THE PRESIDENT'S ADDRESS. 25 



the least separable distance in the image at the focus, to the 

 focal-length, is the same as that of the least separable distance 

 in the object itself to its distance from the telescope ; there- 

 fore, the least separable distance in the object bears the same 

 proportion to its distance from the telescope, as the wave- 

 length does to the diameter of the object-glass.* 



* As it is important that even our most junior members should 

 thoroughly comprehend this exceedingly simple problem, which is a 

 common rule-of-thre i sum, and which precisely resembles, and requires no 

 more mathematical knowledge than, the well-known child's problem abouo 

 the herring-and-a-half which cost three-halfpence, this note is appended. 

 Cut out of a piece of paper two precisely etpaal triangles like A B W (Fig. 

 1), and placing one over the other, so that the lower is exactly covered by 

 the upper, stick a pin through both their corners at A. Now take hold of 

 the lower one at the point D and move it out at one side, as shown by the 

 dotted triangle (Fig. 1). It will then be seen that as the point D is moved 

 to one side, so the point C of the lower triangle will protrude beyond the 

 point W of the upper one. A moment's thought will show that the displace- 

 ment at D is proportional to the protrusion C W, and that this proportion is 

 that of the length of the line A B or AD (the focus of the object-glass) to 

 the length of the line A W (the diameter of the object-glass). For example, 

 if the focal length A B is four times as great as the diameter A W, and if 1) is 

 moved four-tenths of an inch from B, the point C of the lower piece will 

 protrude one-tenth of an inch beyond W the upper. Stating this simple 

 problem mathematically, let us call 5 the displacement of D from B, and \ 

 the protrusion at W, let A C or A W, the aperture, be called A, and A B or 

 A D the focal length, F. Then 



3 : X :: F : A (i) 



and every school-boy knows that to find S we must multiply X by F, and 

 divide the product by A. 



When the dotted line C D (Fig. 1) protrudes one wave-length beyond W, 

 then the remainder, W I), must be shorter than A I) by that one wave- 

 length, and we have seen in Fig* 2 that when the distance from the M side 

 of the object-glass to D is one wave-length shorter than that from the E 

 side, the light will be extinguished at the point D. The distance B D, 

 therefore, represents the minimum visible, when the protrusion of C D 

 beyond W is one wave-length. Now, as the wave-length, the focal length, 

 and the diameter of the object-glass are all known, £, the minimum visible, 

 can be determined. 



Now one word with regard to the proportion between the size of the image 

 at the focus, which we have just been considering, and the size of the object 

 to which the telescope is supposed to be directed. 



