## Abstract

A procedure is described and discussed, by which the requirements imposed on an optical system can be formulated in terms of the unknown surfaces, so that conditional (differential) equations can be written as a basis for the design. This is made possible through a projective transformation by which any ray is represented as a point of a mapping plane, referred by duality to the physical plane of the optical trajectory.

In the first part of the paper, after a general statement of the problem, a summary is given of the principal results of projective geometry that are utilized in the discussion; next, the differential equation of a refracting or reflecting surface is deduced and discussed in dual terms.

The second part of the paper will concern the explicit formulation of a design and the discussion of the principal aberrations, by effecting the ray tracing in the dual plane.

© 1959 Optical Society of America

### References

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1. The geometrical procedure for the construction of an involution given two pairs, or two double points, is based on the properties of the “complete quadrangle” (Desargues).
2. In homogeneous coordinates x/z and y/z, a general circle is (x−az)2+(y−bz)2=r2z2. The intersection with the straight at infinity, z=0, is in x2+y2=0, that is in y/x=±j.
3. It suffices to write y=Ax+B.
4. Lines connecting corresponding points are united in the case of the involution; so that on any one of them two double points exist. If we consider any two of such lines, it becomes obvious that their common point is double (otherwise, every line should contain three double points and the collineation should be an identity). The point which is common to all the lines is the center of the homology, the (rectilinear) locus of the second double points is the axis.
5. The involution is a particular case (m=2) of a cyclic collineation, that is, of a collineation whose m th power is the identity.
6. The coefficients mik could be obtained by inverting (10).
7. While the tangents actually coincide for two infinitely close pairs, the normals are only “parallel,” so that their abscissa is the same.
8. It is seen at once that t=−1/n satisfies the equation(t,n,j,-j)=-1 which is the translation of (15) in the parabolic plane.

#### Other (8)

The geometrical procedure for the construction of an involution given two pairs, or two double points, is based on the properties of the “complete quadrangle” (Desargues).

In homogeneous coordinates x/z and y/z, a general circle is (x−az)2+(y−bz)2=r2z2. The intersection with the straight at infinity, z=0, is in x2+y2=0, that is in y/x=±j.

It suffices to write y=Ax+B.

Lines connecting corresponding points are united in the case of the involution; so that on any one of them two double points exist. If we consider any two of such lines, it becomes obvious that their common point is double (otherwise, every line should contain three double points and the collineation should be an identity). The point which is common to all the lines is the center of the homology, the (rectilinear) locus of the second double points is the axis.

The involution is a particular case (m=2) of a cyclic collineation, that is, of a collineation whose m th power is the identity.

The coefficients mik could be obtained by inverting (10).

While the tangents actually coincide for two infinitely close pairs, the normals are only “parallel,” so that their abscissa is the same.

It is seen at once that t=−1/n satisfies the equation(t,n,j,-j)=-1 which is the translation of (15) in the parabolic plane.

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### Figures (9)

Fig. 1

For the definition of anharmonic ratio.

Fig. 2

Projectivity between points of the same line.

Fig. 3

Polarity.

Fig. 4

Projective form of Snell’s law.

Fig. 5

Reflection.

Fig. 6

Multisurface path.

Fig. 7

Trigonometric Snell’s law.

Fig. 8

Stationarity of optical paths.

Fig. 9

Constancy of the path between stigmatic points.

