Path:  physics insights > basics > simple optics > Tiny Convex Lenses

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:

Page Contents I. On-Axis Behavior of a Small Convex-Convex Lens II.  Off-Axis Behavior of a Small Convex-Convex Lens     Taylor Series for Sine and Cosine     Preliminaries:  Sines and cosines of the angles in the figures     Finding sin(ξ2) as a function of sin(ξ0)     Conversion from sine to tangent in (2-18)     Comparison with the ideal lens equation     Curvature of the Field of a Symmetric Lens III.  What Goes Wrong? IV.  A Small Convex Half-Lens -- A Lens with One Surface     Ideal Scaling Lenses     On-Axis Behavior of a Small Half-Lens     Off-Axis Behavior of a Small Half-Lens     Second Order Shape of the Focal Surface     The Focal Surface of an Eye

I. On-Axis Behavior of a Small Convex-Convex 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 on-axis.  Consequently, a small, thin lens will image a distant scene the same way an ideal lens would.

Figure 1-1  Light entering "tiny" lens

To take advantage of the fact that the lens is "small" and "thin" we will do several things:

1. 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.
   
2. 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).
3. 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 on-axis, we will be further assuming

4. 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 1-1.  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 ξ₀ is the angle between the ray and the lens axis.  The angle of entry of the ray into the lens, θ₀, is given by

1-1)   

By Snell's Law we have

1-2)   

From figure 1-1, we also have

1-3)   

So far everything we've stated is exactly correct.  But now we're going to take small angle approximations, and substitute θ₀ for sin(θ₀) and θ₁ for sin(θ₁).  Substituting (1-2) and (1-1) into (1-3), we'll obtain

1-4)   

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

1-5)   

We now define the curvature, C, to be

1-6)   

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 (1-4) as

1-7)   

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 1-2:  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 1-2).

Note that ξ₁ and φ are the same angles in figure 1-2 as they were in figure 1-1, and, by assumption 2, x has the same value in the two figures.  The angle at which the light hits the interface in figure 1-2, on the other hand, is not the same as the angle at which it left the interface in figure 1-1, so we've called the angles the ray makes with the line which is normal to the surface in figure 1-2  σ₁ and σ₂ rather than θ₁ and θ₂.

From figure 1-2 we have

1-8)   

Using equation (1-7), Snell's law, and the small angle approximations we can now write

1-9)   

Applying (1-7) one more time gives us

1-10)   

Finally, collecting terms, we arrive at

1-11)   

To first order in a small angle the tangent equals the angle, so we can rewrite that as

1-12)

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 convex-convex 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. 3-1?  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

1-13)   

Differentiating,

1-14)   

Differentiating again,

1-15)   

Evaluating at x=0 gives us

1-16)   

This is the curvature of the spherical lens surface at the origin.  Plugging it into (1-12) in place of C we obtain

1-17)   

Comparing this with the ideal lens equation here, we see that the focal length of our tiny lens must be

1-18)

And, much to our pleased surprise, we find that this is identical to equation (Spherical-lenses:3-1) for the center focal length of two glued-together spherical half lenses.  This gives us additional confidence that the result is correct, as the two formulas were found by very different means.

II.  Off-Axis Behavior of a Small Convex-Convex 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 1-3 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

T-1)   

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:

T-2)   

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 φ² 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 1-1 and 1-2, and discarding all second order and higher terms in angle φ.

From figure 1-1, we have:

2-1)   

From figure 1-2, we have:

2-2)   

We now use equations (T-2) to find the sines and cosines of these:

2-3)   


2-4)   


2-5)   


2-6)   

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:

2-7)   

2-8)   

That takes care of the preliminaries.  Now we'll set about finding sin(ξ₂) as a function of sin(ξ₀), and then we'll find the same formula in terms of the tangents.

Finding sin(ξ₂) as a function of sin(ξ₀)

In this section we'll successively substitute expansions for the sines and cosines of the various angles into equation (2-6) until we work our way back to a formula for sin(ξ₂) in terms of ξ₀.

