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,
We can immediately recognize
The distances OF = nR/(n '- n) =
and OF '= (n '/n)nR/(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
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).
OF = 12 cm OF ' = 18 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.
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,
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.
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
We might note that the plane surface is simply a special case in which
and the refraction matrix is
The focal lengths are equal to .
It is easy to show
if s is the object distance measured from O and
s ' is the image
measured from O. For example
s ' represents the "apparent depth"
of an object
a distance s under water (index 4/3) say,
s ' = (3/4) s.
Derive equation (63).
The Thin Lens
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
(assuming O and O ' approximately coincident).
Equation (64) reduces to the well-known lens-makers
n ' = n = 1.00, namely,
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.
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
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 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
the most common optical systems such as the microscope and
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.
The combination of two thin lenses in air, a distance d apart,
in Fig. 25. We may take the vertices of the
system as the lens centres O1 and
The matrix is then, using (66) with f1 =
f1 ', and
f2 = f2 '
Therefore if f ( = f ')
is the focal length of the combination, equation(28)
The focal points are F and F ' such that
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
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).