The General Optical System
The Focal Points and Planes
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
in the focal plane.
That is, one of the linear equations relating y ',
' to y,
must be of the form
Thus, the lower diagonal element of the matrix of equation (22) must
be zero so that,
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
indicates that F ' is to the right of
O '; a negative
value means F ' is to the left of
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
' although the
incident rays all have different angles
at the plane PP, the focal plane.
So one of the linear equations must state
that ' is independent of
(and depends only on y). Equation (22) gives
this relation if the upper diagonal element is zero, that is, if
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
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 ',
at the second focal
plane to an incident ray y, ,
at the first focal plane is given by
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
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
For the moment we can anticipate
the later results by designating the
two non-zero matrix elements of equation (26) as
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
is unity and
the determinant of the refraction matrix
separating two media of index n1 and
n2 say, is
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:
ignoring all the translations as they involve
multiplication by one. For the
optical system |A|OO ' =
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
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.
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.
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 '.
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.
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.
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,
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.
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