In Fig. 14 is illustrated the formation of an image point
at y' of an object point at y by an optical system
between O and O '. The plane PP represents the plane through
object point, a distance x to the left of O; QQ
is the plane through the image point, a distance
to the right of O '. The focal points F and F ' can be
discussed in the previous chapter and the matrix relating a ray
passing from the first focal plane to the second is given by
The planes PP and QQ can be related by using this matrix and two
further translation matrices. Thus a ray y,
, at PP emerges as the ray
', at QQ such that
Since all the rays emerging from y arrive
at y ' regardless of the angle
the upper right hand element of the matrix of (32) must be 0,
in which case
It is clear from Fig. 14 that an object of height y is imaged
as an image of height y '. The lateral magnification,
is defined as the ratio y '/y
which from equation (34) is given by
Equations (33) and (35) are Newton's equations for image formation
and apply generally to any optical system.
An object 2 cm high is placed 40 cm to the left of the vertex O of the
thick lens of the
example of chapter 4. Find the image position and
Solution: Since F is 8 cm to the left of O, x = 32 cm
32 x ' = (12) (12)
x ' = 4.5 cm. (i.e. 4.5 cm to the right of F ').
Or the image is 8 + 4.5 =12.5 cm to the right of O '. The
magnification, m, is given by m = -12/32 = -3/8
and so the image is inverted and is 2 x 3/8 = 3/4 cm
In summary, we are now able to find corresponding image and object
positions for any optical system by:
Determining the product of the refraction and translation
the first vertex O and the last vertex O ',
the product matrix being
designated as [A]OO '.
Obtaining from [A]OO '
the position of the focal points
relative to the vertices, (the distances OF and
O 'F ' from
and (24)) as well
as the focal lengths f and f ' from equations (28) and (30).
Using Newton's equation (33) to compute the positions of the image and
object relative to the focal points and equation (35) to compute the
We shall use the following example to illustrate further these steps and
later use the example to illustrate other cardinal points which are
discussed in the following sections.
The thick lens of Fig. 15 has radii of curvature equal to 1.5 cm
and OO ' is 2 cm. The index of the lens is 1.6, the index to
of O is 1.00 and the index to the right of O ' is 1.30. An
is placed 10.467 cm to the left of the vertex O. Where is the
what is the magnification?
f = 1/.3 = 3.33 cm (from equation 28).
f ' = 1.3 x f = 4.333 cm (from equation 30).
O 'F ' = 0.65/0.3 = 2.167 cm (from equation 23).
OF = 1.25/0.3 = 4.167 cm (from equation 24).
(See Fig. 16. Also illustrated in the Figure are the principal points, H,
which will be discussed in the next section)
The object is placed 10.467 cm
to the left of O so that
x = 10.467 - 4.167,
or x = 6.3 cm.
x '= ff '/x=(4.333)x(3.333)/6.3 = 2.293 cm
or the image is at a distance
2.293 + 2.167 = 4.460 cm
to the right of O '.
m = -f/x =
-3.333/6.3 = -0.529 (from equation (35))
m = -x '/f ' = -2.293/4.333=-0.529