The determinant of the matrix, (AD−BC), is a key parameter. The ratio of the two focal lengths of the system is simply given by the determinant. That is to say the ratio of the two focal lengths is given by:
(1.28)
Inspecting all matrix expressions in Eqs. (1.27a–1.27f), the determinant of the matrix is simply n1/n2, the ratio of the indices in the two media, for all possible scenarios. Since the determinant of a matrix product is simply the product of the individual determinants, then the determinant of the overall system matrix is simply the ratio of the refractive indices in image and object space. Thus:
(1.29)
This relationship was anticipated in the more generalised discussion in 1.3.9. Looking at the relationships for the principal and nodal points, it is clear when the determinant of the system matrix is unity, i.e. object and image space indices are the same, then the principal and nodal points are co-located.
In addition to the principal and nodal points, anti-principal points and anti-nodal points are sometimes (rarely) specified. Anti-principal points are conjugate points where the magnification is −1. Similarly, anti-nodal points are conjugate points where the angular magnification is −1.
1.6.3 Worked Examples
We can now use the foregoing analysis to see how matrix ray tracing might be used in practice. Here we focus on a number of useful practical examples.
Figure 1.19 Thick lens.
Worked Example 1.1 Thick Lens
The matrix for the system is simply as below – note the order:
We have two translations. The first translation represents the thickness of the lens and the second translation, by convention, traces the refracted rays back to the origin in z. This is so that, in interpreting the formulae for Cardinal points, we can be sure that they are all referenced to a common origin, located as in Figure 1.19. Positive axial displacement (z) is to the right and a positive radius, R, is where the centre of curvature lies to the right of the vertex. The final matrix is as below:
As both object and image space are in the same media, there is a common focal length, f, i.e. f1 = f2 = f. All relevant parameters are calculated from the above matrix using the formulae tabulated in Section 1.6.2.
The focal length, f, is given by:
The formula above is similar to the simple, ‘Lensmaker’ formula for a thin lens. In addition there is another term, linear in thickness, t, which accounts for the lens thickness.
The focal positions are as follows:
The principal points are as follows:
Figure 1.20 Hubble space telescope schematic.
Of course, since the refractive indices of the object and image spaces are identical, the nodal points are located in the same place as the principal points. If we take the example of a biconvex lens where R2 = −R1, then:
So, for a biconvex lens with a refractive index of 1.5, then the principal points lie about one third of the thickness from their respective vertices.
Worked Example 1.2 Hubble Space Telescope
The telescope part of the Hubble Space Telescope instrument is made up of two mirrors, a primary and a secondary. Characteristics of the telescope are shown in Figure 1.20. Data is courtesy of the National Aeronautics and Space Administration.
There are four matrix elements to consider here. First, there is a mirror with a radius of −11.04 m (note sign), followed by a translation of −4.905 m (again note sign). The third matrix element is a mirror (M2) of radius − 1.359 m. Finally, we translate by +4.905 m, so that both the input and output co-ordinates are referenced with respect to the same origin. The matrices are as below:
The focal positions are:
The principal points are at:
Since object and image space are in the same media, then the two focal lengths are the same. In addition, the nodal and principal points are co-located. However, when dealing with mirrors, one must be a little cautious. Each reflection is equivalent to a medium with a refractive index of −1, so that the matrix of a reflective surface will always have a determinant of −1. Therefore, any system having an even number of reflective surfaces, as in this example,
