176 Nul Point of Thin Axial Pencils of Light. 



Turn the optical system round (fig. 2) through two 

 right angles so that H 2 comes to H 2 ' and H^ to H/. ceases 

 to be the nul point ; but if the system be displaced to the 

 right a distance a + 20H 2 , H/, and H 2 ' will occupy the 



Fig. 2. 



Hi Hg O H 2 H, 



former positions of H 2 and Hj. will be again the nul 

 point with the same magnification as before. Thus, since 

 a + 20H 2 can be measured, and since a has been already 

 determined, 0H 2 can be found, and consequently the positions 

 of H x and H 2 . 



We can now find easily the positions of F 2 and F 2 , and 

 thus the focal lengths f and f 2 . Through (fig. 1) draw 

 any line intersecting the object and image planes in Q x 

 and Q 2 , and through Q x draw a line parallel to the axis 

 meeting the first principal plane in K 2 . Take H 2 K 2 equal 

 and parallel to HjKx. Then K 2 Q 2 intersects the axis in F 2 . 

 Similarly, by beginning with Q 2 and ending with Q l5 Y l may 

 be found. Also, since F 1 N 1 = H 2 F 2 , and FgN^HiFi, the 

 nodal points N x and N 2 are determined. Or i f l and/ 2 may 

 be found from the formulae 



m Vl ,_ v 2 



Jl — ~~ 1 ~9 /2 — 



1 — ??i' J 1 — m 



The nodal points Ni and N 2 may^ however, be determined 

 directly. We have 



0N 2 = 0H 2 + H 2 N 2 = 3, + H,F, + HxFj 

 = 0H 2 +/i+/, 



ma 



and 



1 — m 

 ON 1 =OH 1+/l +y 2 = r 



m 



Thus ON 2 /ON! = m. When the last medium has the same 

 refractive index as the first, Nj and N 2 coincide with Hi 

 and H 2 , and we have OH 2 /OH : = w. The method described 

 above for determining H x and H 2 becomes very simple in 

 this case. In fact, the distance d through which the system 

 is moved is 



z . H 2 H 1? and. therefore, 



1 — m 



1^=*=^, 0H 2 =^, 0H 1 = 



1 + m ' ' 1 + m' 1 1 + m 



