Diffraction of an Annular Aperture. 5 



In like manner 



cos — ^/FTT;^ =cos?-^ v/Fir^2 . cos {e, COS Q) 



X A, 



- sin — \/^M^ • sin {e, cos 9) 



A- • 



cos %^ </ AM^= cos^ \^F~T¥ . cos (e„ cos 9) 



A. A. 



_ sin— v/F~+~a^. sin (^/^cosfl), 

 A/ 



and these quantities are now to be substituted in the expres- 

 sion above, which multiplies 8 9, and the whole is then to be 

 integrated with respect to 9. 



If our object were to find the intensity of light produced by 

 a portion of the annulus, these expressions could not be made 

 more simple; but in finding the intensity produced by the 

 whole annulus we may introduce an important simplification ; 

 for the expressions will then be integrated, with respect to 9, 

 from 9 = to 9 = 2 TT. Now it is evident that the expansion 

 of sin (e, cos 9) or sin {en cos S) will produce only odd powers 

 of cos 9; and the integral of every one of these, from 9= to 

 6 = 2 TT, is zero. The terms depending on sin [e^ cos 9), 

 sin {Cii cos 9), may therefore be neglected at once ; and thus we 

 have, for this investigation, 



sin?^-v/FT72 = sin^V'FT^- cos (^^ cos 9), 



A. A, 



and so for the others ; and the quantity which we have to in- 

 tegrate with respect to 9 is 



X . 2Tr vt 

 -— . sm — r — 

 2 » X 



(2 TT — 

 sin—— \^ h^ + i* ■ cos {en cos 9) 



27r 



— sin -— \/ A* + a- 

 X 



COS {Ci cos 9) ) 



+ A, . cos ^-^ . f COS V" ^h' +l>' . cos {e, cos fl) 



2 TT X \ X 



— COS "^ \//z^ + a^ . cos {e, cos 9) J . 

 Let 



/ cos {e cos 9j = 2 TT E, / cos {e, cos 6) = 2 tt E„ 



/^cos {e,! cos 9) = 2 »r E„, 