Plugging equations 2-8 (Snell's Law) into equation 2-6, we obtain an equation in terms of σ₁:

2-9)   

Substituting equation 2-5 into (2-9) brings us to a somewhat messy equation in terms of ξ₁:

2-10)   

Looking at just one term from (2-10), we can use equations (2-4) to rewrite it in terms of θ₁:

2-11)   

Discarding the φ² term leaves us with:

2-12)   

We also want the square of this expression.  Squaring and again discarding high order terms leads to:

2-13)   

Substituting (2-12) and (2-13) into (2-10) leads to an equation for sin(ξ₂) in terms of θ₁:

2-14)   

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:

2-15)   

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 (2-7).  We also multiply out the terms with n in them, leading to:

2-16)   

Finally, we use equations (2-3) to write (2-16) in terms of ξ₀.  We also discard terms of order φ², to obtain:

2-17)   

Finally, collecting terms leads to the formula we've been looking for, which expresses sin(ξ₂) in terms of ξ₀:

2-18)

Conversion from sine to tangent in (2-18)

Figure 2-1:

Equation (2-18) 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(ξ₂) as a function of tan(ξ₀).  Reinforcing our rather weak memory of the trig identities by looking at the right triangle in figure 2-1, we start by writing the ones we'll need:

2-19)   

We now plug (2-18) into (2-19a), which gives us a formula for the tangent of ξ₂ in terms of the sine and cosine of ξ₀.   It's a little ugly, unfortunately, even with the second order terms dropped out:


2-20)   

We're now going to express sin(ξ₀) and cos(ξ₀) in (2-20) in terms of tan(ξ₀), using (2-19) (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 non-linear terms:


2-21) 


Combining terms in tan(ξ₀) makes that look a little better:

2-22)   

And now we recognize that the bottom of (2-22) 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:

2-23)   

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:

2-24)

This has the form we wanted.

Comparison with the ideal lens equation

Equation (2-24) 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:

2-25)   

which allows us to rewrite (2-24) in terms of x:

2-26)   

Comparing (2-26) 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

2-27)

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 half-lens 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 2-2:

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 x₁, the distance from the axis (see figure 2-2).  From the figure, we have:

2.2-1)   

We're going to plug that into (2-27) and then differentiate twice and solve for f''.  However, it's going to be awkward to separate f and x₁ after we substitute (2.2-1).  To make our job easier, before we do the substitution we're going to simplify (2-27) by reducing it to second order in ξ₀.

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 (2-27) to first order in tan²(ξ₀).

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 tan²(ξ₀), we can invert it by flipping the sign on tan²(ξ₀).  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 n² out of the square root.  That brings us to:

2.2-2)   

To first order in tan²(ξ₀) -- which is second order in ξ₀ -- we can multiply the tan² term in the square root by 1/2, and discard the radical, leading to


2.2-3)   

By pulling n-1 out of the denominator, we put it in the form of the sum of 1 and a small value.  To first order in tan²(ξ₀) we can flip the sign on the small value and move the denominator into the numerator, leading to:

2.2-4)   

We now multiply it out, and discard all terms higher than second order in tan(ξ₀).  At the same time, we'll also simplify the slightly ugly looking coefficient on the second tan²(ξ₀), by multiplying top and bottom by n, factoring n²-1 into (n+1)(n-1), and cancelling the common factor of n-1.  This all leads to:

2.2-5)   

Finally we multiply it out again, for once dropping no more terms (this is just a rearrangement of 2.2-5), leading us to the form we were aiming at, which is correct to second order in ξ₀:


2.2-6)   


Finally, we plug in equation (2.2-1).  To separate f and x₁, we multiply through by f², and move all terms in f to the left hand side:

2.2-7)   

The left side has a nonzero root at R/2(n-1), which is the focal length when x₁=0; that's where an image exactly on-axis would focus.  We want the curvature at that point.  To find it, we'll differentiate (2.2-7) twice with respect to x₁.  We differentiate the left side using the chain rule, since it's a function of f rather than x₁.  After the first differentiation, we have:

2.2-8)   

The coefficient on f' is nonzero when x₁=0 and  f=R/2(n-1), 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,

2.2-9)   

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 x₁=0, so we drop it out.  We plug in f(0)=R/2(n-1), sum up the terms in R²/(n-1)², rearrange to isolate f'', and we obtain

2.2-10)   

or, multiplying everything out,

2.2-11)   

The focal length at the center is R/2(n-1), so we can rewrite (2.2-11) 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:

2.2-11b)   

Equation (1-16), 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

2.2-12)

Once again, the focal length at the center is R/2(n-1), so we can rewrite (2.2-12) as

2.2-13)

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 double-checked 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 (2.2-12) and (2.2-13) 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:

1. 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.)
2. Rays passing through the center of the lens are not deflected.
3. 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 (2-26) 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:

3-1)   

And that's not what happens here, according to (2-26) -- 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, (2-26) will match (3-1) 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 (2-27).  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 (2-26).  However, that results in a violation of assumption (1) about ideal lenses, and that, in turn, means parallel on-axis 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 (2-27) -- 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 non-ideal!

