Chapter 3
The Basic Matrix Transformations of Geometrical Optics
Translation and Refraction
Now we have two basic matrix operations
which describe the progress of a
ray through an optical system. We need
only apply the laws of matrix multiplication
to determine the transformation between any two planes using the
translation and refraction matrices.
We illustrate this with the following example:
A glass rod of index n ' = 1.5 has a hemispherical
convex surface
of radius 6 cm on one end as in Fig. 5. A ray makes an angle a
equal to 0.1 radians with the plane PP at a height of 1.6 cm
from the optic axis. The plane PP is 8 cm to the left of the
vertex O. Find the height of the ray y '
and its direction '
at the plane QQ 9 cm to the right of O.
Figure 5
Solution:
or
Note that we can check that the determinant
of the product matrix is correct
since
Now
or
The solution to these equations is
y ' = 1.8 cm and '
= .0667 radians.
It is perhaps well to emphasize at this point that we have not yet
mentioned the idea of image formation and y '
of the example is certainly not
the image of y, since a ray originating at
y will go through the plane QQ
at a height y ' which depends on
. If
were .05 radians say, in the last example
then a ray through y = 1.6 cm passes the
plane QQ at the height y ' = 1.3 cm instead
of 1.8 cm. This is illustrated in Fig. 6.
Figure 6.
In anticipation of further sections, we can observe
that I ' in Fig. 5
appears to be an image of I as the ray along the optic axis
(which is undeviated) and
the ray at angle a both originate at I and pass
through I '.
To prove that I ' is indeed an image of I we need to show
that
all the rays
from I pass through I ' regardless of the angle
as in Fig. 7. We
choose the planes through I and I ' as our reference planes
and illustrate this with the help of the following exercise.
Exercise 1
The numbers in this example have been chosen so that I '
is 36 cm to the right of O and I is 24 cm to the left of O.
Figure 7.
Find the matrix connecting the planes through I and I '
and show that
where now y ',
',
represents the ray at the I ' plane
and y,
, the ray at
the I plane.
In this exercise we note that y ' = y
independent of
so that
if y = 0 (point I) then y ' = 0 as illustrated
in
Fig. 7 which shows I ' to
be an image of I.
As can be seen from the above exercise, if the matrix element in the
upper right hand corner
a_{12}
is zero, then the
two planes are such that y '
depends only on y. With this fact we can derive
the usual equation (equation
(21) below) for image formation by refraction
at a spherical surface.
Exercise 2
Let s be the distance OI and s ' be
OI '.
If the index on the left is n, on the
right n ' and R is the radius of curvature, then
Show that:

(21) 
if I ' is an image of I.
Sign Convention
Equation (21) does presuppose the sign convention of
Jenkins and White,
for example, in which the object distance s is
positive if the object is to
the LEFT of O and the image distance s ' is
positive
if the image is to the
RIGHT of O.
In the above example we have used positive image and object distances.
In Figs. 8 and 9 we illustrate negative distances which in these cases
represent a virtual object and a virtual image.
Figure 8.
Figure 9.
Generally we shall be able to extend the sign convention to a complex
optical system in which the relevant distances are measured from
various
cardinal points which we shall define for the system.
The cardinal points are readily obtained using the matrix approach.