On this page we'll analyze the behavior of a lens in the limit as it is
made very small.
As we've noted elsewhere (
here),
the
behavior of a
lens very close to its center is characterized entirely by the second
derivative of the equation which defines its surface. Call the
second derivative the
curvature. If we consider a part of
the lens which is so small that the curvature varies only
insignificantly over the region, then the behavior of the lens within
that region will be (essentially) identical to the behavior of any
other lens with the same curvature at the center.
Consequently, this analysis applies to
any lens, no matter what
the figure: Spherical, paraboloidal, or any other aspheric
surface. The only requirements are that the lens be small, and
that it be symmetric, with the same (convex) curve on both surfaces.
Throughout the following, we'll be using "
n" for the refractive
index of the lens material,
divided by the refractive index of
the surrounding medium. For a glass lens in air,
n may be
taken to be the refractive index of the glass, as air's refractive
index is almost exactly 1.
This is a lengthy page, so we shall include a table of contents:
I. OnAxis
Behavior of a Small
ConvexConvex Lens
As we have also noted
elsewhere,
a
single
surface lens cannot be ideal no matter what "figure" it has.
In this section
we will find that, in contrast, a tiny
symmetric lens, with two
convex surfaces, acts as an ideal lens for rays entering nearly
onaxis. Consequently, a small, thin lens will image a distant
scene the same way an ideal lens would.
Figure 11
Light entering "tiny" lens

To take advantage of the fact that the lens
is "small" and "thin" we
will do several things:
 We'll assume that a line perpendicular to the lens surface has
such a tiny angle with the lens axis that we can use the limits
of the sine, cosine, and tangent of that angle as the angle approaches
zero. That means we can use the angle itself in place of the sine
or tangent, and we can use a constant 1 in place of the cosine.
 We'll assume that the lens is so thin
that a ray passing through
the lens enters and leaves at the same distance from the
axis. Thus, the slope of the entry surface and the slope of the
exit surface have identical magnitude, but opposite signs (as the
curves of the top and bottom surfaces are mirror images of each other).
 We'll assume that the second derivative of the function defining
the surface  i.e., the curvature of the surface  is constant across
the entire lens.
Note that the third assumption  constancy of the curvature  is the
one which breaks down when using a "small" "thin" and "
cheap"
plastic magnifier for lens experiments! To achieve ideality with
a bad lens one would need to mask off all but a tiny pupil in the
center of the lens (and one would need to hope that the optical centers
of the front and back surfaces really were in the same place).
To take advantage of the fact that our scene is nearly onaxis, we will
be further assuming
 The angle between the
entering ray and the line perpendicular to
the lens surface is so small that we can use the limits of the
sine, cosine, and tangent of that angle as the angle approaches zero in
place of the actual values. That means we can substitute
the angle itself for the sine and tangent, and use a constant 1 in
place of the cosine.
We have shown light entering such a tiny lens in
figure
11. In the picture we've grossly exaggerated all the angles
in order to make it possible to see them clearly; we must imagine that
we're actually looking at a much smaller piece of the lens, and all the
paths go
almost straight up and down.
Throughout the following we'll be discarding terms of the second order
or higher in both θ and φ.
φ is the angle between a line normal to the lens surface and the lens
axis, and ξ
_{0} is the angle between the ray and the lens
axis. The angle of entry of the ray into the lens, θ
_{0},
is given by
11)
By
Snell's Law we have
12)
From figure 11, we also have
13)
So far everything we've stated is exactly correct. But now we're
going to take small angle approximations, and substitute θ
_{0}
for sin(θ
_{0}) and θ
_{1}
for sin(θ
_{1}). Substituting (12) and (11) into (13),
we'll obtain
14)
Now we make use of our assumption that the lens surface shape is fully
defined by its first two derivatives near the center of the lens, and
our assumption that its second derivative is constant in this
region. With this assumption, we have
15)
We now define the curvature,
C, to be
16)
which we have assumed is constant over the region we're
considering. With the assumption that we can substitute φ for
tan(φ) close to the lens center, we can rewrite (14) as
17)
This completes our analysis of the
entry of a ray into our tiny
lens. We'll now consider what happens when the ray
exits
the lens.
Figure 12:
Light exiting "tiny" lens

The situation isn't an exact mirror image, because the curvature flips
on the other side of the lens  the lens surface is bending up, not
down, at the point where we've shown it. So we need another image
(figure 12).
