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Mr. L. V. King on the 



the diminution of intensity between the portions S and 

 S + JS of two spherical surfaces of radii r and r + dr cut out 

 bj a small solid angle &>. 

 We then have 



Sdl=* -IdS-ISxdr. 



The first term on the right-hand side expresses the diminu- 

 tion of intensity per unit cross-section due to the increase 

 with r of S to S-h^S, while the second term expresses the 

 diminution of intensity by absorption in a distance dr. 



Since S = cor 2 , we have in a uniform medium (* inde- 

 pendent of r), 



glT 



an o- 



■/I 

 I 



= — 



2 dr 

 r 





1 = 



c 



k dr. 



where C is a constant of integration. 



As long as S remains in the same medium as the point- 

 source s 3 we may write the above in the form 



1 = 



(2) 



where s is the intensify across unit-area at unit-distance when 

 there is no absorption (« = 0), in which case (2) simply 

 expresses the law of inverse squares. 



Fig. 1. 



Suppose s to be in a uniform medium whose coefficient of 

 absorption is k, while the intensity is measured in a medium 

 of coefficient X, the two media being separated by a surface 

 cutting sS at a point P such that sF = R. Then provided 

 we neglect effects of refraction and reflexion at the bounding 

 surface, a condition which is justifiable in the application to 

 be dealt with, we may consider the path of the rays forming 

 the pencil of small solid angle co to be unaltered in crossing 