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 Half-Lens -- 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 single-surface 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 low-index air again.  The front surface of the cornea forms a single-surface lens.

The answer, as we shall see, is that on-axis, a half-lens acts like an ideal lens with magnification.  Off-axis 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 1-1, 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(ξ₁) as a function of sin(ξ₀).  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 (2-1).  The relations between sines and cosines we need, using the small angle approximations, are given in (2-3) and (2-4).  The relation between the sines and cosines of θ₀ and θ₁ which are given by Snell's Law are in equations (2-7).

Our starting point for this analysis will therefore be the expression for sin(ξ₁) from equations (2-4).  Our first step is to use Snell's Law, by substituting equations (2-7) into (2-4) to eliminate θ₁:

4-1)   

From equation (2-3a) we note that, to first order in φ, we have:

4-2)   

We substitute (2-3) and (4-2) into (4-1) to put it in terms of ξ₀:

4-3)   

Finally we collect terms and drop everything which is second order in φ, to obtain:

4-4)   

Now we use (2-19a) to write tan(ξ₁) in terms of (4-4):

4-5)   

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:

4-6)   


Finally we'll use (2-19) (b) and (c) to rewrite (4-6) in terms of tan(ξ₀).  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:

4-7)   

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:

4-8)   

To reduce the noise level a little so we can see what's going on here we'll substitute:

4-9)   

We can now rewrite (4-8) a little more simply:

4-10)   

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:

4-11)   

We now multiply out (4-11) and discard higher order terms, leaving us with:

4-12)   

Unfortunately we still need to undo our substitution of (4-9).  Expanding μ again and multiplying out we get

4-13)   

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(ξ₀).  In one step, this provides us with the on-axis behavior of the lens, and it also lets us compare the result with equation (4-4).  To first order in ξ₀, tangent and sine are equal, and (4-13) and (4-4) should be identical to first order.   We have:

4-14)   

This does, in fact, match (4-4) to first order in ξ₀.  With that bit of evidence, plus some effort checking our algebra, we tentatively conclude that (4-13) is correct.

With the assumption that (4-13) is indeed correct, we will now look at the on-axis and off-axis behavior of our tiny half-lens.  First, however, we need to say something about ideal scaling lenses.

Ideal Scaling Lenses

Figure 4-1:  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:

4.1-1)

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 4-1 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 "f₀" 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 x₁.  For Ray 2 the tangent of the angle at which it emerges is 0, so we must have

4.1-2)   

We've taken the absolute value in (4.1-2) because we've shown x₁ as a length in figure 4-1 rather than a coordinate value.

Since ray 1 enters the lens at the center, with x=0, we can find the tangent of θ₂:

4.1-3)   

We can now find f₀:

4.1-4)   

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:

4.1-5)   

If we just write out x₁+x₀ we see that:

4.1-6)   

which matches (4.1-1).  So ray 0 does arrive at the intersection of ray 1 and ray 2, at distance f behind the lens; since x₀ and θ₀ 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 (4.1-2).  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:

4.1-7)   

And so we see that the front focal  length is also scaled by m.

On-Axis Behavior of a Small Half-Lens

The on-axis behavior of a small half-lens is given by equation (4-14).  We'll rewrite that in terms of the distance from the axis, x, and the radius of curvature, R:

4.2-1)   

Comparing (4.2-1) with (4.1-1) we see that the focal  length of our half-lens is given by

4.2-2)   

We're pleased to note that formula matches the formula we found for the "central" focal length of a spherical half-lens, which was equation 2-10 on our spherical lens page.

The magnification of the image, compared with an ideal lens of the same focal length, is given by

4.2-3)   

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.