Note that ξ
_{1} and φ are the same angles in figure 12 as they
were in figure 11, and, by
assumption 2,
x
has the same value in the two
figures. The angle at which the light hits the interface in
figure 12, on the other hand, is not the same as the angle at which it
left the interface in figure 11, so we've called the angles the ray
makes with the line which is normal to the surface in figure 12 σ
_{1}
and σ
_{2} rather than θ
_{1} and θ
_{2}.
From figure 12 we have
18)
Using equation (17), Snell's law, and the small angle approximations
we can now write
19)
Applying (17) one more time gives us
110)
Finally, collecting terms, we arrive at
111)
To first order in a small angle the tangent equals the angle, so we can
rewrite that as
112) 
This has exactly the same form as the ideal lens equation which we
found
here.
Thus, we can conclude that, for scenes which lie close to the lens
axis, a tiny symmetric convexconvex lens behaves as an ideal lens.
But what is the focal length of this lens  is it what we found in our
analysis of spherical lenses,
eq. 31?
To answer this we need to find the curvature of a spherical lens
surface near the center of the lens. For that, we need the second
derivative of the curve which defines a circle. For a circle of
radius R sitting on the origin, we have
113)
Differentiating,
114)
Differentiating again,
115)
Evaluating at x=0 gives us
116)
This is the curvature of the spherical lens surface at the
origin. Plugging it into (112) in place of
C we obtain
117)
Comparing this with the ideal lens equation
here, we see that the
focal length of our tiny lens must be
118)

And, much to our pleased surprise, we find that this is
identical
to equation (
Sphericallenses:31)
for the
center focal length of two
gluedtogether spherical half lenses. This gives us additional
confidence that the result is correct, as the two formulas were found
by very different means.
II. OffAxis Behavior of a Small
ConvexConvex Lens
We will now redo the analysis of the previous section, but this time
we're going to retain exact values for the functions of the entry and
exit angles of the ray.
Throughout this section, we'll continue to use
assumptions
13 but we're going to drop
assumption 4.
Taylor Series for
Sine and Cosine
The Maclaurin series for both sine and cosine is used pretty frequently
but the full Taylor series is less commonly seen. For any arbitrary
fixed angle λ, and some offset, φ, we have
T1)
We'll be using these series truncated to just the first two terms,
which is accurate to
first order in φ, or to
"O(φ)". This is appropriate when
we know (or assume) that φ is very small:
T2)
We have used the rather peculiar locution "
" to explicitly call out the fact that the second formula
is only accurate to the order of φ. All terms in φ
^{2}
and higher powers of φ have been discarded.
For the rest of this page we'll use that locution only on the specific
steps where we drop terms that are higher order in φ.
Preliminaries:
Sines and cosines of the angles in the figures
We will now take it from the top once more, again using figures
11 and
12, and
discarding all second order and higher terms in angle φ.
From
figure 11, we have:
21)
From
figure 12,
we have:
22)
We now use equations (
T2) to find the sines and
cosines of these:
23)
24)
25)
26)
Finally, using Snell's law, we can connect the angles the ray makes
leaving each surface with the angles it makes as it approaches each
surface:
27)
28)
That takes care of the preliminaries. Now we'll set about finding
sin(ξ
_{2}) as a function of sin(ξ
_{0}), and then we'll
find the same formula in terms of the tangents.
Finding sin(ξ_{2})
as a function of sin(ξ_{0})
In this section we'll successively substitute expansions for the
sines and cosines of the various angles into equation (
26) until we
work our way back to a formula for sin(ξ
_{2}) in terms of ξ
_{0}.
Plugging equations
28 (Snell's Law) into
equation 26, we obtain an equation in terms of σ
_{1}:
29)
Substituting equation
25
into (29) brings us to a somewhat messy equation in terms of ξ
_{1}:
210)
Looking at just one term from (210), we can use equations (24) to
rewrite it in terms of θ
_{1}:
211)
Discarding the φ
^{2} term leaves us with:
212)
We also want the square of this expression. Squaring and again
discarding high order terms leads to:
213)
Substituting (212) and (213) into (210) leads to an equation for
sin(ξ
_{2}) in terms of θ
_{1}:
214)
This isn't as bad as it looks, because the term in φ inside the square
root is actually second order in φ, due to the factor of φ outside the
radical. It can therefore be discarded, leading to:
215)
We've made it back to the point where the ray has just entered the
lens. We'll now move back across the first surface, and out to
the point where the ray hasn't yet entered the lens, by applying
Snell's Law once more, in the form of equations (
27).
