Chapter 5

Image Formation

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, alpha, at the plane through Z goes through the plane at Z ' with coordinates y ', alpha '. This can be expressed as

$$\bmatrix{y \cr \alpha}_Z=\bmatrix{1 & -z \cr 0 & 1}
\bmatrix{0 & -f' \cr 1/f & 0}_{FF'}\bmatrix{1 & -z' \cr 0 & 1}
\bmatrix{y' \cr \alpha'}_{Z'}$$

or
$$\bmatrix{y \cr \alpha}_P=
\bmatrix{-z/f & z'z/f-f' \cr 1/f & -z'/f}_{ZZ'}
\bmatrix{y' \cr \alpha'}_{Z'}.\eqno(36)$$ (36)

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 ') alpha '. Thus we must have zz '/f - f ' = 0 and -z/f = 1 for the positive principal points H and H '.

Then
$$  z = F H =-f, \eqno(37)$$ (37)

$$  z'= FH'=  - f' , \eqno(38) $$ (38)

and
$$\bmatrix{y \cr \alpha}_H=
\bmatrix{1 & 0  \cr 1/f & -f'/f}_{HH'}
\bmatrix{y' \cr \alpha'}_{H'}.\eqno(39)$$ (39)

The first and second focal lengths are generally defined as the distances FH and F 'H ' respectively. We see then from equations (37) and (38) 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.

Figure 17.

The principal point, H ', is a distance f ' to the left of F ', the principal point H a distance f to the right of F 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, H ' may lie to the left of H.

Figure 18.

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 ', alpha ' at the image is related to the ray y, alpha, from the object by

$$\bmatrix{y \cr \alpha}_s=
\bmatrix{1 & -s \cr 0 & 1}\bmatrix{1 & 0 \cr 1/f & f'/f}_{HH'}
\bmatrix{1 & -s' \cr 0 & 1}\bmatrix{y' \cr \alpha'}_{s'}$$

or
$$\bmatrix{y \cr \alpha}_s=
\bmatrix{1-s/f & -s'+ss'/f-sf'/-f \cr 1/f & (-s'+f')/f}_{}
\bmatrix{y' \cr \alpha'}_{s'}.\eqno(40)$$ (40)

As stated before in deriving Newton's equations, the upper right hand element of the matrix must be zero if y ' is an image of y, that is, if the rays through the image plane at height y ' are to be independent of alpha '. Thus we have
$$f'/s'+f/s=1\eqno(41)$$ (41)

which makes the upper right hand element zero and gives y = ((f - s)/f)y ' or the magnification, m = y '/y = f/(f -s). From equation (41), (f - s) = -sf '/s ' and so
$$m=-fs'/f's=-ns'/n's,\eqno(42)$$ (42)

and since the determinant equals f '/f equation (40) becomes
$$\bmatrix{y \cr \alpha}_s=
\bmatrix{1/m & 0 \cr 1/f & mf'/f}_{}
\bmatrix{y' \cr \alpha'}_{s'}.\eqno(43)$$ (43)


Example 3

Use (41) and (42) to compute the image positions of the object in example 2.

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 and

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.
Also m = -(1.0/1.30) X (6.627/9.633)= -0.529 as before.


If the medium has the same index on either side of the optical system, f = f ' and these equations take the very familiar forms,
$$1/s'+1/s=1/f,\eqno(44)$$ (44)

and
$$m=-s'/s.\eqno(45)$$ (45)

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
$$ z'= F'H'(-) = f', \eqno(46)$$ (46)

and
$$  z  F H (-)  = f ,\eqno(47) $$ (47)

and
$$\bmatrix{y \cr \alpha}_{H(-)}=
\bmatrix{-1 & 0 \cr 1/f & -f'/f}_{H(-)H'(-)}
\bmatrix{y' \cr \alpha'}_{H'(-)}.\eqno(48)$$ (48)

Figure 19.

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
$$f/u+f'/u'=1.\eqno(49)$$ (49)

and
$$  m = f u'/f'u . \eqno(50)$$ (50)