THE ABSOLUTE INTENSITY OF INTERFERING LIGHT. 321 



CASE 2. When the centre of gravity of the triangle falls in the line with the 

 focus of the lens. 



TT T M. / 2ccosa\ / 2ccosa\ , 



Here our limits are y = ( x+ ^ 1 tan a, y = ( * + - j tan a ; 



2 c cos a _c cos a 

 3 '* 3~' 



Integrating with respect to y we get 



\ba C f 27T/ _ p x q 2, c \ 



~ rr / dx - cos -^ I v t B + -T T tan a x + -~- cos a 1 



2-7T/ # o 2~c \ 1 



cos c- I v t B + -T- + I tan a a; + -5- cos a ) 

 A V b b 3 / J 



\ 6 \ba f . 2 < 7T c / cos a \ 2 'TT c 2 cos a "I 



.-. M = ~ pr . s j N \ sm-^-r I e s 9 sm a ) + sm -= r -^-5 f 



27ryD 2 TT ( j y tan a)^ A.o\<J / A.6 



2 IT c fp cos a \ 2 TT c 2 cos a 1 



sm -^-j- I - h q sin a I 4- sm -^-: 



Ao\d / A.O o J 



\6 \6a f 27Tc/jcosa \ 2 ire 2jcosa\ 



]^ t . . < cos I q sm ct I cos -^ . ^ 



2 *7r o* D ' 2 TT (/> q tan o) l.A6v3 / A6 3J 



If we adopt the notation previously used, this gives 



f . (x \ . 2x . 



A b a c cos r 



{. /x \ . 2x . (x \ . 2x\ 

 sm^g-^j +sm-g- sm (g + yj +sm ~3- I 

 xy x+y ) 



27TO _ 



xy x+y 



2 a c 2 sin a cos a f x x . 2x 



3 C S y~* S 3 Sm 3 



2 a c 2 sin a cos a f * x . 2x \ 



N = (# -y") y D 1 y cos 3 cos y + * sm 3 sm y - y cos T J 



4 a 2 c 4 sin 2 a cos z a f n 2* a^ . 



-y sm a? sm y + -^ sm 2 y 



4 sin 2 a cos 2 a f . x . \ 



(x 2_ y ^ { (^s *- cos y) 2 + (sm x -- sm y) 2 j 



_4o 2 c 



As this expression is precisely the same as that in the last case, it is unnecessary 

 to integrate it. 



PROB. IV. Every thing the same, except that the aperture is a circle concentric 



rvith the lens. 



The vibration at the point M, whose distance from the focus of the lens is p, due 

 to an element of the front of the wave at P, whose distance from the centre is r, 