We also multiply out the terms with
n in them, leading to:
216)
Finally, we use equations (
23) to write (216)
in terms of ξ
_{0}. We also discard terms of order φ
^{2},
to obtain:
217)
Finally, collecting terms leads to the formula we've been looking for,
which expresses sin(ξ
_{2}) in terms of ξ
_{0}:
218)

Conversion from
sine to tangent in (218)
Figure 21:

Equation (218) describes the behavior of a tiny convex lens.
However, to compare it with the
ideal lens equation,
we need to rewrite it to express tan(ξ
_{2}) as a function of
tan(ξ
_{0}). Reinforcing our rather weak memory of the
trig identities by looking at the right triangle in figure 21, we
start by writing the ones we'll need:
219)
We now plug (
218) into (219a), which gives us
a formula for the tangent of ξ
_{2} in terms of the sine and
cosine of ξ
_{0}. It's a little ugly, unfortunately,
even with the second order terms dropped out:
220)
We're now going to express sin(ξ
_{0}) and cos(ξ
_{0}) in
(220) in terms of tan(ξ
_{0}), using (219) (b) and (c). We're
also multiplying the top and bottom by
to eliminate some of the mess. This looks pretty
horrible, but bear with me; we'll simplify it a lot when we drop out
the nonlinear terms:
221)
Combining terms in tan(ξ
_{0}) makes that look a little better:
222)
And now we recognize that the bottom of (222) is very close to the
square root of 1, since φ is assumed to be very small. So, to
first order, we can eliminate the square root and pull the denominator
up into the numerator, like this:
223)
This still looks pretty bad but it's about to implode, because most of
the mess is second order in φ. We multiply it out and drop all
second order and higher terms, and we are left with:
224)

This has the form we wanted.
Comparison
with the ideal lens equation
Equation (224) is in terms of φ, the angle between a line
perpendicular to the lens surface and the lens axis, but we need it in
terms of
x, the distance from the axis at which the ray enters
the lens. If
R is the radius of curvature of the lens
surface, then to first order in φ and
x, we have:
225)
which allows us to rewrite (224) in terms of
x:
226)
Comparing (
226) with the ideal lens equation,
we see that the focal length of our tiny lens
varies depending
on the angle at which the rays enter the lens, and we must have
227)

This is certainly not ideal, for the focal length
varies
depending on the incidence angle. However, we can see that
parallel rays at any angle are indeed brought to a point.
Nonetheless, the focal "plane" is not flat. It curves toward the
lens as we move away from the axis.
Curvature of
the Field of a Symmetric Lens
We'll find the shape of the focal surface of a halflens
later on this page.
That will actually be somewhat harder than finding the curvature of the
focal surface of a
symmetric lens, which we'll be doing in this
section.
Figure 22:

The curvature of the focal surface is determined by the second
derivative of the focus with respect to the distance from the center of
the lens. So, we need to find the focal length as a function of
x_{1},
the distance from the axis (see figure 22). From the figure, we
have:
2.21)
We're going to plug that into (
227) and then
differentiate twice and solve for
f''. However, it's
going to be awkward to separate
f and
x_{1}
after we substitute (2.21). To make our job easier, before we do
the substitution we're going to simplify (227) by reducing it to
second order in ξ
_{0}.
To
second order, tan(ξ)=ξ (the tangent function has an
inflection point at 0, so its
second derivative is zero at
zero, and its first derivative at zero is 1). So, what we're
actually going to do is reduce (227) to first order in tan
^{2}(ξ
_{0}).
We're going to start by addressing the two main factors in the
demoninator separately. The first is the sum of 1 and a small
value, and, to first order in tan
^{2}(ξ
_{0}), we can
invert it by flipping the sign on tan
^{2}(ξ
_{0}).
