APPENDIX
Examples 1 to 5 below illustrate the determination of optical axes in a variety of optical systems.
The first three show that the methods give results one already knows. Examples 1 and 3
represent systems with an infinity of optical axes. In Example 2 there is no optical axis.
Example 4 considers a simple astigmatic eye; it has an infinity of optical axes. Example 5 is an
eye with astigmatic refracting elements that are relatively decentered and whose principal
meridians are obliquely crossing. It turns out to have a unique optical axis.
Reference to right and left is for the observer looking along the longitudinal axis in the positive
sense, that is, in the general direction the light is traveling. In the case of eyes the viewpoint is
that of the optometrist looking at the patient’s eye.
The calculations make use of the following results: The 5 5 transference of an optical system
takes the form9
S
T   T
o


1 
(A1)
where S is the 4  4 transference defined by Equation 4 and  the 4  1 matrix of Equation 5.
Thus
A

T C
oT

B
D
o
T
e

 .
1 
(A2)
The dioptric power (as dioptric power matrix) is given by13
F : C .
(A3)
For a homogeneous gap of reduced width 
 I

TO
oT

I
I
o
T
o

o
1 
(A4)
and for a thin system
 I

T C
oT

O
I
o
T
o


1 
(A5)
where O is the 2  2 null matrix.
Consider a system S  S1S2S3 , that is, a system S compounded of component systems
S1, S2, S3, etc, in order. The transference T of S is given in terms of the transferences T1 , T2 ,
T3 , etc, of the component systems by9
T  T3T2 T1 .
(A6)
Example 1: A Homogeneous Gap
Consider a homogeneous gap of reduced width  . We expect the analysis to show that the gap
has an optical axis and that every longitudinal axis is, in fact, an optical axis.
The system has 4  4 transference
 I I 

S  
O I 
and   o (Equations A1 and A4). From Equation 22
O O
 .
P  
O O
Hence
O O
 .
P   
O O
Equation 25 is clearly satisfied. Hence an optical axis exists. Equation 27 reduces to
d*Z0  g .
Thus d *Z0 is arbitrary. In other words every longitudinal axis is an optical axis as was expected.
Example 2: A Thin Prism
A thin prism deflects all rays. Hence one expects the analysis to reveal that a prism has no
optical axis.
o
Consider a thin prism of deflectance  . It has transference S  I and     (Equations A1

and A5). It follows that P (Equation 22) is the same as in Example 1.
Now Equation 25 fails in general. Hence, as expected, there are no optical axes.
Example 3: A Flat Plate
The analysis should show that a flat plate of transparent material in a homogeneous medium has
an infinity of optical axes all of which are perpendicular to its surfaces.
Consider a flat plate of plastic (index n 1 ) of thickness t and in air. From Equation A4 it has
4  4 transference
 I I 

S  
O I 
where   t / n1 is the reduced thickness. For longitudinal axis Z orthogonal to the surfaces
  o . Then from Equation 22
 O n1  1I 
 .
P  
O
O

Application of the Penrose conditions17-21 shows that
O
O

 .
P   





I
/

n

1
O
1


That the system has optical axes is shown by the fact that Equation 25 is satisfied. Equation 27
reduces to
 I O
g
d *Z0  
O O
or
 y *Z0   g1 

 
 a*   o 
 Z0   
which shows that every axis parallel to Z, that is, perpendicular to the plate, is an optical axis.
Example 4: A Simple Eye
Consider a simple eye with single spherical refracting surface (the cornea) of power F and
reduced length  . The refracting surface is centered on the longitudinal axis Z. The entrance
plane T0 is immediately in front of the cornea and the exit plane T immediately in front of the
retina.
The 4  4 transference is
 I I  I O 


S  
O
I

F
I



 I  F I 

 
I 
 F
and   o . For an eye in air n 0  1 . From Equation 22 one obtains
 F
P  
F
n  1I 
.
n  1I 
It is immediately apparent that the top block-row is a scalar multiple of the bottom block-row.
Because its rows are not linearly independent P is singular.
Equation 25 is obviously satisfied. Hence the eye has an optical axis. All of them are given by
(Equation 27)


d*Z0  I  P  P g .
Because P is singular P  P is singular and I  P  P  O . g  o generates one solution, namely,
d *Z0  o , among an infinity of others. Thus Z is one among an infinity of optical axes.
In this case it follows from Equation 24 that the inclination and transverse position of an optical
axis at the entrance plane are related by
a *Z0  
1
Fy *Z0 .
n 1
Suppose the cornea had principal meridians along the horizontal and vertical, that the
corresponding principal powers were 58 and 62 D respectively, that the length of the eye was 24
mm, and that the index was 4/3. Then the reduced length is  
0
0.0060
0 
1.0440


