with Special Reference to the Microscope, 185 



tinues to diminish until it also finally disappears when v 

 drops below ir. The field is then uniformly illuminated, 

 showing no trace of the original structure. The case v = ir is 

 that of fig. 4, and curve iii. shows that at a stage when an 

 infinite series shows no structure, a pair of luminous points 

 or lines of the same closeness are still in some degree 

 separated. It will be remembered that v = 7r corresponds to 

 e = J\/sina, e being the linear period of the original object 

 and a the semi-angular aperture. 



We will now pass on to consider the case of a grating or 

 row of points perforated in an opaque screen and illuminated 

 by plane waves of light. If the incidence be oblique, the 

 phase of the radiation emitted varies by equal steps as we 

 pass from one element to the next. But for the sake of 

 simplicity we will commence with the case of perpendicular 

 incidence, where the radiations from the various elements all 

 start in the same phase. We have now to superpose ampli- 

 tudes, and not as before intensities. If A be the resultant 

 amplitude, we may write 



A/ N sinw sin(w-fv) , sin(w— v) 



A(u) = 1 ; H + 



u u+v u—v 



=A + A!cos + ...+A r cos 1- (28) 



When v is very small, the infinite series identifies itself 

 more and more nearly with the integral 



IT 



viz. — , 



V 



1 C +co smu 



In general we have, as in the last problem, 

 . ir + ™sinuj 2( +co sinw 2irru . 



so that A =7r/v. As regards A r , writing s for 2irr/v, we 

 have 



*-a 



1 f+» sin(l+s)w + sin(l— s)u , tr .„ ,. 

 rf« = -(l±l), 



the lower sign applying when (1— s) is negative. Accord- 

 m gty> 



A(u) = — < 1 + 2 cos— +2 cos +....{- . (30) 



the series being continued so long as 2irr < v. 