The second term contains the square root of a sum of a large value and
a small value; we'll start working on it by pulling
n^{2}
out of the square root. That brings us to:
2.22)
To first order in tan
^{2}(ξ
_{0})  which is second
order in ξ
_{0}  we can multiply the tan
^{2} term in
the square root by 1/2, and discard the radical, leading to
2.23)
By pulling
n1 out of the denominator, we put it in the form of
the sum of 1 and a small value. To first order in tan
^{2}(ξ
_{0})
we can flip the sign on the small value and move the denominator into
the numerator, leading to:
2.24)
We now multiply it out, and discard all terms higher than second order
in tan(ξ
_{0}). At the same time, we'll also simplify the
slightly ugly looking coefficient on the second tan
^{2}(ξ
_{0}),
by multiplying top and bottom by
n, factoring
n^{2}1
into (
n+1)(
n1), and cancelling the common factor of
n1.
This all leads to:
2.25)
Finally we multiply it out again, for once dropping no more terms (this
is just a rearrangement of 2.25), leading us to the form we were
aiming at, which is correct to second order in ξ
_{0}:
2.26)
Finally, we plug in equation (
2.21). To
separate
f and
x_{1}, we multiply through
by
f^{2}, and move all terms in
f to the left
hand side:
2.27)
The left side has a nonzero root at
R/2(
n1), which is
the focal length when
x_{1}=0; that's where an image
exactly onaxis would focus. We want the curvature at that
point. To find it, we'll differentiate (2.27) twice with respect
to
x_{1}. We differentiate the left side using
the chain rule, since it's a function of
f rather than
x_{1}.
After the first differentiation, we have:
2.28)
The coefficient on
f' is nonzero when
x_{1}=0
and
f=
R/2(
n1), so we must have
f'(0)=0.
That's what we expected, of course, since it means the center of the
focal surface is perpendicular to the lens axis.
Differentiating again,
2.29)
We're looking for
f''(0). Since
f'(0)=0, we know
that the first term in the sum on the left side is
zero at
x_{1}=0,
so we drop it out. We plug in
f(0)=
R/2(
n1),
sum up the terms in
R^{2}/(
n1)
^{2},
rearrange to isolate
f'', and we obtain
2.210)
or, multiplying everything out,
2.211)
The focal length at the center is
R/2(
n1), so we can
rewrite (2.211) in a form which may be more useful working with real
lenses, where the focal length is typically better known than the
radius of curvature:
2.211b)
Equation (
116), which was the second
derivative of the curve of a
circle, tells us that the radius
of curvature of the focal surface is the inverse of the second
derivative. So we have
2.212) 
Once again, the focal length at the center is
R/2(
n1),
so we can
rewrite (2.212) as
2.213) 
For typical glass with an index of refraction of about 1.5, the radius
of curvature of the focal surface is about 1/4 the focal length.
That's very far from flat!
NB  I find this result highly
surprising  the focal surface is far more curved than I
expected.
I've doublechecked the algebra in this
section (and this entire page), and I've done some very
crude experiments involving a flashlight, a long hall, and a pair of
+1.25 diopter reading glasses
which seemed to confirm a rather strong curvature of the focal
surface.
However, I
haven't yet verified equations (2.212) and (2.213) against any other
sources, and at this time I don't have the equipment here to test the
results quantitatively (though one of these days I may get some decent
lenses and try to do that).
So, treat these formulas with a little caution; it is quite possible
that they are not correct. 
III. What Goes Wrong?
We have seen that a tiny convex lens does not function as an ideal
lens. But what goes wrong? What assumption regarding "ideal
lens behavior" is violated? Recall that, in our analysis of a
symmetric
ideal lens, we did
not
assume the focal plane was flat. Rather, we
proved
it, based on our far simpler assumptions. So, which of those
assumptions doesn't hold here? We assumed:
 Rays passing through the front focus exit the lens on paths
parallel to the axis. (For a symmetric lens, 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.
 The lens produces an image of a scene at infinity. In other
words, parallel rays entering at any angle are brought to a
focus (somewhere). We didn't assume anything about where
the focus of a particular bundle of parallel rays must lie; merely that
it had one.
Assumption (2) is true trivially. Assumption (3) certainly seems
to be true also. So, the breakdown must be in assumption (1).
If we look at (
226) again, we realize that,
indeed, assumption (1) is false. For rays which pass through the
front focus before entering the lens
must be bent according to
the
ideal lens formula:
31)
And that's not what happens here, according to (226)  it only
matches the ideal lens formula for rays entering nearly parallel to the
lens axis.
Note, however, that assumption (1) can't be tested with a tiny
lens!
With a sufficiently small lens, all rays which pass through the front
focus and enter the lens are traveling nearly parallel to the axis, and
for those "focal" rays, (226) will match (31) as precisely as we
like.
