Page 321 - Modern Optical Engineering The Design of Optical Systems

300 Chapter Thirteen

pupil and a 1-in-diameter eyelens, the 4-in eye relief R limits us to an
apparent field as follows:
1
u 4u (eyelens dia. pupil dia.)
e o
2R
1
(1 0.25) 0.09375
2 4

u 0.0234 ( 1.3°)
o
To determine the spacing and powers of the components, we note
that the length will be

L f s s f
o 1 2 e
and the magnification will be

f os 2
M
f s
e 1
We can combine these expressions and derive equations for s 1 , s 2 ,
and f r in terms of M, L, f o , and f e as follows:

f (L f f )
e
o
o
s
1
(Mf f )
e o
Mf e (L f o f e)
s Mf e
1
s
2
f o (Mf f )
e o
s 1s 2 Mf e f o (L f f )
o
e
f
r s s 2
1 2 (Mf e f o)
At this point, we are faced with a situation which is very common in
the layout stages of optical design. We can elect to proceed algebrai-
cally to find an expression for f o and f e which will yield a scope with the
desired eye relief R, or we can proceed numerically. In general, for a
one-time solution, the numerical approach is usually the better choice,
especially if the system under consideration is well understood. If one
is likely to design a number of systems of the same type with various
parameters, or if one is “exploring” and wishes to locate all possible
solutions, the often tedious labor of an algebraic solution may be well
repaid.
The preceding equations indicate that we have two choices (or
degrees of freedom) which we can make, namely f o and f e , and arrive at