1.1160
0
0.006 
 0
P  
58
0
0.3333
0 


 0

62
0
0
.
3333


approximately. Making use of MATLAB we obtain
0
17.2352
0 
 0.3102


0
0.2902
0
16.1233 

3 
.
P  10 
0.0018
0
0.0991
0 


 0
0.0016
0
0.0867 

Then
d *Z0
0
0.005747
0
 0.000033
 g1 

 
0.000029
0.000000 0.005376  g 2 
 0.000000

 g 
0.005747
0
0.999967
0

 3 
 0.000000 0.005376 0.000000
0.999971  g 4 

or
d *Z0
 0.000033g1 0.00574g 3 


 0.000029g 2 0.005376g 4 

0.005747g1  0.999967g 3 


 0.005376g  0.999971g 
2
4

approximately. If we choose g1  g 2  g 3  0 this becomes
d *Z0
0




 0.005376g 4 

.
0


 0.999971g 
4 

0.024
 0.018 m and
4/3
If , in addition, we choose g 4  0.1 then we have generated an optical axis in a vertical plane
through Z and with an inclination of about 0.1 radians or 5.7 up which intersects the entrance
plane at a point about 0.54 mm below Z.
Example 5: A More Complicated Eye
Consider the astigmatic model eye used before10. Its single-surface cornea has power
4818043 , that is, the principal meridians are horizontal and vertical and the corresponding
principal powers along them are 48 D and 43 D respectively. The lens has front and back
surface powers 6309 and 116015 respectively. The reduced distance between the cornea and
the lens is 2.7 mm and between the lens and the retina 12.5 mm. The lens has reduced thickness
2.6 mm. We suppose further that all the refracting surfaces are centered on longitudinal axis Z
except that the cornea is decentered 1 mm to the right. The index of the lens is 1.416 and the
index of the rest of the eye is 1.333.
The length of the eye is
z  1.333  0.0027  1.416  0.0026  1.333  0.0125  0.02395 m.
Undecentered the cornea would have transference
0
 1

1
 0

T1   48 0

 0  43
 0
0

0 0 0

0 0 0
1 0 0 .

0 1 0
0 0 1 
Decentration of 1 mm to the right is equivalent to displacing the longitudinal axis 1 mm to the
left. That is, from Equation 11,
d *Z0
  0.001


 0 
.

0 


 0 


From Equation 22
0
0
0 
 0


0
0
0 
 0
P1  
 48
0
0.333
0 


 0


43
0

0
.
333


for the cornea. Thus, from Equation 21, the decentered cornea has
1  P1d *Z0
 0 


 0 

0.048 


 0 


and, hence, transference
0
 1

1
 0

T1   48 0

 0 43
 0
0

0 0
0 0
1 0
0 1
0 0
0 

0 
0.048  .

0 
1 
The anterior chamber has transference
1

0
T2   0

0
0

0 0.0027
1
0
0
1
0
0
0
0
0

0.0027 0 
0
0 .

1
0
0
1 
0
The transference of the first surface of the lens is
1
0


0
1


T3  6.7500 1.2990

 1.2990 8.2500

0
0

0 0 0

0 0 0
1 0 0 ,

0 1 0
0 0 1 
the transference of the body of the lens is
1

0
T4   0

0
0

0 0.0026
1
0
0
1
0
0
0
0
0

0.0026 0 
0
0 ,

1
0
0
1 
0
and the transference of the second surface of the lens is
1
0


0
1


T5  14.0000 1.7321

 1.7321 12.0000

0
0

0 0 0

0 0 0
1 0 0 .

0 1 0
0 0 1 
Finally the transference of the posterior portion of the eye is
1

0
T6   0

0
0

0 0.0125
1
0
0
1
0
0
0
0
0

0.0125 0 
0
0 .

1
0
0
1 
0
It follows that the transference of the eye is (Equation A6)
T  T6 T5 T4 T3T2 T1 ,
that is, the eye’s transference is
0.0331
 0.0709

0.0116
 0.0324

T  64.0947 2.4109

 2.3604 59.3247

0
0

0.0166 0.0002 0.0008 

0.0002 0.0167 0.0000 
0.9083 0.0125 0.0436  .

0.0125 0.9148 0.0006 
0
0
1 
Hence
 0.0008 


 0.0000 
.

0.0436 


 0.0006 


From Equation 22 we obtain
0.0331 0.0073 0.0002 
 1.0709


0.9884 0.0002 0.0073 
 0.0324
P
64.0947 2.4109 0.4247 0.0125 


 2.3604 59.3247 0.0125 0.4182 


which has rank 4 and so is nonsingular. Hence P   P 1 .
It follows that Equation 25 is satisfied and so an optical axis exists. In fact Equation 28 applies
from which we obtain
d *Z0
 0.001160 


 0.000162 
,

0.2780 


 0.0197 


representing a unique optical axis. It intersects entrance plane T0 at a point with position vector
 1.160 
 0.2780 
 . Thus displacing the cornea 1 mm to the
 mm and inclination a *Z0  
y *Z0  
 0.0197 
 0.162 
right has shifted the optical axis about 1.16 mm to the left and 0.16 mm up in the entrance plane
and turned it about 0.278 radians or 15.9 to the right and 0.020 radians or 1.1 down.