THE ABSOLUTE INTENSITY OF INTERFERING LIGHT. 



By developing the sines and cosines, these expressions become, 



319 



M = 



- tan 



P 



a 



cos 



2% 



sm 



27T 



27T 



p . /27T \ 72^ \ 



+ sm I ^ , p c cos a ) sin ( ^-7 q c sin a 1 



? \X ? / \ A6 / 





- - 



Abbreviate -j pccosa by a:, and -^- y c sin a by y ; then - = - tan a ; 



!f !? 





M 2 + N 2 , or the intensity at the point whose co-ordinates are jt>, q, is 

 4 a 2 c 4 .. 2 



4 a 2 c 4 f .. 2 # . ir 2 . ) 



na - ^ sm 2 acos 2 a 1 1 2 cos x cos y -- smzsiny + cos 2 y + -^ sm 2 ^ J- 



U (* y j i y y ' 



If then M = total intensity, we have 



4o 2 c 2 \ 2 .6 2 sinacosa 



- ST-RP - 



7r 2 D 2 



2 a; . 

 5 - %r<>^l 2cos;rcosw -- smarsin y4- 



2 22 y 



y 



We proceed now to find the value of this integral. To eifect this we assume the 



- f^acosqxdz TT __ ,, ^ 



following as proved,^ q2+a . 2 =-3 e 



Dividing by a, and then differentiating with respect to a, we get, 



/" COS '#?#_ TTya + l ja .. . 



a? + 3? ~~ T ~^^ 



Differentiating this with respect to q 



/' a xsmqxdx_'7rg a 

 2 -~ 





ty ty o 



Now 1 + cos 2 y + -g sin 2 y2 + 3*- sin 2 y. 



_ 



Put a=V-l.y,?=Oi 



TK.J. , -i 'n 



VOL. XV. PART II. 



/JJ- 



4 K 



