Chapter 6

Practical Optical Systems

The methods developed in the preceding chapters may now be applied to a number of common optical systems. In each case we shall compute the matrix [A]OO ' from which all the cardinal points are obtainable. We shall mainly use equations (23), (24), (28), and (30) which give the positions of the focal points and the focal lengths. Then Newton's equations (33) and (35) may be used for image formation, or alternatively equations (41) and (42) which make use of the principal points given by equations (37) and (38).

The Single Refracting Surface

For a single spherical surface separating two media of indices n and n ', the matrix [A]OO ' is simply the refraction matrix, namely,
$$\bmatrix{1 & 0 \cr (n'-n)/nR & n'/n}.$$

We can immediately recognize
$$f=nR/(n'-n), \eqno(61)$$ (61)

and
$$f'= (n'/n)f=n'R/(n'-n). \eqno(62)$$ (62)

The distances OF = nR/(n '- n) = f and OF '= (n '/n)nR/(n '- n) = n 'R/(n '- n) = f '. Since (38) gives F 'H ' = -f ' and (37) gives FH = -f, the principal points H and H ' are here coincident at the vertex O of the refracting surface.

Example 1

As an example, consider exercise 1 of chapter 3 in which an object is placed 24 cm to the left of a vertex O of a convex spherical surface separating media of index 1.00 and 1.50. The radius of curvature of the surface is 6 cm (see Fig. 5). Solution:
$$f'=(1.5)6/(.5)   =  18\ {\rm cm} $$

$$  f =(1/1.5)  (18)=  12\ {\rm cm}$$

OF = 12 cm OF ' = 18 cm
$$  x =  24 - 12 = 12\ {\rm   cm}  $$

$$x' =  (18)(12)/12 = 18\ {\rm cm}. $$

The image distance (from O to the image point) is 18 + 18 = 36 cm. In this example the image and object positions are in fact the negative principal points since m = -x '/f' = -18/18 = -1. The nodal point N is a distance f ' to the right of F and ON = f '- f = n 'R/(n '- n) - R/(n ' - n) = R = 6 cm. Of course N ' is coincident with N, as H ' is coincident with H This example is illustrated in Fig. 23.

Figure 23.

Example 2

We can do example 1 in chapter 5 by repeated application of the above equations for a single refracting surface. We had a thick biconvex lens, index 1.5 in air, 10 cm thick, with radii of 10 cm. An object was placed 40 cm in front of the first vertex O.

For the first refracting surface, f ' = (1.5)(10)/.5 = 30 cm, f = (1/1.5)(30) = 20 cm, and 30/s ' + 20/40 = 1 gives s' = 60 cm. This gives an image 50 cm to the right of the second vertex O ' so that the image distance for refraction at the second surface is s = -50 cm. For the second surface, f ' = (1)(-10)/(-.5) = 20 cm, f = (1.5) 20 = 30 cm, and 20/s ' + 30/(-50)= 1 gives s' = 12.5 cm in agreement with our previous result. And the magnification, m = -(60/40)(-12.5)/(-50)= -3/8 as before.

Exercise 1

Find the first and second focal points of the lens of example 2 in chapter 5 by repeated application of the equation for image formation by a single refracting surface. We might note that the plane surface is simply a special case in which R = infinity and the refraction matrix is
$$\bmatrix{1&0\cr 0&n'/n}.$$

The focal lengths are equal to infinity. It is easy to show that
$$n'/s' + n/s = 0  \eqno(63)$$ (63)

if s is the object distance measured from O and s ' is the image distance also measured from O. For example s ' represents the "apparent depth" of an object a distance s under water (index 4/3) say, such that s ' = (3/4) s.

Exercise 2

Derive equation (63).

The Thin Lens

Figure 24.