What this seems to show, however, is that we
cannot make a large
lens which sharply focuses parallel rays from any direction.
The central area of the lens will have a
curved focal surface,
which is formed according to equation (
227).
For the outer
regions of the lens to share the same focal surface as the central
region, they would need to deflect light according to equation
(
226). However, that results in a
violation of assumption (1)
about ideal lenses, and that, in turn, means parallel onaxis rays will
not all be brought to a common focus.
This is a remarkable result, because it applies to a simple lens with
any
figure whatsoever, as long as it's symmetric, and made from a
material with a fixed refractive index.
There is, however, an issue here which prevents this from being an
airtight proof regarding all simple lenses: We assumed that our
tiny lens is not only of small diameter, but is also
thin.
In fact, when one enlarges a convex lens, the center of the lens
naturally becomes
thicker  and as the center becomes thicker,
our
second assumption becomes less and
less accurate. In particular, for rays at a large angle to the
axis, the thickness of the center of a large lens becomes very
significant, and our model is no longer applicable, because it is no
longer the case that φ is the same angle at entry and exit of the
ray. Consequently, for a lens with a
thick center, we
cannot conclude that the focal surface will be described by (227) 
it is, therefore, conceivable that one could build a tiny but
thick
lens which had a flat focal surface. None the less, we can still
conclude that any symmetric
Fresnel lens must be nonideal!
Whether we could in fact generalize the proof to cover all thick
lenses, or complex ensembles of lenses, or lenses
with graded refractive indices, is (far) beyond the scope of this page.
IV. A Small Convex HalfLens  A Lens
with One Surface
As we've already noted, both
above and on our
spherical
lenses page, a lens with just one surface does not behave as an
ideal lens. But that begs the question, does such a lens form an
image, and if so, what sort of image? This is particularly
interesting since
eyes use singlesurface lenses  the light
passes from air, with a refractive index of 1.000..., into the front of
the eye, with a refractive index of about 1.3, and never returns to the
lowindex air again. The front surface of the cornea forms a
singlesurface lens.
The answer, as we shall see, is that onaxis, a halflens acts like an
ideal lens with
magnification. Offaxis it turns out to
have a curved focal surface, just as we found for a symmetric lens,
above.
We'll once again be using
figure 11, and
we'll proceed just as we did in part II: We'll find the
relationships between angles which we'll be needing from the figure,
we'll use first order Taylor expansions of the sums of sines and
cosines to take advantage of the fact that φ is very small, we'll use
Snell's Law to connect the entry and exit angles, and then we'll
combine the expressions for the angles we've found to obtain sin(ξ
_{1})
as a function of sin(ξ
_{0}). Finally we'll rewrite the
final equation in terms of tangents, which will give us an equation we
can use to determine how the lens behaves.
In fact, most of the analysis which we need was already done in
Part II, and we won't repeat it here. The
relations between angles we need are given in equations (
21). The relations between sines and
cosines we need, using the small angle approximations, are given in (
23) and (
24). The
relation between the sines and cosines of θ
_{0} and θ
_{1}
which are given by Snell's Law are in equations (
27).
Our starting point for this analysis will therefore be the expression
for sin(ξ
_{1}) from equations (
24).
Our first step is to use Snell's Law, by substituting equations (
27) into (24) to eliminate θ
_{1}:
41)
From equation (
23a) we note that, to first
order in φ, we have:
42)
We substitute (
23) and (42) into (41) to put
it in terms of ξ
_{0}:
43)
Finally we collect terms and drop everything which is second order in
φ, to obtain:
44)
Now we use (
219a) to write tan(ξ
_{1})
in terms of (44):
45)
This one doesn't look like much fun, but before we go any farther we'll
simplify it a bit. We multiply out the square term in the
denominator and drop the nonlinear terms in φ to obtain:
46)
Finally we'll use (
219) (b) and (c) to rewrite
(46) in terms of tan(ξ
_{0}). At the same time we'll
multiply the top and bottom of the fraction by
to clean up some of the mess. This brings us to:
47)
What a mess, eh? But most of it's going away. We'll start
by multiplying by
n/
n to clear some of the fractions,
and we'll combine some terms, to obtain:
48)
To reduce the noise level a little so we can see what's going on here
we'll substitute:
49)
We can now rewrite (48) a little more simply:
410)
With the reduced noise level we realize that the denominator is the
square root of the sum of a large value and a small value. By
dividing through by μ/μ we put it into a form that lets us eliminate
the square root in the denominator and pull the denominator up into the
numerator, as:
411)
We now multiply out (411) and discard higher order terms, leaving us
with:
412)
Unfortunately we still need to undo our substitution of (49).
