of the Eye-pieces of Telescopes. 25 



and, as KD is perpendicular to the second surface, they will, after the 

 second refraction, converge to M. If, then, a plane be drawn through M 

 perpendicular to the axis of the lens, and cutting it in O, OC is the distance 

 which we are to find. 



Draw LN, DP, perpendicular to the axis of the lens : as before, lei 

 BF=a, and let AB = b, CE = c; also, let NB = a, QD = x, CO = z: and 

 let B, C, A, Z, be the values of b, c, a, z, when a = o. Also, let F = focal 



length of the lens, f=^, 



if If 



and let ^ =^- + e, then •7,='^ — e : 

 B 2 C 2 



If If 



and let j^'^ + g, then ■g=''- - g- 



Let r and s be the radii of the first and second surfaces, and let 



- = —r^ — "T + V, then - = —j^ -v. 



r 2(m— 1) '; s 2(n— 1) 



Now, NL approximately = , a ; 



.•. BQ.— — ; — .—y—a; .•. Ht = T-. — ; — -a. 

 r + a b b{r + a) 



Now, by the formula which we have before used, 

 1 n-1 1 QF^ n-i 



QD nr 71. QL 



UF' n-1 yl J_\" /I n+l^ 

 "^ 2 w' A; ^ QL) \r "^ QL ) 



But HL= \/j^q:71^+^^ «'} =a + r+^ 



a-b] 



.-. QL = a + 



6-(a + r) ' 2 ' 

 a — 61 a' 1 1 a - 61' o' 



b'{a + r)' 2' ■ ■ QL~a a^b' {a +r) ' 2 ' 

 hence, (neglecting always a^) 



1 n-1 _1_ a- b\ 



QD nr na na'b-{a+r)'2 

 Fol. 111. Parti. D 



