206 



THE AMERICAK MONTHLY 



[Kovember, 



focal length of the eye being taken as 

 ten inches) , and a lens of elev^en 

 inches, or more, would even reduce 

 the apparent size of the object. But 

 this, we know, is not the case. 



A lens shows an object just the size 

 that the unassisted eye would see it if 

 the focal length of the latter were 

 equal to the focal length of the mag- 

 nifier and the eye. Now, let ni repre- 

 sent the magnifying power of the lens, 

 V the distance of distinct vision, or 

 focal length of the eye, f the focal 

 length of the magnifier, and we have 

 the formula : 



m = — ^- 



According to which the magnification 

 of a lens of one-inch focal length, for 

 instance, the focal length of the eye 

 being taken as ten inches, would be 

 eleven diameters, not ten diameters, as 

 is commonly supposed. The magni- 

 fying power of a 2-inch lens would be 

 6 diameters, that of a lo-inch lens 

 would be 2 diameters, and a 20-inch 

 lens i^ diameters. 



A half-inch eye-piece magnifies the 

 image of the objective 2i times (lin- 

 ear), a one-inch ii times, and a 

 two-inch 6 times. 



A similar mistake is also made in 

 determining the magnifying power of 

 objectives. It is commonly supposed 

 that by simply dividing the tube- 

 length, or rather the distance of the 

 image from the objective, by the 

 equivalent focal length of the latter, 

 that we get the magnifying power of 

 an objective. If this were so, then, 

 for instance, a five-inch objective 

 would, with ten inches tube-length, 

 give a magnified image of two diam- 

 eters ; while, in fact, it does not 

 magnify at all. That is, the image 

 will be just equal in size to that of the 

 object. The size of an image formed 

 by a lens or an objective is to that of 

 the .object as the distance between the 

 image and lens to the distance be- 

 tween lens and object. Now, if a 

 true five-inch objective had to be just 

 five inches from the object to form an 



image at ten inches distance, then the 

 magnifying power would certainly 

 be two diameters. But, to forin an 

 image, the distance of a lens from an 

 object has to be greater than the focal 

 length of the same. To form an im- 

 age at two focal lengths of the lens it 

 has to be at an equal distance from the 

 object. Therefore, a five-inch objec- 

 tive will, with ten inches tube-length, 

 have to be just ten inches distant from 

 the object to be in focus ; and conse- 

 quently the image formed by it will 

 be equal in size to the object. If fn 

 be the magnifying power, ^ the tube- 

 length, andy the focal length of the 

 objective, then we have this formula: 



According to which a one-inch ob- 

 jective, for instance, will, with ten 

 inches tube-length, give an image 

 magnified nine diameters, and not ten, 

 as is commonly supposed. A two- 

 inch will magnify four diametei's and 

 not five. 



Having now found the correct for- 

 mulae for the computation of the mag- 

 nifying powers of both the objective 

 and the eye-piece, it is a simple mat- 

 ter to properly combine them to a 

 complete and correct formula for the 

 computation of the magnifying power 

 of the compound microscope, which 

 is — 



In which m is the magnifyingpower 

 of the compound microscope, j^i the 

 equivalent focal length of the objective, 

 f- the equivalent focal length of the eye- 

 piece, t the tube-length, v the distance 

 of distinct vision or focal length of the 

 eye. According to this formula, a 

 combination of a one-inch objective 

 and a one-inch eye-piece will, at ten 

 inches tube-length, give a magnifica- 

 tion of 99 diameters, not loo, as is 

 commonly supposed. A four-inch 

 objective combined with a two-inch 

 eye-piece will magnify 9 diameters, 

 not 13.5. 