Expanding μ again and multiplying out we get
413)
This is not very pretty, and we suspect that it may not be
correct. As a quick check, we're going to discard all but the
first order terms in tan(ξ
_{0}). In one step, this
provides us with the
onaxis behavior of the lens, and it also
lets us compare the result with equation (44). To first order in
ξ
_{0}, tangent and sine are equal, and (413) and (44) should
be identical to first order. We have:
414)
This does, in fact, match (44) to first order in ξ
_{0}.
With that bit of evidence, plus some effort checking our algebra, we
tentatively conclude that (413) is correct.
With the assumption that (413) is indeed correct, we will now look at
the onaxis and offaxis behavior of our tiny halflens. First,
however, we need to say something about
ideal scaling lenses.
Ideal Scaling Lenses
Figure 41:
Ideal Scaling Lens

"Scaling lenses" is what I'm calling the objects discussed in this
section. I have no idea if there's a commonly accepted name for
them.
The ideal lens formula, which we derived
here, is
Suppose that we write a slightly different formula, which we'll call
the
ideal scaling lens formula:
4.11)

This certainly is not an "ideal lens", since it violates
assumption 2 for ideal
lenses. We'll now determine if it produces an image of a scene at
infinity (
assumption 3),
and if so, where the image is, how large it is, and what shape the
lens's focal surface has.
In
figure 41 we've shown a scaling lens
with three parallel rays entering it. Ray 1, which passes through
the center of the lens, and Ray 2, which leaves the lens traveling
parallel to the lens axis, will certainly intersect. If the lens
is to form an image, then Ray 0 must intersect rays 1 and 2 at the
point where they cross. We'll now determine whether it does
so. We'll also determine how long the line I've labeled as "
f_{0}"
actually is; that's the effective focal length of this lens.
The distance from the center at which Ray 2 must hit the lens in order
to emerge parallel to the axis is
x_{1}. For Ray
2 the tangent of the angle at which it emerges is 0, so we must have
4.12)
We've taken the absolute value in (4.12) because we've shown
x_{1}
as a
length in figure 41 rather than a coordinate value.
Since ray 1 enters the lens at the center, with x=0, we can find the
tangent of
θ_{2}:
4.13)
We can now find
f_{0}:
4.14)
which shows that the focal length is not changed by adding a scale
factor. But does the lens actually focus? Does ray 0 arrive
at the common intersection? It does so only if:
4.15)
If we just write out
x_{1}+
x_{0} we see
that:
4.16)
which matches (4.11). So ray 0 does arrive at the intersection
of ray 1 and ray 2, at distance
f behind the lens; since
x_{0}
and
θ_{0} were arbitrary this shows that all parallel
rays will arrive at one
point, and that the point lies in the focal plane, which is indeed flat.
The difference between our "scaling" lens and an ordinary ideal lens is
the factor of
m in (
4.12).
That tells us that the image
size is scaled by a factor of
m
compared with an ideal lens; hence our name "ideal scaling lens".
Finally, note that the ideal scaling lens is
not
symmetric: The back focus is
f units behind the lens, but
the distance from the front focus to the lens is:
4.17)
And so we see that the front focal length is also scaled by
m.
OnAxis
Behavior of a Small HalfLens
The onaxis behavior of a small halflens is given by equation (
414). We'll rewrite that in terms of the
distance from the axis,
x, and the radius of curvature,
R:
4.21)
Comparing (4.21) with (
4.11) we see that the
focal length of our halflens is given by
4.22)
We're pleased to note that formula
matches the formula we found
for the "central" focal length of a spherical halflens, which was
equation 210 on our
spherical lens page.
The
magnification of the image, compared with an ideal lens of
the same focal length, is given by
4.23)
Of course the focal plane is flat  but this is a first order
approximation to the lens's behavior, and to first order
everything
is flat. In the next section we'll learn more about the "real"
shape of the focal surface.
