40S 



Transactions of the Society. 



image of e t as of a conjugate point at infinite distance from which 

 a beam of light along the axis e x ... c fell upon the aperture. The 

 axial distance of ifr will be ?/i = sin tf r by equation (10) App. I., 

 p. -429. This then will be a point within the antipoint formed by the 

 aperture when fitted with a lens focussing at the distance r from 

 the optical centre c. Similarly ei will be the corresponding point in 

 the plane of the screen in the diffraction fringe formed by the un- 

 focussed beam. The point ei cannot, however, as a rule be directly 

 observed because, lying within the projection of the beam itself, it is 

 drowned in the superior effulgence of the direct light. If then we 

 wish to speak oi the antipoint as the image of a diffraction fringe, 

 it must be of the diffraction fringe formed at infinite distance. 

 For the purpose of determining its magnitude, however, we may 



(. r >j. r ^ 



Fig. 94. 



compare it with the point to which the central ray of the equi- 

 phasal beam would be deflected on a screen at a distance from the 

 aperture equal to the distance of the principal focal plane, and then 

 we shall have the following numerical relation : — 



Semi-diameter of the antipoint __ sin 6 _ rf 

 diameter of the fringe tan 6 



From this equation a very important inference may be drawn. 

 For the diameter (axial distance) of any particular phase zone F in 

 the fringe is, as we have seen, 



w / \ in sin 6 



1 s (6. . .€i) = r tan 6 = — s r. 



v cos 6 



But sin = f ' if we write K for the semi-diameter of the 

 '2 R 



