434 



Transactions of the Society. 



for comparison any conjugate areas in the object and image fields and 

 we shall find it convenient to choose for a criterion a radial line equal 

 to half the diameter of the aperture and having one end on the axis. 

 "With this datum we construct the following diagram (fig. 106). 



Here C is the optical centre ; C . . . v the optical axis ; A . . . ^ is 

 a line drawn parallel to the axis from the edge of the aperture. It 

 denotes, therefore, the projection of the aperture upon all the fields and 

 wave-fronts through which it passes. We will select for comparison two 

 fields distant r x and r v respectively from the aperture. If now we 

 describe the spherical fields situate at these distances and draw the 

 position angles X and Or, to the points where these two fields are cut by 

 the edge line A ... rjr,, we shall evidently have two position angles equal 



Fig. 106. 



to the two divergence angles u x and u n each to each. We already know 

 in general how these different image scales are related to their position 

 angles, and as we have now obtained an equation between position 

 angles and divergence angles we can extend the law to divergence 

 angles. 



Assume, then, two points vi and ^ positioned by the angles X and 

 V respectively in the r x and r v images. The images will be proportional 

 to r x and r^, and therefore the magnitude A . . . C or its equivalent (which 

 we will call /J x ) in the two image planes respectively will have scale values 



respectively proportional to — and — . 



Now 



r-. 



fi x = sin X r, 

 = sin 6 V r v 

 = &c. 



(16) 