Possibly the most useful equations of geometrical optics are those for image formation by a thin lens. A thin lens of index µ, is shown in Fig. 24 with media of indices n and n ' to the left and right of O and O ' respectively. The thickness, t, of the lens is assumed to be small compared to the focal lengths of the lens such that we can set t = 0 in the translation matrix from O to O '. Thus

$$[A]_{OO'}=
\bmatrix{1 & 0 \cr (\mu-n)/nR_1 & \mu/n}
\bmatrix{1 & 0 \cr 0&1}
\bmatrix{1 & 0 \cr (n'-\mu)/\mu R_2 & n'/\mu}.$$

or
$$[A]_{OO'}=
\bmatrix{1 & 0 \cr (\mu-n)/nR_1 +(n'-\mu)/nR_2& n'/n}_{OO'},$$

(assuming O and O ' approximately coincident).

So,

$$1/f = (n'-\mu)/nR_2-(n-\mu)/nR_1$$

or
$$n/f =n'/f'= (n'-\mu)/R_2-(n-\mu)/R_1.\eqno(64)$$ (64)

Equation (64) reduces to the well-known lens-makers equation when n ' = n = 1.00, namely,
$$1/f =1/f'= (\mu-1)(1/R_1-1/R_2).\eqno(65)$$. (65)

Exercise 3

Show that the points H and H ' coincide at the assumed approximate coincidence of O and O ' which we take to be the lens centre of the thin lens.

Exercise 4

Find the positions of the nodal points of the thin lens.

Newton's equations apply here using (64) for f and f '. It is also convenient to use (41) since s and s ' are measured from the centre of the thin lens, that is, f '/s '+ f/s = 1 or n '/s ' + n/s = n/f = n '/f ' which for a thin lens in air becomes 1/s + 1/s '= 1/f =1/f ' = (µ - 1)(1/R1 - 1/R2). Also m = - ns '/n 's ( = -s '/s if n ' = n)

The matrix representing a thin lens has the particularly simple form
$$[A]_{OO'}=
\bmatrix{1 & 0 \cr 1/f& f'/f}\eqno(66)$$ (66)

Question

A zero in the upper right hand corner of the matrix means that the matrix relates an image plane and an object plane. What does the zero mean in Equation (66)?

Combination of Two Thin Lenses

The combination of two thin lenses in air forms the basic arrangement for the most common optical systems such as the microscope and astronomical refracting telescope. Also, most oculars are combinations of two lenses as corrections for some of the defects of images can be reduced in this way. The defects are classed in terms of five monochromatic aberrations arising as correction terms to the first order theory, plus two so-called chromatic aberrations arising from the dependence of the index of refraction on frequency.

Figure 25.

The combination of two thin lenses in air, a distance d apart, is shown in Fig. 25. We may take the vertices of the system as the lens centres O1 and O2. The matrix is then, using (66) with f1 = f1 ', and f2 = f2 '

$$A_{O_1O_2}=
\bmatrix{1 & 0 \cr 1/f_1 & 1}\bmatrix{1 & -d \cr 0 &1}
\bmatrix{1 & 0 \cr 1/f_2 & 1}$$
$$=\bmatrix{1 -d/f2& -d \cr 1/f_2+(f_2-d)/f_1 f_2
& 1-d/f_1}_{O_1 O_2}.$$

Therefore if f ( = f ') is the focal length of the combination, equation(28) gives,
$$1/f=1/f_1+1/f_2-d/f_1f_2\eqno(67)$$ (67)

The focal points are F and F ' such that
$$  O_1 F = f (1 - d/f_2) \eqno(68)$$ (68)

and
$$  O_2 F= f (1-d/f_1) \eqno(69)$$ (69)

using (23) and (24).

One simple special case of interest is that in which two thin lenses are cemented together to form a doublet. In this case d = 0 and 1/f = 1/f1 + 1/f2. Such a doublet is usually constructed to reduce longitudinal chromatic aberration. The lenses are made of different glasses which have different dispersive powers (i.e. the indices of the glasses depend differently on the frequency of the light) and so the combination may be designed so that the focal length f will have a minimum dependence on the frequency of the light (Jenkins and White, p. 157). It is also possible to reduce the dependence of the focal length, f, on the frequency when using two lenses of the same index of refraction by choosing a separation d = (f1 + f2)/2. This may be readily shown using (67) and (65) and solving for a minimum dependence of f on µ, the index of the glass of the lenses (Jenkins and White, p. 163).