*The concept of hyperfocal distance (or hfd), is known and used by many photographers.When a camera is focused at the hfd, depth-of-field is maximized and everything from half that distance to infinity is acceptably sharp.The depth-of-field scale found on many lenses with a self-contained focusing barrel is easily used to set the lens at its hfd by aligning the infinity distance mark with the index for the f-stop in use. It is not so easy with view cameras, which may use any of several different lenses and different film formats.The darkness of the view screen at the apertures where hfd is commonly desired may put the observer’s eyesight in the low-acuity, dark-adapted mode. In such a case, the best strategy is to find or place a focusing target at the hfd, focus wide open, and then stop down to the proper aperture. But how do you know what distance to use?*

I will show you one lazy man’s approach to solving this problem. (“Lazy,” as any lazy person knows, is a synonym for “efficient.”)

**Basics**

Calculations of depth-of-field are based on principles of geometrical optics, according to which the rays entering a lens from a point in the object space are focused through a point in the image space, forming a cone of light. If film or a viewing screen is placed at the apex of the cone, the amount that the point may be moved in the object space before its image becomes objectionably large is the depth-of-field. The amount the viewing screen may be moved is the depth-of-focus. The disk formed by an out-of-focus point is called the “circle of confusion.”

The limit of depth-of-field is reached when the circle of confusion becomes unacceptably large to a human observer. Defining “unacceptably” for use in artistic photography therefore requires knowledge of the resolving power of the average human eye, because a photograph will be judged unsharp wherever its circle of confusion equals or exceeds that of the judge.

The human eye is an auto-focusing device with about a 1o field of clear vision. It accommodates its focus to make what is in that 1o as sharp as possible. Its maximum resolving power in that area is about one arc minute, which is 1⁄60 degree or 0.000336 radians, which it reaches at about a 2mm pupil. While the maximum resolving power of any lens increases (the angle resolved decreases) as the diameter of its entrance pupil increases, the eye is limited by the granularity of the retina when its pupil is greater than 2mm. The iris is used mostly for rapid accommodations to changes of light intensity.

**The ingredients**

Hardy and Perrin in The Principles of Optics, McGraw-Hill, 1932, derived the following equation for hfd: hfd=R• F/z’wherez’istheradiusof the largest circle of confusion that can be tolerated, R is the radius of the lens opening, and F is the focal length. The usual small-angle assumptions were made in the derivation, but the error caused by them would not have any effect on practical photographic uses. These are indeed very small angles.

The ratio z’ / F is dimensionless when z’ and F have the same dimension. It has the physical significance in this case of an angle measured in radians. It seems reasonable to relate that angle to the maximum angular resolving power of an average human eye for the purpose of defining the allowable circle of con- fusion. In that case: z’/ F = 0.000366 radians, and z’ = 0.000336 • F.

If we substitute this value of z’ and useD/2forRintheequation,the focal length drops out and: hfd = D / (0.000672) (2), where D = diameter of entrance pupil. Note that if D is in millimeters, the hfd will be in millimeters. Multiply by 0.003281 to get feet. Let us modify equation (2) to include this conversion. Thus, for hfd in feet: hfd = 4.88 • D; and for hfd in meters: hfd = 1.49 • D. D is still measured in millimeters. If you do not know the diameter of the f-stop, substitute F / f # for D.

You may have noticed that so far I have tacitly assumed that the viewing distance is equal to F because the allow- able circle of confusion is that which would subtend a viewer’s angular resolving power at a distance of F. Enlarging the negative enlarges the proper viewing distance and z’ by the same factor, thus keeping the angular resolution constant.

The focal length of a “normal” lens is considered to be about equal to the diagonal dimension of the negative. If you view a photograph from a distance equal to its diagonal, you will see that the corners are about as far as you can scan without head movement. It is natural to view a print from about that distance. When you are using a telephoto lens or are planning to crop as if you had, the effect will be as if you gave the viewer of the print a magnifier with a power equal to the ratio of the focal lengths of telephoto to normal lenses, or the ratio of full-sized to cropped print dimensions. The viewer’s acuity will be as if amplified by the power of the magnifier. Just as small grain becomes more apparent when you view a print with a magnifying glass, so will smaller focusing error become apparent.

