How wide is your lens' view?

PSA Journal, Dec, 2002 by Philip G. Engstrom

Most of us who own 35mm cameras now have an assortment of zoom lenses to cover a convenient range of focal lengths. But many of the longer focal lengths are of the fixed-focus variety. And even within the range of focal lengths covered by our zooms, there may remain a fixed-focus lens we use occasionally. So from time to time we still find ourselves using fixed-focus lenses.

A question which sometimes arises as we reach into the camera bag for the appropriate lens is, "Which focal length do I need to capture adequately the subject I want to photograph?" There is one focal length that will serve best to fill the frame with the subject I have in mind.

The purpose of this article is to present a simple, helpful rule that relates focal length to angle-of-view. With it, given the angle subtended by the subject field, we can find the focal length best suited for the task.

A little trigonometry and the diagram shown below in Figure 1 will help our development here. The diagram represents a cross section of the cone of light within a camera. The lens is at C and the diagonal of the film is represented by the line AB.

[FIGURE 1 OMITTED]

If the image of the subject being photographed is to fall entirely on the film, then clearly the angle subtended by that subject can be no larger than the angle R formed by the lines AC and BC. If it is larger, then part of the image will lie off the film. We need here a simple relationship between the distance AB, the distance F between the lens and the film, and the angle R. That relationship is provided by the tangent function. Because the triangle AEC is a right triangle, we can write

(1) tan R/2= [bar]AE/F

It would be helpful to simplify this equation, since most of us do not carry about with us a table of values of the tangent function. The simplification is achieved by using a well-known property of the tangent function--provided we measure the angle R in radians. Visualize a circle. If a length equal to its radius is laid off along the circumference, then the central angle of the circle so determined has to measure one radian. One radian is precisely 180/p (approximately 57.3) degrees. For reasonably small values of x (measured in radians), the values of x and tan(x) differ only slightly. In other words, when x is reasonably small, we can approximate tan(x) with x. This property is shown graphically in Figure 2 where the two lines (upper curved and lower dotted) are nearly coincident in the left third of the figure.

[FIGURE 2 OMITTED]

Using this idea, we replace tan(R/2) with R/ 2 in Equation 1 to obtain

R/2 = [bar]AE/F or R = 2[bar]AE/F

This is a much simpler equation but still not simple enough, since most of us think in terms of degrees rather than radians. We'd prefer the above equation in terms of degrees. As suggested above, one radian is equal to 180/p (approximately 57.3) degrees. If D is the measure in degrees of the angle R, then by making this substitution in the last equation, we obtain

(2) [pi]D/180 = 2[bar]AE/F or D = (180/[pi])[bar]AB/F

We've used here the fact that AB = 2AE,

Let us apply Equation 2 to the 35mm camera. The diagonal AB of a frame of 35mm film is about 43mm. So the value of (180/ p)AB is about 2460. Without doing any serious harm--and for the sake of convenience--we may round this value up to 2500 to obtain

(3) D = 2500/F

Equivalently,

(4) F = 2500/D

Thus, provided D is not too large, we can easily find the focal length of the lens that will "cover the subject" by simply dividing the number 2500 by the angle (in degrees) subtended by the subject.

There is an easy way to estimate the angle subtended by a photographic subject. If I hold my thumb and forefinger, separated by a distance of 1 centimeter, at a distance of about 22.5 inches in front of me, then the separation of thumb and forefinger will determine an angle of one degree. Happily, 22.5 inches is about the distance between my eye and my fingers when my arm is extended in front of me. Thus, suppose the subject to be photographed is a bird at a distance of, say 30 feet. I visualize a circle which "comfortably" contains the bird and which, measured by my thumb and forefinger at arm's length, is about an inch in diameter. One inch is about 2.5 centimeters (D = 2.5 degrees) and 2500/2.5 = 1000. So a 1000mm lens will place the image of the bird "comfortably" on the film.

In a similar way, since 2500/1000 = 2.5, I know that a 1000mm lens will satisfactorily capture on film that subject which fits comfortably within my one-inch circle.

How adequate is the equation we've been using here? Below is a table which, for various values of D (degrees), compares the values F1 obtained from Equation 1 (exact values) and values F2 from Equation 2 (approximated values).

What we have said above can also be applied to other formats. A brief calculation will show that, for 2-1/4 x 2-1/4 inch format, equations 3 and 4 may be replaced by

(5) D = 4600/F and F = 4600/D

The above table illustrates that, even for angles-of-view as large as 60 degrees, the focal length determined by Equation 2 (or Equation 4) is reasonably accurate. Equations 3 and 4 provide us with an easy way to relate the focal length: of a lens to the angle-of-view associated with it.