Chapter 5

Image Formation

Newton's Equations
Figure 14.

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 the object point, a distance x to the left of O; QQ is the plane through the image point, a distance x ' to the right of O '. The focal points F and F ' can be determined as discussed in the previous chapter and the matrix relating a ray passing from the first focal plane to the second is given by

$$\bmatrix{0 & -f' \cr 1/f & 0}_{FF'}.$$

The planes PP and QQ can be related by using this matrix and two further translation matrices. Thus a ray y, alpha, at PP emerges as the ray y ', alpha ', at QQ such that
$$\bmatrix{y \cr \alpha}_P= \bmatrix{1 & -x \cr 0 & 1}_{PF}\bmatrix{0 & -f' \cr 1/f & 0}_{FF'} \bmatrix{1 & -x' \cr 0 & 1}_{F'Q} \bmatrix{y' \cr \alpha'}_Q\eqno(31)$$ (31)

or
$$\bmatrix{y \cr \alpha}_P= \bmatrix{-x/f & x'x/f-f' \cr 1/f & -x'/f}_{PQ} \bmatrix{y' \cr \alpha'}_Q.\eqno(32)$$ (32)

Since all the rays emerging from y arrive at y ' regardless of the angle alpha, the upper right hand element of the matrix of (32) must be 0, in which case
$$ x x' = f f' \eqno(33)$$ (33)

and
$$ y = -(x/f)y'{\rm(independent\ of} \alpha) \eqno(34)$$ (34)

It is clear from Fig. 14 that an object of height y is imaged as an image of height y '. The lateral magnification, m, is defined as the ratio y '/y which from equation (34) is given by
$$ m = y'/y = -f /x = -x' /f' . \eqno(35) $$ (35)

Equations (33) and (35) are Newton's equations for image formation and apply generally to any optical system.

Example 1

 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 image size.

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 size.

In summary, we are now able to find corresponding image and object positions for any optical system by:
  1. Determining the product of the refraction and translation matrices between the first vertex O and the last vertex O ', the product matrix being designated as [A]OO '.
  2. Obtaining from [A]OO ' the position of the focal points relative to the vertices, (the distances OF and O 'F ' from equations (23) and (24)) as well as the focal lengths f and f ' from equations (28) and (30).
  3. 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 magnification.

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.

Example 2
Figure 15.

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 the left of O is 1.00 and the index to the right of O ' is 1.30. An object is placed 10.467 cm to the left of the vertex O. Where is the image and what is the magnification?

Solution:


  1. $$[A]_{OO'}=\bmatrix{1 & 0 \cr 0.4 & 1.6} \bmatrix{1 & -2 \cr 0 & 1}\bmatrix{1 & 0 \cr .125 & .8125},$$

    or
    $$[A]_{OO'}=\bmatrix{1.25 & -1.625 \cr 0.3& 0.65}$$
  2. 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, H ', which will be discussed in the next section)
  3. 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))
    or
    m = -x '/f ' = -2.293/4.333=-0.529
Figure 16.