common in elementary DoF equation derivations to eliminate the
diameter of the stop by the substitution

<var>d</var> = <var>f</var> /<var>N</var>

where <var>N</var> is the <var>f</var>-number, <var>f</var> is the focal
length, and <var>d</var> the diameter of the stop (assuming, in essence
that the lens is symmetrical). However, for an asymmetrical lens for which

<var>N</var> = <var>f</var> /<var>D</var>

and

<var>p</var> = <var>d</var>/<var>D</var>

where <var>p</var> is the pupillary magnification, <var>d</var> is the
diameter of the exit pupil, and <var>D</var> is the diameter of the
entrance pupil, the corresponding substitution

<var>d</var> = <var>p</var><var>f</var> /<var>N</var>

necessarily includes the pupillary magnification <var>p</var> in the resulting
equation.