OffAxis
Behavior of a Small HalfLens
Let's start by rewriting (
412) in terms of
x,
the distance from the axis. We'll also clean up some rather
confusing signs at the same time; in (412) some terms which were
subtracted were actually negative:
4.31)
where μ is as defined in equation (
49); we
repeat the definition here:
The first thing to notice is that μ is a function only of ξ
_{0}
and the refractive index of the glass. Consequently, for fixed ξ
_{0},
μ is fixed. So, for any particular
fixed incidence angle,
(4.31) has the form of the ideal scaling lens formula, (
4.11). Consequently, when parallel rays
arrive at the lens,
μ will be the same for all of them, and
for those rays the lens
will behave as an ideal scaling lens; we can therefore conclude that
parallel rays will be brought to a single focal point. So, the
lens
does form an image of a distant scene.
μ is an increasing function of ξ
_{0}. So, we can see from
(4.31)
that the magnification
decreases as we move away from the
axis. Parts of the scene far offaxis will therefore be imaged at
smaller scale than parts onaxis, with the magnification going as
1/tan(ξ
_{0}) for points far from the axis. The extent of
the focal surface will be the integral of the magnification times tan(ξ
_{0}),
integrated over the angle from 0 to π/2. In fact that appears to
be finite, indicating that the entire focal surface will be
bounded.
What of the shape of the focal surface? The focal length for a
particular incidence angle is given by
4.32)
By inspection, near the axis this is a decreasing function of ξ
_{0}.
In other words, the focal plane curves toward the lens for parts of the
scene lying offaxis. For parts of the scene which are far off
axis the higher order terms in the denominator express themselves, and
the behavior is less clear. Let's look at the limit as tan(ξ
_{0})
becomes large. First, we have:
4.33)
Plugging that into (4.32) and again discarding all but the largest
terms we obtain
4.34)
which can be simplified quite a bit to obtain
4.35)
or, simpler yet,
4.36)
This says that for points far off axis, the focal surface moves to a
fixed distance from the lens. For a very small refractive index
(n≈1) the focal surface actually approaches the lens surface; for
larger refractive indices it's farther from the lens surface. In
the limit for a large refractive index the focus is fixed at
R,
as we would expect: With a large enough refractive index, all
rays entering the lens follow a path that's (nearly) perpendicular to
the surface, and they all arrive at a single point, which is the center
of the sphere defining the lens surface.
So,
for finite
n, the focal surface appears to be a nearly flat
sheet with a bulge in the
middle  but keep in mind that, due the decreasing magnification as we
move away from the axis, the focal surface is actually bounded, and
doesn't really extend off to infinity as a flat sheet.
Second Order
Shape of the Focal Surface
We'd like to know what the curve of the focal surface looks like close
to the axis. This is not simply due to an abstract itch to solve
messy equations; rather, it's because, as previously noted, an eye uses
a halflens  and the retina is certainly not
flat! I'd
like to know if the focal surface of a halflens is a natural match for
the spherical curve of the retina.
For this, we need to find the function for the focal surface to second
order (but no higher).
Unfortunately this is a little tricky, due to the variable
magnification; we need to determine that to second order also in order
to find the focal length for a given angle as a function of
distance
of the image point from the axis.
As we'll be doing everything to second order, the first thing we need
are the trig functions to second order. Since the sine and
tangent both have inflection points at 0, they both also have a zero
second derivative there; consequently they're the same to second order
as to first order. The cosine, on the other hand, is concave down
at 0, and is certainly not constant to second order. We have:
4.41)
To second order in ξ
_{0}, we can rewrite
μ
as:
4.42)
To second order, the focal length for a given incidence angle, equation
(
4.32), is:
4.43)
We now multiply the top and bottom by
which, evaluating only to second order in ξ
_{0}, produces this:
4.44)
In case you're wondering whether I really just work this stuff out as I
type up the page, as my conversational style might seem to imply, the
answer is "sometimes", and in this case that's exactly what I've been
doing. I thought (4.44) was going to be a lot simpler. It
looks pretty raunchy, but we'll see if we can eliminate a few terms and
get something we can deal with more easily.
Multiplying out a few things, collecting our ξ
_{0}^{2}
terms, throwing away higher powers, and dividing through by
n^{2}/n^{2},
we cut the noise level a bit and get something we can almost read:
4.45)
We'll now divide the bottom by
n^{2}n, and combine a
couple more terms in ξ
_{0}^{2}, which brings us to
4.46)
Since we're only working to first order in ξ
_{0}^{2},
we're now in a position to pull the denominator up into the numerator,
multiply it out, and once again discard higher order terms, leading us
to:
4.47)
We're almost done
(I hope). Looking back at figure (
41), to obtain the curvature of the focal
surface, we need to find
f as a function of
x_{1}.
