### 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 a12 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.