### Equations (57)

$P R = O P sin ( p r ) sin O R P$
$( P Q R ) = P R Q R = O P O Q sin ( p r ) sin ( q r ) .$
$( P Q R S ) = ( P Q R ) ( P Q S ) = sin ( p r ) sin ( q r ) ÷ sin ( p s ) sin ( q s ) = ( p q r ) ( p q s ) = ( p q r s ) ,$
$( n 1 n 2 n 3 n 4 ) = n 3 - n 1 n 3 - n 2 ÷ n 4 - n 1 n 4 - n 2$
$x ′ z ′ = a x + b z c x + d z$
$x ′ = a x + b c x + d ,$
$( A B C D ) = ( A B C ′ D ′ )$
$( A B C C ′ ) = ( A B D D ′ ) .$
$a - c x 2 a - c x 1 = d + c x 1 d + c x 2 = a x 1 + d x 2 a x 2 + d x 1 ,$
$x = - d x ′ + b c x ′ - a$
$x ′ = a x + b x - a , x = a x ′ + b x ′ - a$
$( A B C C ′ ) = - 1.$
$( i 1 i 2 p q ) = ( + j , - j , m , n ) = m + j m - j · n - j n + j = 1 + m n - j ( n - m ) 1 + m n + j ( n - m ) = e - 2 j α .$
$α = 1 2 j ln ( i 1 i 2 p q ) ; ( Laguerre )$
$ρ x ′ = a 11 x + a 12 y + a 13 z , ρ y ′ = a 21 x + a 22 y + a 23 z , ρ z ′ = a 31 x + a 32 y + a 33 z ,$
$x ′ = a 11 x + a 12 y + a 13 a 31 x + a 32 y + a 33 , y ′ = a 21 x + a 22 y + a 23 a 31 x + a 32 y + a 33 ,$
$x ′ = ( a 11 + a 12 A ) x + ( a 13 + a 12 B ) ( a 31 + a 32 A ) x + ( a 33 + a 32 B ) ,$
$| a 11 - ρ a 12 a 13 a 21 a 22 - ρ a 23 a 31 a 32 a 33 - ρ | = 0$
$x 0 a + y 0 b + 1 = 0.$
$x 0 a + y 0 b + z 0 c = 0$
$σ u = k 11 a + k 12 b + k 13 c , σ v = k 21 a + k 22 b + k 23 c , σ w = k 31 a + k 32 b + k 33 c ,$
$ρ x = k 11 p + k 21 q + k 31 r , ρ y = k 12 p + k 22 q + k 32 r , ρ z = k 13 p + k 23 q + k 33 r .$
$p = m 11 x + m 12 y + m 13 m 31 x + m 32 y + m 33 , q = m 21 x + m 22 y + m 23 m 31 x + m 32 y + m 33$
$m 11 x 2 + m 22 y 2 + 2 m 12 x y + 2 m 13 x + 2 m 23 y + m 33 = 0$
$sin ( p n ) sin ( q n ) = ν 2 ν 1 .$
$sin ( p b ) sin ( q b ) = - 1 ,$
$( p q n b ) = - ν 2 ν 1 .$
$( p q n b ) = - 1.$
$( r i - 1 , r i , n i , b i ) = - ν i ν i - 1 [ r 0 = p , r n = q ] ,$
$( r i - 1 b i i 1 i 2 ) = ( b i r i i 1 i 2 ) ,$
$( b i ′ b i ″ i 1 i 2 ) = - 1.$
$( n i t i i 1 i 2 ) = - 1.$
$( x 1 x 2 x 3 x 4 ) = x 3 - x 1 x 3 - x 2 · x 4 - x 2 x 4 - x 1 .$
$u = k 11 a + k 12 b + k 13 k 31 a + k 32 b + k 33 ; v = k 21 a + k 22 b + k 23 k 31 a + k 32 b + k 33 .$