To find
x_{1}, we need to find ξ
_{1}, which is
the angle followed a ray of light entering the
center of the
lens at angle ξ
_{0}. Of course, x
_{0}=0 at the
center of the lens, so, from equation (
4.31)
the items we need are given by
4.48)
To plug into (4.47) we actually need these turned around, as:
4.49)
Expanding μ in the second equation in (4.49), and retaining everything
only to second order in ξ
_{0}, we have
4.410)
This isn't exactly what we had in mind. However, since we're only
taking everything to second order in ξ
_{0}, we can invert the
expression in ξ
_{0} which is on the right just by flipping its
sign, after which we move it to the other side and obtain
4.411)
But, again, we're only interested in second order and lower terms in ξ
_{0},
so we can drop the
third order term we see here and finally get
to something reasonable:
4.412)
We plug this into (
4.47), multiply through by
f^{2}, collect terms in
f, and we obtain:
4.413)
We factor the polynomial in
f on the left, and combine some
terms on the right, which leaves us with:
4.414)
Before we continue on, let's look at this equation when
x_{1}=0.
The polynomial in
f has two solutions in that case, but only
the nonzero solution makes sense here. So, according to (4.414),
the focal length for onaxis images (for which x
_{1}≈0) must be
4.415)
This
matches equation (
4.22) for the
onaxis focal length of a halflens. This is good.
To find the curvature we need the second derivative of
f with
respect to
x_{1}. We only have
f defined
implicitly in (4.414); however, that's good enough. Let's
differentiate both sides of (4.414). The left side is a function
of
f so we need to use the chain rule; we get:
4.416)
By inspection we can see that, when f=Rn/(n1) as in (4.415), the
first factor on the left is
nonzero. Consequently, at x
_{1}=0,
we must have
f' = 0. In other words, the center of the
focal plane is perpendicular to the lens axis  as we certainly
expected!
Now, for the curvature, we differentiate 4.416, once again using the
chain rule on the left side, to obtain:
4.417)
We're only interested in this at x
_{1}=0. At that point,
we also have
f'=0, so the first term in the sum on the left
drops
out. We know from (4.415) that at x
_{1}=0,
f=Rn/(n1);
if we plug that in, and divide through by the coefficient on
f'',
we obtain what we wanted. This is liable to get messy, however,
so I'm going to do it in two steps. First, we substitute in
(4.415) to obtain:
4.418)
And now we substitute that into 4.417, dividing through to get
f''
by itself on the left:
4.419) 
And that, believe it or not, is what we were looking for: It's
the curvature of the focal surface at the center.
n
varies from 1 to infinity; for all values in that range, this is
negative
 but we've defined
f as increasing as we move away from the
lens, so this is correct: the surface curves up toward the lens.
It is certainly
not flat.
From (
116) we can find its radius of
curvature, which is the inverse of the second derivative of the
function defining the surface:
4.420)

The minus sign results from the fact that the surface is concave up; in
our rather upside down coordinates, concave down would get a positive
sign.
The Focal Surface of an
Eye
As one final little thing, without doing any real analysis on it, let's
look at the curvature of the focal surface of an eye.
Most of the focusing of an eye is done by the surface of the
cornea. For our purposes here we'll just assume that the cornea
has the same radius of curvature as the eye itself  this is false but
will get us into the right ballpark, or so we hope.
The index of refraction of the contents of an eye is very roughly
similar to that of water. For our calculations here, we'll assume
it's
1.35.
We'll totally ignore the contribution by the crystalline lens.
A human eye is about an inch in diameter, which gives it a radius of
curvature of about 12 mm.
Putting that all into equation (
4.420), the
radius of curvature of the focal surface of an eye should be about
4.51)
Since the radius of curvature of the eyeball itself is about 12 mm, a
curvature of 10mm for the focal surface is an a very good fit!
Given our rather casual choice of dimensions, this is a startlingly
close match.
Page
created on 01/18/2009. Updated on 1/21/2009, with addition of
discussion ideal scaling lenses and halflenses. Major
corrections, and completion of halflens section, 1/27/2009.
Addition of curvature of focal surface of a symmetric lens, 2/1/2009.