The general equation then becomes for hfd in feet:hfd=4.9• M• F/f#.For hfd in meters:hfd=1.49• M• F/f#, where M is the ratio of focal length of taking lens to that of a normal lens. The use of a wide-angle lens reduces hfd because M is then less than 1.

Equation (2) illustrates that hfd is independent of focal length, but if you don’t know the diameter of your f-stops, you can use the following equation: hfd = 4.88• M• F/f# orhfd=1.49• M• F/f#.

**Tasting the chowder**

Now let us see how well this equation fits those lenses I have that are equipped with depth-of-field scales on their focusing mounts. A 50mm normal lens at ƒ/16 has a pupil of 3.125 mm diameter. Equation (2) predicts an hfd of 15.25 feet, with total depth of field from 7.6 feet to infinity. The depth-of-field scale of my Leica Summicron 50mm lens shows depth of field from 8 feet to infinity at ƒ/16, with an hfd of 16 feet. This same rationale also appears to have been used on an ancient Nikkor-QC 50mm uncoated lens.

The magnification of my 90mm Tele Elmarit over the normal 50mm lens is 90/50. Thus, the hfd for that lens should be:hfd=1.8• 4.88• F/f#.Thehfdat ƒ/16shouldbe:hfd=1.8• 4.88• 90/ 16, so hfd = 49 feet.

That is, in fact, the reading of the scale on the lens barrel as nearly as I can read it.

I computed 44 feet for an 85mm telephoto lens. Here again, the computed value agrees with the scale on my Canon 85mm, 1.8 as closely as I can read it.

To test the premise that hfd is dependent only on pupil size, I photographed a scene with 35mm and 5×7-inch film using the same aperture size for both.

The 35mm camera was a Canon Elan IIe with 50mm lens set for manual operation at ƒ/11. I calculated the aperture to be 4.55 mm in diameter. The 5×7 camera was equipped with a rapid rectilinear lens of about 200 mm focal length. I set its diaphragm by removing the front element and closing the diaphragm on a plastic tube that I had measured to be 4.5 mm in diameter, since I was unsure of the accuracy of its f-stop scale. The scale indicated close to ƒ/45. A normal lens of 4.5mm aperture should have an hfd of 22 feet by my equation. I prefocused each camera at that distance and then shot the landscape you see in figure 1. The distant ridge is about as far as you can see in my part of West Virginia before a mountain gets in your way. The power line in the foreground is about 20 feet from the camera. I printed small portions scanned from each negative side by side for comparison. The 5×7 was scanned at 600 ppi, the 35mm at 2400 ppi. The major difference is due to the fact that the rapid rectilinear lens is uncoated and shows the effects of flare. It is about 100 years old. The modern Canon lens has it beaten for contrast, but the fact that there is no significant difference in hfd shows through the flare.

**Conclusions**

You have seen that the factors affecting hfd are the diameter of the lens aperture of the camera, the ability of the human eye to judge relative sharpness, and the desired perspective of the photograph. A lens aperture diameter of, say, 4 mm will give the same hfd, about 20 feet, no matter what the focal length, as long as the lens is used as a normal lens. The exposure time will vary with focal length, of course. The hfd can be computed for any lens without the assumption of an arbitrary allowable circle of confusion.

Once the negative is made, the proper viewing distance for any print made from it is the focal length of the taking lens multiplied by the enlargement factor between negative and print. If that rule were followed, the M factor in the equation would always be 1. The viewer ordinarily has no way of knowing that a telephoto or wide angle was used, so I made allowance for viewing at a distance about equal to the diagonal of the print. The effect, then, of telephoto or wide-angle lenses is to increase or decrease the ability of the viewer to judge sharpness when the viewing dis- tance is equal to the diagonal of the print, thereby increasing or decreasing the hfd with respect to that of a normal lens. The only way to satisfy extremely close inspectors (technically known as “grain sniffers”) that there is no fuzz or grain is to make the film out of unobtainium and make the depth-of-field extend from zero to infinity. But I believe that the use of angular resolution of the eye in place of an arbitrarily fixed circle of confusion has not only simplified the calculation of hfd, but has made it easier to understand.