We already discussed the basic properties of ideal lenses in
Ideal Lenses and Mirrors.
However, all we learned there was what happens to rays which are
entering the lens parallel to the axis, or are traveling on a path
which passes through the front or back focus of the lens. On this
page we'll determine the path followed by any ray passing through the
lens, and along the way we'll show that an ideal lens has a
flat focal plane. (In our
earlier work we
assumed there was a flat focal plane; we'll be
using much weaker assumptions on this page.)
We will proceed by assuming that an ideal positive lens produces an
image (of some sort) of a scene at
infinity. In particular, this implies that parallel rays must be
brought to a focus at a single point, no matter what angle they are
traveling at relative to the lens axis. We don't need to assume
anything beyond that about the details of the image produced.
Formally, we'll assume the following:
 The lens has a "front focus", which is
a particular point located somewhere on the axis of the lens.
Rays which enter the lens after passing through the front focus exit
the lens on paths
parallel to the axis. (For a symmetric lens, due to the
reversibility of light paths, this is
equivalent to assuming the lens brings rays entering parallel to the
axis to a common focus.)
 Rays passing through the center of the
lens are not deflected.
 Parallel rays entering at any
angle are brought to a common
focus (somewhere). In other words, any collection of parallel
rays which enter the lens will be bent such that their paths will all
intersect at a single point, somewhere behind the lens. We won't
assume anything about where
the "focus" of a particular bundle of parallel rays must lie; merely
that
it must have one.
Ideal Positive Lenses
In our initial discussion we'll be considering a cross section of the
lens, and we'll assume that all rays lie in the plane of the cross
section.
Figure 1
 A ray passing through an ideal lens:

In figure 1 we show an ideal lens which lies on the
x
axis. The center of the lens is on the
y
axis, and the "front" focus is shown with a blue dot, above the lens on
the
y
axis.
We show a red ray, "ray 1", passing through the lens. It hits the
lens with angle
θ_{0} from a line perpendicular to the
surface of the lens, and after passing through it emerges with angle
θ_{1}
from the line perpendicular to the lens. The point at which the
ray enters the lens is at distance
x_{0} from the center of the
lens. We want to find
θ_{1} as a function of
θ_{0}
and
x_{0}.
We know how rays which pass through the center of the lens are
deflected (they just go straight), and we know how rays which pass
through the front focus are deflected (they emerge from the lens
traveling parallel to the axis of the lens). So, we've drawn rays
2 and 3 parallel to ray 1, with ray 2 passing through the center of the
lens, and ray 3 passing through the front focus. We also know
that rays which enter the lens traveling parallel to each other will
all converge to a single point, so we know that ray 1 must be bent such
that it arrives at the intersection of rays 2 and 3, and that is how
we've shown it in figure 1.
The distance from the lens to the point at which the rays focus
is
f_{2}.
Comparing Triangle T1 with Triangle T2, we can see that
1)
Consequently the focal point of the parallel rays is at distance
f
from the lens, regardless of the angle at which the rays enter the
lens. So, the focal plane of an ideal lens is, indeed, a
plane.
From Triangle T1, we can also see that
2)
And from equation (1) and Triangle T3 we can see that
3)
from which we obtain the result we wanted:
4) 
This fully characterizes an ideal positive lens (for rays which are
coplanar with the axis).
The Z Axis
We have said nothing about the
z
axis. We showed a cross section of an ideal lens, and assumed
that all
the rays we were discussing lie in a single plane, which contains the
axis of the lens. What about rays which are skew to the axis?
We won't prove this rigorously (at least, not at this time) but skew
rays can be understood by drawing a line from the center of the lens to
the point where the ray enters the lens. That line, along with
the
lens's axis, define a plane; call that the
axis plane for the ray. That
plane contains the lens axis, along with a cross section of the lens.
Project the ray into the axis plane. Now the picture in that
plane
looks exactly like what we've been discussing, and in fact the behavior
of the ray, projected into that plane, goes according to equation (
4).
Now, what about the leftover component of the ray? The ray also
had a
component of motion which was perpendicular to the axis plane.
What
happens to it? Answer: Nothing. Imagine another
plane, perpendicular
to the axis plane, which contains the point where the ray enters the
lens. Call it the "tangential plane" (it's tangential to a circle
drawn on the lens which is centered on the origin). If we project
the
ray into the tangential plane, we find that its projection in that
plane does not bend  it is not deflected in that plane by passing
through the lens.
With this final argument, we really have fully characterized an ideal
positive
lens.
While this last section really is
straightforward enough, it would benefit a great deal from some
pictures and some clearer reasoning. Unfortunately the 3d
pictures
required would take substantial time to draw, and the reasoning, while
not arcane,
is hard to state clearly. I may eventually add some images and a
clearer argument.
Ideal Negative Lenses
Not done yet.
Negative lenses also follow formula (
4).
However, I have not shown that on this page, and for the time being I'm
not going to  if I find the time I may come back and add another
image and some more trig to complete the proof. (In fact the
formulas are essentially identical with the positive case; only the
diagram needs to be redrawn.)
Page
created on 01/14/2009. Updated, with clarified assumptions, 20
Jan 2009