All these facts appear very quickly if one explores the image from an objective with a slip of ground glass and a pocket lens. An ordinary camera tells the same story. A distant object which covers 1° will cover on the ground glass 1° reckoned on a radius equal to the focal length of the lens. If this is equal to the ordinary distance of clear vision, an eye at the same distance will see the image (or the distant object) covering the same 1°.
The geometry of the situation is as follows: Let o Fig. 5, Chap. 1, be the objective. This lens, as in an ordinary camera, forms an inverted image of an object A B at its focus, as at a b, and for any point, as a, of the image there is a corresponding point of the object lying on the straight line from A to that point through the center, c, of the objective.
A pair of rays 1, 2, diverging from the object point A pass through rim and center of o respectively and meet in A. After crossing at this point they fall on the eye lens e, and if a is nearly in the principal focus of e, the rays 1 and 2 will emerge substantially parallel so that the eye will unite them to form a clear image.
Now if F is the focal length of o, and f that of a, the object forming the image subtends at the center of the objective, o, an angle A c B, and for a distant object this will be sensibly the angle under which the eye sees the same object.
If L is the half linear dimension of the image, the eye sees half the object covering the angle whose tangent is L/F. Similarly half the image ab is seen through the eye lens e as covering a half angle whose tangent is L/f. Since the magnifying power of the combination, m, is directly as the ratio of increase in this tangent of the visual angle, which measures the image dimension
m = F/f, as before
Further, as all the light which comes in parallel through the whole opening of the objective forms a single conical beam concentrating into a focus and then diverging to enter the eye lens, the diameter of the cone coming through the eye lens must bear the same relation to the diameter of o, that f does to F.
Any less diameter of e will cut off part of the emerging light; any more will show an emergent beam smaller than the eye lens, which is generally the case. Hence if we call p the diameter of the bright pencil of light which we see coming through the eye lens then for that particular eye lens,
m = o/p
That is, f = pF/o which is quite the easiest way of measuring the focal length of an eyepiece.