The position of optically conjugate points in variable-magnification optical systems

T. V. Tochilina

St. Petersburg State University of Information Technologies, Mechanics, and Optics, St. Petersburg�Submitted December 5, 2005�Opticheski� Zhurnal 73, 48–49 �August 2006�

Relationships are obtained that determine the position of optically conjugate points in one-, two-,and three-component, variable-magnification optical systems. It is shown that, when the imagemagnification is varied in a two-component system, the distance between pairs of optically con-jugate points changes. © 2006 Optical Society of America.

The positions of the axial object and image points rela-tive to a one-component optical system are determined bythe thick-lens equation in the form 1/a0�−1/a0=1/ f0�. Thedistance between the axial object and image points in thiscase equals

L0 = − a0 + dHH� + a0� = dHH� −a0

2

a0 + f��1�

When a component is discretely displaced along the op-tical axis by distance b, we have a=a0−b. In this case, thedistance is

L = dHH� −a2

a + f��2�

Setting the right-hand sides of Eqs. �1� and �2� equal, weobtain an equation of the form

a02 − �b − 2�a0 − b = 0. �3�

Here the linear quantities are written in the scale of thefocal length of the component. The solution of this equationcan be written in the form1

a0k = −1

2�2 − b − �− 1�k�b2 + 4� �4�

and, accordingly,

ak = a0k − b = −1

2�2 + b − �− 1�k�b2 + 4�, k = 1,2. �5�

Note that the ratio a1 /a01=1/2�b+�b2+4��0, while the ra-

tio a2 /a02=1/2�b−�b2+4��0; i.e., in this case, when a lon-gitudinal displacement occurs, the component passes throughthe object plane, with a virtual image of it being formed.Moreover, the transverse magnification of the image formedby the optical component is

V0k =1

1 + a0= −

1

2�b − �− 1�k�b2 + 4� , �6�

Vk =1

1 + a=

1

2�b + �− 1�k�b2 + 4� . �7�

In this case, the falloff of the magnification is deter-mined as

533 J. Opt. Technol. 73 �8�, August 2006 1070-9762/2006/08

�k = 1 +1

2b�b + �− 1�k�b2 + 4� . �8�

It is easy to convince oneself that �1�2=1. It can beshown that the distance between the axial object and imagepoints is

L0k = Lk = dHH� + 2 − �− 1�k�b2 + 4.

We assume that the optical system under considerationconsists of two components separated by a finite air gap. Thefocal power of the system in this case is �=�1+�2−�1�2d,while the focal segments are sF=−1−�2d /� and sF�

� =−�−1−�1d /��. The distance between the principal planes is

dHH� = −�1�2

�d2. �9�

When �1=−�2=�0, we have �=�02d, sF=−1−�2d /�,

sF�� =1−�1d /�, and dHH�=d.

We transform Eq. �1� into the equation a02− �dHH�

−L0�a0− �dHH�−L0�=0, the solution of which can be writtenin the form

a0 =1

2�dHH� − L0 ± ��dHH� − L0��dHH� − L0 + 4�� . �10�

It follows from Eq. �9� that, when the distance d betweenthe components is varied, distance dHH� will vary, and seg-ment a0 will vary in accordance with Eq. �10�, and conse-quently the transverse magnification of the image formed bya two-component system. However, in this case, the differ-ence of the segments, determined by Eq. �10�, is a01−a02

=��dHH�−L0��dHH�−L0+4�. It follows from this that, whendistance dHH� varies and L0=const, the distance between thepairs of optically conjugate points will vary.

The distance from the first component to the axial objectpoint and from the second component to the axial imagepoint are determined by the corresponding relationships ofthe form

s1 = a0 + sF + f�, �11�

s2� = a0� + sF�� − f�. �12�

It follows from Eq. �10� that, when dHH�=L0, one seg-ment is a =0, and the other segment accordingly is a�=0.

0 0

5330533-02$15.00 © 2006 Optical Society of America

Let �1=−�2=�0. In this case, �=�0�2−�0d�, sF�� =−sF

= �1/�0��1−�0d /2−�0d�, and dHH�=d. Substituting thesequantities into Eqs. �11� and �12�, we get

s1 = sF + f� =d

2 − �0d, �13�

s2� = sF�� − f� = −

d

2 − �0d. �14�

In this case, s1�=−s2=1/2d. It follows from this that,when the optical system under consideration is modified byintroducing a third component between the first two and atan equal distance from each of them, the path of the axialpencil of rays is not disturbed. When the end components aredisplaced along the optical axis by distance �0, the focusingof the image is not disturbed if the focal power of the endcomponents is �0=1−V0, while the focal power of themiddle component is �k=−�2/1−V0�V0

3 / �1+V0�2−V02�0

2.Here all the linear quantities are given in the distance scaleof d0=1/2d, while V0 is the transverse magnification of theimage formed by the first component in the originalposition.2 As a result, we get a three-component variable-magnification optical system with discrete compensation ofimage defocusing. In such a system, with the components inthe original positions, the distances from the first componentto the axial object point are determined by relationships ofthe form

s011 =1

, �15�

V0

534 J. Opt. Technol. 73 �8�, August 2006

s012 =1 + V0�1 − V0��1 + �0

2�2V0

2 − 1. �16�

The corresponding distances between the optically con-jugate points are

L01 = − 21 − V0

V0, �17�

L02 = 22�1 − V0

2� + V0�1 − V0��1 + �02�

1 − 2V02 . �18�

Thus, for example, when V0=0.5 and �0=1, the dis-tances are L01=−2d0=−d and L02=8d0=4d.

Knowing the position of the optically conjugate pointsdetermines the possibility of a competent choice of the use ofvariable-magnification optical systems.

This work was carried out with the financial support ofthe Administration of St. Petersburg under the grant compe-tition for young candidates of sciences from institutions ofhigher education and academic institutes, located on the ter-ritory of St. Petersburg, Grant No. PD05-2.0-201.

a�Email: tvtochilina@mail.ru

1S. A. Zhurova and V. A. Zverev, “Bases for designing variable-magnification optical systems,” J. Opt. Technol. 66, No. 10, 68 �1999� �J.Opt. Technol. 66, 898 �1999��.

2T. A. Ivanova and V. K. Kirillovski�, Design and Monitoring of the Optics

of Microscopes �Mashinostroenie, Leningrad, 1984�.

534T. V. Tochilina