26 the president's address. 



In order, however, to accurately represent the area of an 

 objective we must inscribe a circle in our square (Fig. 2), and we 

 shall see that the rectangles H and I in the centre of the circle 

 very nearly represent the true area of that portion of the objec- 

 tive, but at the opposite sides of the square the rectangles E 

 and M are ever so much larger than the corresponding portions 

 of the circular object-glass, therefore the effect of the light 

 passing through that portion of the objective represented by E 

 is insufficient to neutralize the portion passing through I. The 



In Fig. 3 let O B be the object, and I M the image, A C the object glass, 

 A being the point where the lines O M and B I intersect. Because B I and 

 O M are straight lines the alternate angles O A B, I A M are equal, and the 

 triangles O A B, I A M are similar, therefore O B bears the same proportion 

 to B A that I M does to M A. But B A is the distance of the object from 

 the telescope, and A M is the focal length ; therefore we have 



object : distance : : image : focus. 

 Let us call for brevity the object O, its distance D, the image (supposed to 

 be a minimum visible) 8, and the focal length F, then 



O d 



D F. 



Next let us transpose our previous proportion (i.) thus — 



5 : F : : A : A, 

 S \ 

 that is — = — ... ... ... (ii) 



F A, 



8 O 



but we have just seen that — = — 



F D, 



5 A O 



therefore — = — = — (iii) 



F AD. 

 Now because the angles in question are very small (about 5") these ratios 





 express the angles themselves, therefore — becomes the anule the object 



D 

 subtends at the object-glass (1) being the distance between the object 



x 

 and the telescope), and — the angle one wave-length subtends at a distance 



A 

 equal to the diameter of the object-glass {i.e., A the aperture), therefore the 

 statement above is correct which says that the smallest object 

 that can be seen with a telescope is that which subtends an 

 angle at the object-glass equal to that subtended by one wave- 

 length at a distance equal to the diameter of the object- 

 glass. 



In Fig. 4, A C or A W is the diameter of the objective, 

 and C W is one wave-length, then equation (ii) shows that 

 the angle C A W (Fig. 4) is equal to the angle I A M (Fig. 3) 

 for a minimum visible, but the angle I A M is equal to the 

 angle O A B, therefore the angle () A B is equal to the angle 

 C A W, provided that the image is a minimum visible, equa- 

 tion (iii). 

 Fig. 4 



