The Cardinal Points
It is often convenient to discuss image formation in terms of other
positions than the focal points. These other points are the positive
and negative principal points and the positive and negative nodal points.
These, together with the focal points, are all defined as the cardinal
points of a lens system and the planes perpendicular to the optic axis
through these points are the cardinal planes. From the basic definitions
of principal points and nodal points, it is a simple matter to find
their positions using the matrices already described.
The positions of the principal points and nodal points are most readily
specified in terms of their positions relative to the focal points F
and F '. Let us assume that a cardinal point Z, say,
is a distance z from F and the cardinal point
Z ' is a distance
z ' from F '.
Then the ray y, , at the plane
through Z goes through the plane at Z ' with
'. This can be
We may now define the principal and nodal points and the definitions
will impose conditions on the matrix of (36) which will give the
positions z and z ' of the points in question relative
to F and F '.
Positive Principal Points
The planes through the positive principal points, H and
H ', are the
positive principal planes and are defined as planes of unit positive
magnification, that is, planes for which
y ' = y.
But from (36) y = -(z/f)y +
(zz '/f-f ')
Thus we must have zz '/f - f ' = 0
and -z/f = 1 for
the positive principal points H and
The first and second focal lengths
are generally defined as the distances
F 'H ' respectively. We see then from equations (37) and
that we are justified in having written equations (28) and (29)
where we introduced the symbols f and
f ' into
the matrix connecting the first and second focal planes.
The principal point,
H ', is a distance f ' to the left
F ', the principal point H a distance f to the right
for a positive focal length system as illustrated in Fig. 17.
Depending on the system the points may lie on either
side of O or O ' and even may be "crossed," that is,
may lie to the left of H.
In Fig. 18 is illustrated a ray diagram showing the formation of an
image using the properties of the principal planes and of the
focal points. (Equation (34),
y = -x y '/f, also
follows from the diagram.) The object distance
as measured from H is s, the
image as measured from H ' is s '.
The ray y ',
at the image is related to the ray y,
, from the object by
As stated before in deriving Newton's equations, the upper right hand
element of the matrix must be zero if y ' is an image
y, that is, if the rays through the image plane at height
y ' are to be independent of
Thus we have
which makes the upper right hand element zero and gives
y = ((f - s)/f)y '
the magnification, m = y '/y =
From equation (41), (f - s)
= -sf '/s '
and since the determinant equals f '/f
equation (40) becomes
Use (41) and (42) to compute the image
positions of the object in
Solution: From the results of example 2,
(see Fig. 16) H is 3.333 cm to the
right of F or (4.167 - 3.333) = .834 cm to
the left of O. H ' is 4.333 cm to
the left of F ' or (4.333-2.167) = 2.167 cm
to the left of O '. The object
distance s = 10.467 - .834 = 9.633
4.333/s ' + 3.333/9.633 = 1.
so s' = 6.627 cm.
Measured from O ' the image
is 6.627 - 2.167 = 4.46 cm (to the right of O ') in
agreement with Newton's equations of the previous example.
m = -(1.0/1.30) X (6.627/9.633)= -0.529
If the medium has the same index
on either side of the optical system,
f = f '
and these equations take the very familiar forms,
For a thin lens in air, the situation
is even simpler as the image and object
distances s ' and s can be measured from
the same point, the centre of the lens,
since for a thin lens the principal
planes are approximately coincident at the
lens centre. This and other special
cases will be discussed in the next chapter.
Negative Principal Points
The planes perpendicular to the optic axis
through the negative principal
points, H(-) and H '(-), are the negative
principal planes and are defined as
planes of unit negative magnification,
that is, y '= -y. It is left to the
reader to show that
Thus in the positive system illustrated in Fig. 19,
an object placed at the
distance f to the left of F will give rise to an
inverted image, the same size
as the object, at a distance f ' to the right of F '.
It is left as a further exercise to the reader
to show that the equation
for image formation may be expressed in terms
of the object distance, u from
H(-) and the image distance, u ', from
H '(-) as