$u = - a / b , v = 1 / b$
$( p , b , j , - j ) = ( b , q , j , - j )$
$b = p q - 1 ± ( 1 + p 2 ) 1 2 ( 1 + q 2 ) 1 2 p + q = - ( 1 + q 2 ) 1 2 ∓ ( 1 + p 2 ) 1 2 p ( 1 + q 2 ) 1 2 ∓ q ( 1 + p 2 ) 1 2 = p ( 1 + q 2 ) 1 2 ± q ( 1 + p 2 ) 1 2 ( 1 + q 2 ) 1 2 ± ( 1 + p 2 ) 1 2$
$n = ν 1 p ( b - q ) + ν 2 q ( b - p ) ν 1 ( b - q ) + ν 2 ( b - p ) = ν 1 p ( 1 + q 2 ) 1 2 - ν 2 q ( 1 + p 2 ) 1 2 ν 1 ( 1 + q 2 ) 1 2 - ν 2 ( 1 + p 2 ) 1 2 .$
$n = - tan γ = ν 1 sin α - ν 2 sin β ν 1 cos α - ν 2 cos β$
$t = - ν 1 ( 1 + q 2 ) 1 2 - ν 2 ( 1 + p 2 ) 1 2 ν 1 p ( 1 + q 2 ) 1 2 - ν 2 q ( 1 + p 2 ) 1 2 .$
$t = b = tan α + β 2 = - cos α + cos β sin α - sin β = sin α + sin β cos α + cos β = - ( 1 + q 2 ) 1 2 - ( 1 + p 2 ) 1 2 p ( 1 + q 2 ) 1 2 - q ( 1 + p 2 ) 1 2 = p ( 1 + q 2 ) 1 2 + q ( 1 + p 2 ) 1 2 ( 1 + q 2 ) 1 2 + ( 1 + p 2 ) 1 2$
$t = - ν 1 ( 1 + u 2 2 ) 1 2 - ν 2 ( 1 + u 1 2 ) 1 2 ν 1 u 1 ( 1 + u 2 2 ) 1 2 - ν 2 u 2 ( 1 + u 1 2 ) 1 2$
$| u 1 v 1 1 u 2 v 2 1 t s 1 | = 0.$
$t ( v 1 - v 2 ) - s ( u 1 - u 2 ) + u 1 v 2 - u 2 v 1 = 0.$
$( v 2 - s ) u ˙ 1 - ( u 2 - t ) v ˙ 1 = ( v 1 - s ) u ˙ 2 - ( u 1 - t ) v ˙ 2 .$
$( u 2 - t ) [ ( v 1 - v 2 ) u ˙ 1 - ( u 1 - u 2 ) v ˙ 1 ] = ( u 1 - t ) [ ( v 1 - v 2 ) u ˙ 2 - ( u 1 - u 2 ) v ˙ 2 ] ,$
$- u 1 x + y + v 1 = 0.$
$- ( u 1 + u ˙ 1 d ξ ) x + y + ( v 1 + v ˙ 1 d ξ ) = 0 ;$
$ν 1 { v 2 - v 1 u 2 - u 1 - v ˙ 1 u ˙ 1 } ( 1 + u 1 2 ) 1 2 + ν 2 { v ˙ 2 u ˙ 2 - v 2 - v 1 u 2 - u 1 } ( 1 + u 2 2 ) 1 2 .$
${ ν 1 u 1 ( 1 + u 1 2 ) 1 2 - ν 2 u 2 ( 1 + u 2 2 ) 1 2 } 1 ( u 2 - u 1 ) 2$
$ν 1 〈 P A 〉 + ν 2 〈 A R 〉 = stationary - ν 2 〈 B R 〉 + ν 3 〈 B S 〉 = stationary - ν 3 〈 C S 〉 + ν 4 〈 C Q 〉 = stationary .$
$ν 1 〈 P A 〉 + ν 2 〈 A B 〉 + ν 3 〈 B C 〉 + ν 4 〈 C Q 〉 = stationary ,$
$1 u 1 - t = 1 u 2 - u 1 [ u 2 - t u 1 - t - 1 ] ,$
$1 u 1 - t = 1 u 2 - u 1 { ( v 1 - v 2 ) u ˙ 2 - ( u 1 - u 2 ) v ˙ 2 ( v 1 - v 2 ) u ˙ 1 - ( u 1 - u 2 ) v ˙ 1 - 1 } .$
$u ˙ 1 u 1 - t = u ˙ 2 - u ˙ 1 u 2 - u 1 - u ˙ 2 x 0 - v ˙ 2 u 2 x 0 - y 0 - v 2 = d d ξ ln u 2 - u 1 u 2 x 0 - y 0 - v 2 .$
$v ˙ 1 u 1 - t = d d ξ v 1 - v 2 u 1 - u 2 .$
$(t,n,j,-j)=-1$