Chapter 4

The General Optical System

The Focal Points and Planes
Figure 12.

If a beam of parallel rays enters an optical system the emergent beams will converge toward a point (or diverge away from a point for a negative system). This point will lie in the second focal plane. If the rays are parallel to the optic axis of a system the point is the second focal point. The focal plane passes through the focal point and is perpendicular to the optic axis. This is illustrated in Fig. 12.

We shall now determine the position of the second focal point, F ', in terms of the elements of the matrix [A]OO '. To do this let us take our second reference plane QQ to be the second focal plane in Fig. 12. Then D ' = O 'F '. The first reference plane PP can be anywhere to the left of O. We now note that every ray of the parallel beam passing through PP no matter what its y coordinate is, passes through the same point y ' in the focal plane. That is, one of the linear equations relating y ', alpha ' to y, alpha must be of the form

$$<i>y' ={\rm (constant)} \alpha \ \ {\rm(independent\ of}\ \alpha'). $$

Thus, the lower diagonal element of the matrix of equation (22) must be zero so that,
$$ D' = O'F'= a_{22} /a_{21} \eqno(23) $$ (23)

and thus
$$\alpha=a_{21}y'.$$

The position of the second focal point, F ', relative to the second vertex O ' is thus determined in terms of the matrix elements relating O' to O. A positive value of D ' indicates that F ' is to the right of O '; a negative value means F ' is to the left of O '.

Figure 13.

Rays emerging from a point in the first focal plane are illustrated in Fig. 13. A parallel beam emerges from the optical system. We can take the first reference plane PP through F and QQ the second reference plane anywhere to the right of O '. Now we note that the rays through Q ' pass this plane at constant angle alpha ' although the incident rays all have different angles alpha at the plane PP, the focal plane. So one of the linear equations must state that alpha ' is independent of alpha, (and depends only on y). Equation (22) gives this relation if the upper diagonal element is zero, that is, if
$$D = O F = a_{11} /a_{21}. \eqno(24)$$ (24)

Here D is positive if F is to the left of O, negative if F is to the right of O. Then we have from equation (22) that
$$ y =(a_{12}-D a_{22})\alpha'\eqno(25)$$ (25)

Thus the focal points may be easily found from (23) and (24) once the matrix [A]OO ' is computed.

The Matrix Relating the Focal Planes

It is extremely useful to choose the reference planes PP and QQ to be the first and second focal planes respectively. The matrix equation relating the emergent ray y ', alpha ', at the second focal plane to an incident ray y, alpha, at the first focal plane is given by
$$\bmatrix{y\cr \alpha}_F= \bmatrix{{\rm matrix} }_{FF'}\bmatrix{y'\cr \alpha'}_{F'}$$

From the discussion above, the matrix has a particularly simple form, since both diagonal elements are zero. Using (23) and (24) in (22), we obtain
$$\bmatrix{y\cr \alpha}_F= \bmatrix{0&(a_{12}-D a_{22})\cr a_{21} & 0}_{FF'} \bmatrix{y'\cr \alpha'}_{F'}\eqno(26)$$ (26)

The significance of the two non-zero elements of the matrix connecting the two focal planes is not evident at this point. However, as will be shown later the elements are related to the first and second focal lengths, f and f ', defined as the distances from the first and second principal points to the first and second focal points respectively (see Fig. 17). The principal points are defined later as well. In the familiar case of the thin lens, the two principal points are coincident at the lens centre. For the moment we can anticipate the later results by designating the two non-zero matrix elements of equation (26) as
$$a_{21}=1/f\eqno(28)$$ (28)

and
$$(a_{12}-Da_{22})=-f'\eqno(29)$$ (29)

A very general relationship between f and f ' may be easily shown by using the theorem on the product of determinants of matrices. First let us note that the determinant of a translation matrix,

$\bmatrix{1&-D\cr0& 1}$,

is unity and the determinant of the refraction matrix separating two media of index n1 and n2 say, is
$$\bmatrix{1&0\cr (n_2-n_1)/n_1R&n_2/n_1}=n_2/n_1.$$

Now if there is a series of refracting surfaces separating media with indices of refraction n, n1, n2, ...n ', starting with the medium of index n on the left and ending with the medium n ' on the right then the determinant of the product matrix is given by:

$$ (n/n_1) (n1/n_2)(n_2/n_3)\ldots (n_i/n_{i-1})(n'/n_i)=n'/n,$$

ignoring all the translations as they involve multiplication by one. For the optical system |A|OO ' = n'/n, and since a translation from the planes through the vertices O and O ' of the optical system to the first and second focal planes respectively does not change the value of the determinant then
$$\detr{0 & a_{12}-D a_{22}\cr a_{21}&0}=\detr{0&-f\cr1/f&0} =f'/f=n'/n\eqno(30)$$ (30)

The ratio of the first and second focal lengths is seen to be equal to the ratio of the index on the left of O to the index on the right of O '. Also f and f ' must have the same sign.

Equation (30) illustrates the power of the matrix method as the result is difficult to demonstrate in other ways for even a single thick lens and a general proof is not usually attempted for more complicated systems.

It is also perhaps well to note that the theorem on matrices constitutes a useful check on the accuracy of multiplication in computing [A]OO '. In the example of Fig. 10, the final product matrix (involving 7 matrices) should have a determinant equal to unity since we had n = n ' = 1.

Example

A thick biconvex lens in air has a radius of curvature of 10 cm at each surface and the surfaces are 10 cm apart. Find the focal points and the focal lengths of the lens if the index of the glass of the lens is 1.50.

Solution:

$$[{\bf A}]_{OO'}=\bmatrix{1 & 0 \cr (1.5-1)/10&1.5/1} \bmatrix{1 & -10 \cr 0&1}\bmatrix{1 & 0 \cr (1-1.5)/(-10)&1/1.5}$$

$$=\bmatrix{2/3 & -20/3 \cr 5/60&2/3}$$

f = 60/5 = 12 cm = f ' since n = n ' = 1.

OF = (2/3)×(12)= 8 cm. O 'F ' = (2/3)×(12)= 8 cm.

F is 8 cm to the left of O, F ' is 8 cm to the right of O '.

Exercise 1
Take the lens of the example to be 1 cm thick and show that f = 10.17 cm, OF = 9.83 cm, O 'F ' = 9.83 cm.

Exercise 2
Take the lens to be a biconcave thick lens with radii of 10 cm, index 1.5, thickness 1 cm. Show that f' = f = -9.84 cm, OF = -10.16 cm, O 'F ' = -10.16 cm.

Exercise 3
A thick lens is concavo convex, the first radius is 10 cm, the second radius is 20 cm. If the lens is in air and has an index = 1.5 and the thickness = 3 cm, show that f' = f = 36.36 cm,

Exercise 4
A biconvex lens of index 1.50 has radii 10 cm and 20 cm, a thickness of 10 cm. Air is to the left of O and water (of index 4/3) is to the right of O '. Show that f' = 24 cm, f = 18 cm, O 'F ' = 16 cm, OF = 17 cm.

Exercise 5
Find the matrix [A]OO ', and the [matrix]FF ' for a thin lens (that is, the thickness of the lens is taken to be zero), with index n on the left, n ' on the right and the index of the lens, µ. Take the radii of curvature as R1 and R2.

The above exercises may be worked using the computer programme in the appendix.