Highly Magnified Images. By J. W. Gordon. 9 



Let e in fig. 4 be a point on the stage of a Microscope. Let 

 P be the principal plane of the objective. Let A be an aperture 

 which limits the diameter of the transmitted beam, and let rj be 

 the point in the image plane conjugate to e. 



The question is to find an expression for the diameter of any 

 given zone in the antipoint formed about ij by the aperture A. 



Fig. 4. 



Helmholtz' solution of the problem may be stated thus : 

 Let A in the following diagram (fig. 5) be the given aperture, 

 liaving rectilinear boundaries; a. . a, the axis of collimation, and 

 rj the "point upon that axis to which the transmitted wave-front 



converges 



Also let a l . . ,a 1} intersecting the axis a. . .a at an angle 6, be 

 the axis along which the diffracted beam in question would be 



Fig. 5 . 



deflected if the wave-fronts passing the aperture were not spheri- 

 cal but plane wave-fronts. Then, by the known law of diffraction, 



sin 6 = y> the symbol </> being used to express in terms of X the 



retardation of the most retarded ray in the diffracted beam. Now 

 it is plain that the axis a lt . .a x will intersect the perpendicular 

 plane through r) — which may be called the focal plane — at a point 

 rji so situated that its axial distance 



7} . . . r)i = tan (e . . . rj) . . . , (3) 



