The Helmholtz Theory of the Microscope. By J. W. Gordon. 431 



also that 

 and 



a e = r f 



tie = 0, 



sin u v (a. v ... rf) = sin Ut, r v = sin w e r e 

 .-. Multiplying (10) by (11) 



c sin tie = t] sin Ut, .... 



(11) 

 (12) 



In this proof it has been tacitly assumed that we have the same 

 refractive index behind as before the aperture. It is obvious that if 

 the refractive index underwent a change between the two images there 

 would be a corresponding change in (12), and from what has been already 

 said (see above, p. 396) we know that the general expression is 



Me Sill We € = ?l v Sill U v rj 



(13) 



which is Helmholtz' law. 



The foregoing proof has proceeded upon the assumption that the 

 image-forming aperture is capable of transmitting a plane wave-front. 

 This is not necessarily the case. The image may be formed by a pin- 

 hole, in which case only a minute axial pencil of the plane wave-front 

 could pass it and no such image as is shown in fig. 99 of an extended plane 

 wave-front could be formed. Yet a plane image can in that way be 

 formed of a plane object which gives off beams of light with a finite 

 divergence angle. The law of image scale in such a system remains 

 therefore to be investigated. 



Fig. 104 will serve for this purpose. 



Fig. 104. 



It is obvious at a glance that in this unfocussed system the law is 

 now not the sine law, for writing, as before, r f for the distance c . . . C 

 and r n for C ... 77, we have for the positions of e x and rj 1 respectively, 



e = tan re ; 



■r] l = tan r v . 



The images are still proportional to r e and r v , for tan 6, like sin 6, is 



2 f 2 



