A great many optical systems consist of just two separated compo-nents. The following expressions can be used to handle any and all
u3y2 y3u2
u1y2 y1u2 y3u1 u3y1
u1y2 y1u2
components as well as “thin lenses.” Also, these expressions are valid for components of any degree of complexity. For thick lenses the spac-ings are measured from their principal points. For thin lenses the spacings are simply the lens-to-lens distances.
Given. The powers (or focal lengths) of the components and their spacing.
Find. The efl, bfl, and fflof the combination. See Fig. 1.11.
Power:
ab a b dab (1.27) efl:
fab (1.28)
bfl:
B (1.29)
ffl:
FF fab(fb d) (1.30)
fb
fab(fa d)
fa fafb
fa fb d
FF d B
fab
φa φb
F1
P2
F2
Figure 1.11 Two components with the object at infinity, showing the spacing, the sec-ond principal point, the effective focal length, the back focus distance B, and the front focus distance FF.
Given. The efl, d, and B of the combination.
Find. The focal lengths or powers of the components. See Fig. 1.11.
fa (1.31)
fb (1.32)
Finite conjugate systems. See Fig. 1.12.
Given. The component locations (defined by the object distance s, the image distance s′, and the spacing d) and the magnification m h′/h
u/u′.
Find. The component powers.
a (1.33)
b (1.34)
Given. The component powers, the object-to-image distance, and the magnification m h′/h u/u′.
d ms s′
ds′
ms md s′
msd
1
b
fab B ddB
1a
dfab
fab B
d s'
φa φb
T h
h'
(--) s
u u'
Figure 1.12 Two components working at finite conjugate distances, showing the
spac-Find. The component locations (i.e., s, s′, and d). Solve this quadratic for d:
0 d2 dT T(fa fb) (1.35)
using x
Then:
s (1.36)
s′ T s d (1.37)
Sample calculation
Find the Gauss points of a system whose first lens has a focal length of 200 mm and whose second lens has a focal length of 100 mm, where the separation between lenses is 100 mm. See Fig. 1.13.
Power by Eq. (1.27):
ab 0.005 0.01100•0.005•0.01 0.01 冢f 100.0冣
Focal length by Eq. (1.28):
1
(m 1)d T
(m 1) mda
b 兹b苶2苶苶 4苶a苶c苶
2a
(m 1)2fafb
m
100
100 50
f = 200 f = 100
P1
P2
F1 F2
Figure 1.13 Showing the system used for the sample calculation of the cardinal points of a two component system.
fab 100.0
Back focus by Eq. (1.29):
B 50.0
Front focus by Eq. (1.30):
FF 0.0
Find the component powers necessary to produce an erect image 15 mm high from a distant object which subtends an angle of 0.01. The system should be 250 mm long from front lens to image. See Fig. 1.14.
For an erect image the focal length must be negative. [Consider Eqs. (1.7) and (1.8).] Its magnitude [from Eq. (1.7)] is 15.0/0.01 150
100(100 100)
Figure 1.14 A widely separated system of two positive components with the second com-ponent acting as an erecting relay lens. This system forms a real, erect image and has a negative focal length. Note that the second principal point is to the right of the second focal point. The combination has a focal length equal to the focal length of the first lens
mm. Thus fab 150. The sum of the space d and back focus B must
Obviously we are free to select the value of B (or d). Arbitrarily set-ting B 50, we get
fa 150.0 a 0.00666…
fb 25.0 b 0.040
Additional study project: Calculate the component powers for sever-al additionsever-al vsever-alues of B. Plot a, b, and (|a|+|b|) against B to find the minimum power required to do the job.
Sample calculation
Find the Gauss points for the system in the previous calculation. See Fig. 1.14.
Since we know the focal length is 150 mm and that B is 50, F2
must be 50 mm to the right of component b. P2is 150 mm to the right of F2because the system focal length is negative; therefore, P2is 200 mm to the right of the rear component.
Using Eq. (1.30) we find FF (150)(25200)/25 1050, which indicates that F1 is 1050 mm to the left of component a. This puts P1 at (1050150) 1200, or 1200 mm to the left of the first lens.
Sample calculation
We need a magnification of 2 in a distance of 0.9 m, using two components which are evenly spaced between object and image.
Determine the necessary component powers. See Fig. 1.15A.
By the “evenly spaced” requirement we mean that s, d, and s′ are all the same size. Then, by our sign convention, s 300, d 300, and s′ 300.
By Eq. (1.33):
a 0.00833…
fa 120.0 By Eq. (1.34):
b 0.0133…
f 1 75.0
300 2(300) 300
300•300
1a
2(300) 2•300 300
2(300)300
300 300 300
f = 120 f = 75
h h'
(a)
300 300 300
f = 200 φ = 0
h
h'
(b)
Figure 1.15 Two solutions to the problem of producing a two times magnification in an object-to-image distance of 900 mm. One solution has a positive magnification (and an erect image) and the other has a negative magnification (and an inverted image). It is common to find two different systems which produce images of the same size in the same place, one producing an erect image, the other an inverted image, if you use plus and minus magnifications in the equations.
However, if we use a magnification of 2, so that the image is inverted, we get the following. See Fig. 1.15B.
By Eq. (1.33):
This indicates that if image orientation were not a concern, this par-ticular task could be handled with just one lens rather than the two which are needed when the image is required to be erect. This illus-trates the importance of considering both erect and inverted imagery if one is not preferred over the other.
Sample calculation
Lay out a Cassegrain mirror system with a focal length of 100, a mir-ror separation of 25, and an image distance of 30. Use Eqs. (1.31) and (1.32) with fab +100, d 25, and B 30. See Fig. 1.16.
Note that when we raytrace the Cassegrain system, both the spac-ing and the index between the mirrors are considered to be negative (because the secondary mirror is to the left of the primary, and because light is traveling from right to left). The equivalent air dis-tance is the spacing divided by the index, so, as explained in Sec. 2.13, we use 25/1.0 +25 for d in Eqs. (1.31) and (1.32).
Remembering that a mirror radius is twice its focal length, and that concave mirrors have positive (convergent) focal lengths, the primary mirror has a focal length of 35.714 and a concave radius of 71.428;
the secondary mirror has a focal length of 16.667 and a convex radius of 33.333. Note that the raytrace sign of the radius is deter-mined by the location of its center of curvature, not by whether the mirror is concave or convex.