But if the objective has a focal length of 100 inches the image, as we have just seen, is already magnified 10 times as the naked eye sees it, hence with an objective of 100 inches focus and a 1 inch eyepiece the total magnification is 100 diameters. And this expresses the general law, for if we took the normal seeing distance of the naked eye at some other value than 10 inches, say 12½ inches then we should have to reckon the image as magnified by 8 times so far as the objective inches is concerned, but 12½ times due to the 1 inch eyepiece, and so forth. Thus the magnifying power of any eyepiece is F/f where F is the focal length of the objective or mirror and f that of the eyepiece. The focal distance of the eye quite drops out of the reckoning. 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.