at a Cylindrical Surface. 47 



and the locus of the })oint g will be the focal line. A plane 

 containing the points o, a, c will contain an incident ray and 

 the normal, and will therefore also contain the corresponding 

 retracted ray. This plane will cut the plane of the figure (1) 

 in the line oca, and the refracted ray will in general cut 

 this line ocg in some point g. Again, a plane containing 

 the points o, h, c will contain the corresponding symmetrical 

 incident ray, normal, and refracted ray, and will cut the 

 plane of the figure in the line occ/. From considerations 

 of symmetry it follows that the two refracted rays will inter- 

 sect in the same point g on the line ocg. If we consider 

 only a thin slice of the cylinder parallel to the plane of the 

 paper in fig. 1, the elevations of the lines oa, ac, and ag may 

 be taken as their true lengths. 



Let the vertical aperture = //, 



The angle of inci 



dence = 0. 



„ of refraction = (/>, 



„ age 



= •3, 



„ com 

 and the length eg 



r=d. 



Then we have 



I3=f-ct> 





smd = /jLsm(l> 





d sin 3 = ?'sin(/) 





^j^rVa' + h' 

 h Got(f> — a' 



from which we get 





d= 



r^a^ + h' 



^f,\a-ry±(fj,:'^l)k'-a ' ' ^^^ 

 Taking the radiant-point as origin, the equation to the locus 



d' = a + d cos -x/^, 



is 



or 



x=:a + 



>^fjL\a''ry+{fi'-i)/i 

 and putting in the value of 



00 



we net finallv 



..2_ ^i^ S {ooc^-ra-a'Y -{y,[a-r){x-a)Y ^ 