Off-Axis Behavior of a Small Half-Lens

Let's start by rewriting (4-12) in terms of x, the distance from the axis.  We'll also clean up some rather confusing signs at the same time; in (4-12) some terms which were subtracted were actually negative:

4.3-1)   

where μ is as defined in equation (4-9); we repeat the definition here:



The first thing to notice is that μ is a function only of ξ₀ and the refractive index of the glass.  Consequently, for fixed ξ₀, μ is fixed.  So, for any particular fixed incidence angle, (4.3-1) has the form of the ideal scaling lens formula, (4.1-1).  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 ξ₀.  So, we can see from (4.3-1) that the magnification decreases as we move away from the axis.  Parts of the scene far off-axis will therefore be imaged at smaller scale than parts on-axis, with the magnification going as 1/tan(ξ₀) for points far from the axis.  The extent of the focal surface will be the integral of the magnification times tan(ξ₀), 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

4.3-2)   

By inspection, near the axis this is a decreasing function of ξ₀.  In other words, the focal plane curves toward the lens for parts of the scene lying off-axis.  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(ξ₀) becomes large.  First, we have:

4.3-3)   

Plugging that into (4.3-2) and again discarding all but the largest terms we obtain

4.3-4)   

which can be simplified quite a bit to obtain

4.3-5)   

or, simpler yet,

4.3-6)   

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 half-lens -- and the retina is certainly not flat!  I'd like to know if the focal surface of a half-lens 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:

4.4-1)   

To second order in ξ₀, we can rewrite μ as:

4.4-2)   

To second order, the focal length for a given incidence angle, equation (4.3-2), is:

4.4-3)   

We now  multiply the top and bottom by



which, evaluating only to second order in ξ₀, produces this:

4.4-4)   

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 (4.4-4) 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 ξ₀² terms, throwing away higher powers, and dividing through by n²/n², we cut the noise level a bit and get something we can almost read:

4.4-5)   

We'll now divide the bottom by n²-n, and combine a couple more terms in ξ₀², which brings us to

4.4-6)   

Since we're only working to first order in ξ₀², 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:

4.4-7)   

We're almost done (I hope).  Looking back at figure (4-1), to obtain the curvature of the focal surface, we need to find f as a function of x₁.  To find x₁, we need to find ξ₁, which is the angle followed a ray of light entering the center of the lens at angle ξ₀.  Of course, x₀=0 at the center of the lens, so, from equation (4.3-1) the items we need are given by

4.4-8)   

To plug into (4.4-7) we actually need these turned around, as:

4.4-9)   

Expanding μ in the second equation in (4.4-9), and retaining everything only to second order in ξ₀, we have

4.4-10)   

This isn't exactly what we had in mind.  However, since we're only taking everything to second order in ξ₀, we can invert the expression in ξ₀ which is on the right just by flipping its sign, after which we move it to the other side and obtain

4.4-11)   

But, again, we're only interested in second order and lower terms in ξ₀, so we can drop the third order term we see here and finally get to something reasonable:

4.4-12)   

We plug this into (4.4-7), multiply through by f², collect terms in f, and we obtain:

4.4-13)   

We factor the polynomial in f on the left, and combine some terms on the right, which leaves us with:

4.4-14)   


Before we continue on, let's look at this equation when x₁=0.  The polynomial in f has two solutions in that case, but only the nonzero solution makes sense here.  So, according to (4.4-14), the focal length for on-axis images (for which x₁≈0) must be

4.4-15)   

This matches equation (4.2-2) for the on-axis focal length of a half-lens.  This is good.

To find the curvature we need the second derivative of f with respect to x₁.  We only have f defined implicitly in (4.4-14); however, that's good enough.  Let's differentiate both sides of (4.4-14).  The left side is a function of f so we need to use the chain rule; we get:

4.4-16)   

By inspection we can see that, when f=Rn/(n-1) as in (4.4-15), the first factor on the left is nonzero.  Consequently, at x₁=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 4.4-16, once again using the chain rule on the left side, to obtain:

4.4-17)   

We're only interested in this at x₁=0.  At that point, we also have f'=0, so the first term in the sum on the left drops out.  We know from (4.4-15) that at x₁=0, f=Rn/(n-1); 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 (4.4-15) to obtain:

4.4-18)   

And now we substitute that into 4.4-17, dividing through to get f'' by itself on the left:

4.4-19)

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 (1-16) we can find its radius of curvature, which is the inverse of the second derivative of the function defining the surface:

4.4-20)

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 (4.4-20), the radius of curvature of the focal surface of an eye should be about

4.5-1)   

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 half-lenses.  Major corrections, and completion of half-lens section, 1/27/2009.  Addition of curvature of focal surface of a symmetric lens, 2/1/2009.