The President's Address. By E. M. Nelson. 165 



For equal deviation at each lens the conjugate of p must bear the 

 same ratio to the equivalent focus of the eye-piece that the distance 

 between the lenses does to the focus of the eye-lens ; thus 



P d r -\ 



In addition to these there are the well-known equations, viz. that for 

 equivalent focus, 



and that for magnifying power. 

 From these we obtain 



P r \ 



d= m - f .f>. (v.) 



Now, as p and m are given, and as a is the coefficient of/' in the 

 last formula, by combining and simplifying, we obtain the following 

 working equations : — 



f = a P f = a P • d = a f 



J TO + a ' J ~ (2 a - 1) m + (a - 1) a ' J ' 



m 



a = 2 + 



2_£T 



Example 1 . — An eye-piece of 10 power is required for a short-tube 

 Microscope. Find the foci of the lenses, and the distance between 

 them. 



Data: m = 10, p = 160 mm. = 6*3 in. 



fl -2 4. 10 -2-1%- /•_ 2-126 x 6-3 _ 13-394 _ 



a - 2 + 79-38~ 2 126 ' /_ 10 + 2-126 -12026 -1 ^ ' 



13 • 394 __ 1 3-394 = Q . 38 ' 36 . 

 1 3 • 252 m + 2 ; 394 " 34-914 " 



d = 2-126 x 0-3836 =0-816. 



A trigonometrical trace of a ray incident upon the field-lens at 

 a distance h = - 3 in. from the axis shows that S = 15° — 22' and 

 &' = 15° — 20', which is sufficiently accurate ; and as the value of d, 

 viz. 0*816, exactly satisfies equation (i.), the required foci and distance 

 have been found (fig. 43). 



Example 2. — An eye-piece of 5 power is required for the long 

 tube. Find the foci and lens distance. 



