Focal Lines of Cylindrical Lenses. 67 



before the diagonal bj of the parallelogram bigj will represent 

 in end view the ray after emergence from both lenses whilst 

 the line off (fig. 8) will represent the ray in front view. 



We shall now find what must be the relation between the 

 angles X and 2 when the two emergent rays af and bj 

 intersect on the line oof or intersect in the central plane of 

 which oo' is the trace in fig. 7 or the plane of the paper 

 in fig. 8. 



We have from fig. 7, 



ac = hi sin Y — li 2 cos ly 



f f 



ae — j ad = j- (1i x sin 2 -f A 3 cos # 2 ), 



fa ./ 2 



bg = h x sin X + h 2 cos U 



bi = § bli =--*§ (A x sin 2 - /i 2 cos 2 ). 



/2 ,/2 



From these values we can readily obtain the components 

 of the diagonals af and bj resolved along and at right angles 

 to the line oo'. 



Let the resolved component of af along oo = k. 



., „ „ af at 90° to oo' = I. 



„ ,, ,, bj along oo' = ?/?. 



., „ „ bj at 90° to oo' = n. 



Then 



k=h x sin 2 0!-A 2 sin t cos ^ + & (h x sin 2 <9 2 + /% sin 2 cos 8 ), 



./2 



/=-/<! sin X cos 6>! + h cos 2 6\ + 4( li x sin <9 2 cos 2 + h 2 cos 2 <9 2 ) 



2 A 



m = h x sin 2 # L + h 2 sin #j cos # 2 + J -j (h x sin 2 # 2 — h 2 sin # cos # 3 )> 



/2 



?? = A x sin X cos ^! -f h 2 cos 2 ^ — *^ (7*1 sin 2 cos 2 — h 2 cos 2 N 



/i 



Now, if k = m, l = n, and P + ^ 2 =??i 2 + n 2 , the two diagonals 

 will be equal in length and equally inclined to the line oo\ 

 and as the points a b are symmetrical with respect to tho 

 line oo', the two diagonals will intersect at some point p on 

 this line. 



¥2 



