The Numerical Solution Method

This approach might be called “cut and try, with differential adjust-ment.” It is beloved by those who detest algebra. It works well provid-ed that the type of system neprovid-edprovid-ed is reasonably well understood by the person doing the layout. And it is probably the most popular and widely used approach to optical system design. In essence one more or less arbitrarily selects one or two parameters (i.e., powers and/or spaces) and then adjusts the balance of the system to satisfy one or two of the system requirements. One parameter is then selected as a free variable. It is changed by a small increment, and the balance of the system is again adjusted. The change in an uncontrolled (but required) system characteristic is noted, and a differential solution for the free variable is made.

Sample calculations

As an excruciatingly trivial and simple example, let’s lay out a kepler-ian telescope with a magnification of 4 and an overall length of 10 in. The system is sketched in Fig. 5.2. Of course, it doesn’t take an Einstein to figure out the correct answer: f_a 8 in and fb 2 in.

But let’s demonstrate the “cut, try, and adjust” process. We can,

fa fb

L

(a) (b)

with reference to Fig. 5.2, start by choosing f_a, then determine f_b to satisfy either the length or magnification requirement; let’s go with magnification. Starting with a (deliberately bad) guess of f_a +5 in, fb

must equal fa/(4)  +1.25 in to get a magnification of 4. This gives us a length of f_a+f_b 6.25 in. Next we choose fa 6 in and determine that f_b must be 1.5 in, giving us a length of 7.5 in. The length has changed 1.25 in for a 1-in change in fa, so (L/fa)≈1.25/1.0  1.25. We need a 10-in length, 2.5 in longer than our second trial; the solution is to increase f_aby 2.5/1.25 2 in to a value of f_a 8 in. Then fbwill be 2 in and the length will be 10 in, as required. Of course, all this is already worked out in Chap. 2 as Eqs. (2.4) and (2.5).

Sample calculations

In order to make this exercise a bit more interesting, let’s make our telescope a lens-erecting type, again 10-in long, with a magnification of 4, and let’s also require an eye relief of 4 in. This type of scope is sketched in Fig. 5.3. Again we play the same sort of game, arbitrar-ily selecting a focal length for the objective f_a +4 in, and also a focal length for the eyelens f_c +1 in. This pair has a magnification of 4/1

 4, so we need a magnification of m  1.0 from the erector to get a power of 4 for the total scope. The objective and eyelens focal lengths add up to a distance of f_a+f_c 4  1  5, leaving a space of 105  5 in for the erector, which at m  1.0 must have s  2.5 in and s′  +2.5 in. Equation (1.4) can be solved for the erector focal length f_b +1.25 in. Now we can trace a principal ray through the center of the objective lens, using Eqs. (1.19) and (1.20) [or perhaps Eqs. (5.4) and (5.5)]. We find that this ray intersects the axis at 2.05 in from the eyelens; our eye relief is short by (42.05)  1.95 in.

Next we try f_a +4 (again) and fc +1.25 (fc +0.25, or 
c

0.2). This pair has a power of MP  4/1.25  3.2 and uses up a length of 4 1.25  5.25 in, leaving 4.75 in for the erector to produce a magnification of 4/(3.2)  1.25. Using the relationships m s′/s  1.25 and ss′  4.75, we find that s  2.111… and s′  +2.63888…, and again Eq. (1.4) gives us the erector focal length as f_b

 1.172840 in. Tracing the principal ray with Eqs. (1.19) and (1.20), we now calculate an eye relief of 2.565789; an increase of 0.515789 in the eye relief (ER) has been produced by an increase of eyelens focal length of 0.25 in.

So we have ER/fc≈+0.515/+0.25  +2.063154 (or ER/
c≈

2.578943). Since we need an increase in the eye relief of (4.02.565784)  +1.434211 from our next try, we want a change in

faSS'fcER

(a)(b)

(c) Exit pupil Schematic lens-erecting telescope.

the eyelens focal length of 1.434/+2.063  +0.695155 to fc  1.945155. [Alternately, we could take a power change of  1.434/(2.578943)  0.556123 to get
c +0.243876 and fc  +4.100437. There is a significant difference between the result we obtain using power compared to that using focal length. This is due to the nonlinearity of the relationships involved. Ordinarily using power works out better, but in this case your author’s a priori knowledge indicates that the focal length result is the better one, so we will go with that.] Again, f_a/f_c 4 in/1.945 in  2.056391; this leads to an erector magnification of 4/2.056  1.945155 in a distance of (1041.945)  4.054845 in. The erector conjugates are s 

1.376785 in and s′  +2.678060 in, and fb +0.909310 in. The prin-cipal raytrace indicates an eye relief of l_c′  +4.334309 in. We’re get-ting closer.

Our last change in f_c was  0.695155 in, and it produced a change in ER of 1.768525 in; thus we get ER/fc≈2.544073. To change the ER by 0.334 in, we need a change in fcof0.334/2.544  0.131407 in, for a new f_cof 1.813748 in. This process can be carried on until the eye relief is close enough to the desired 4-in value. (The exact solution is f_a 4.0 in, fb +0.9529 in, fc 1.8298 in.) Note that when the final value to be determined (the eye relief in this case) is one where an exact value is not required, the process can be terminated as soon as a reasonably close result is obtained.

In the above calculations we arbitrarily chose the focal length of the objective lens as f_a +4.0 in. This is obviously an unused variable parameter, and we could, if desirable, use it to control an additional characteristic of the system. Thus we could repeat the above process for several additional values of f_a and choose from among the solu-tions the one that best suited our application. In this case we could, as indicated at the end of Sec. 5.1, select a solution on the basis of minimizing the component power sum.