Mathematical shape for a lens with no spherical aberration

1. Dec 18, 2008

Hobnob

I'm trying to construct a lens model, and having trouble because I want to portray a perfect lens rather than one with spherical aberration. It's proved impossible to find via Google, although I did find an earlier thread on this forum that was helpful.
From what I can gather, the lens needs to be hyperbolic, but I've tried a hyperbolic lens and it was worse than a parabola. I guess it needs to be a specific hyperbola, not any old thing. Can anyone tell me a formula I can use to make a perfect lens?

Thanks

2. Dec 18, 2008

Antti

3. Dec 18, 2008

Andy Resnick

There is no singlet lens persciption that has zero spherical aberration. Recall there is 3rd order spherical, 5th order spherical, 7th order spherical... I'd have to pull down some of my design books, but IIRC the existence of spherical aberration is related to having a non-zero numerical aperture.

4. Dec 18, 2008

Staff: Mentor

With a mirror, you can eliminate spherical aberration for a particular pair of object and image locations by using an ellipsoidal mirror that has those two locations as its foci. I suspect you can do something similar with a lens, probably using hyperbolic surfaces as you've tried, but it would work perfectly for only one particular pair of object and image locations, and for only one wavelength because of chromatic aberration.

5. Dec 18, 2008

Hobnob

Thanks to all. Chromatic aberration isn't a problem in this case, as it's a purely mathematical model rather than a physical lens (although I'm allowing it as an option).

The irony is that the only reason I'm trying to solve this is to get my model to behave like the picture kids get taught in physics lessons (lies-to-children) - that is, I'm trying to let them look at concepts like focal length, image and object, etc, without worrying about real-life details. Using your aspheric lens (cool though it is) wouldn't be an option either because it also doesn't look like the picture.

I'm not sure about higher-order aberrations - to be honest I don't know anything about the theory at all, only the practical effect. All I really want is for all parallel rays to come to the same focal point; I don't care about other details.

I'm fairly sure I'm going to have to fake it - unfortunately that's likely to bring its own problems. But having spent a day making parabolas, spherical arcs and hyperbolas, I think I'm pretty close to giving up.

I did read somewhere that Huygens solved the problem, but couldn't find any more details - anyone have a reference?

Thanks again
Hob

6. Dec 18, 2008

mgb_phys

There isn't a single surface that produces a perfect image, a parabola does for a collimated beam.
Optics design programs have a 'perfect lens' you can insert but it - cheats it uses a different function for different input rays.

7. Dec 18, 2008

Hobnob

A parabolic reflector works, but a parabolic lens doesn't. The closest I've found was a parabola on one side and flat on the other, but it's still not perfect.

I'm interested to hear more about how optics programs fudge the perfect lens - the real problem I'm having is that I can't work out what it *should* look like - that is, if I point an arbitrary ray at the lens, where should it end up? The best option I can think of is to force parallel rays to hit the focal point, and let any other rays follow the model mathematically. But of course that will create discontinuities.

8. Dec 18, 2008

cesiumfrog

The ideal lens is one that, for each specific point on the object (focal) plane, every ray originating at that point and passing through the aperture is focussed to a common point on the image plane. That should be obvious if you think about how film (or your retina) works, and casts doubt on your fudges-to-children comments.

9. Dec 18, 2008

Hobnob

Well, for my purposes I think the two descriptions are equivalent. The lie-to-children is the picture which shows a bunch of parallel rays coming in and focusing to a single focal point on the other side, and similarly the 'imaging' picture. (In fairness, I've been concentrating on the first, but I assume that the second will come with it)

I don't know enough about the real-life uses of lenses to know how your description and mine differ in the details, right now I'm only interested in the practical problem.

10. Dec 19, 2008

Hobnob

I read your comment at midnight last night, but when I was lying in bed I realised the point you were making. You're right, that should solve my problem - kind of obvious, thanks!

Merry Christmas...
Hob

11. Dec 19, 2008

turin

I thought that the shape isn't the problem so much as the finite thickness. Why do you call it "lie-to-children"? I am unaware of a single physical law that is represented exactly in practice, even in "ideal" laboratory conditions. The thin lens model always worked quite well in classroom demonstrations.

12. Dec 20, 2008

There are several lens defects and these are reduced by using compound lenses ,.eg spherical aberration is reduced by using two separated plano convex lenses the plane faces being opposite eachother,this shares the deviation over four lens faces thereby reducing the aberration.

13. Dec 20, 2008

Redbelly98

Staff Emeritus
Yes, that's right.

In practice, aspheric lens surfaces are defined as one of the conic surfaces, plus correction terms in powers of r^2. I.e.,

z(r) = Conic(r) + A2r2 + A4r4 + A6r6 + ...​

where r is the distance to the center of the lens, and z(r) is the "height" of the surface at r.

Conic(r) could be an ellipse, parabola, or hyperbola.

Reference: go here and click the "lens equation" tab:
http://www.thorlabs.com/NewGroupPage11.cfm?ObjectGroup_ID=800 [Broken]
I'm pretty sure my equation is equivalent (aside from some notational differences), but I could be wrong.

EDIT: if you click on one of the product numbers at the bottom half of the Thorlabs link, you'll get a page showing the parameters for that lens, and so you'll have the equation for an actual lens. These tend to be optimized (minimal aberration) with either the object or image at infinity.

Last edited by a moderator: May 3, 2017
14. Dec 20, 2008

jambaugh

I think there is also a problem with the definition of a "perfect lens" in that a lens which will focus a plane wave to a point (or vis versa) will not perfectly focus an object at a given finite distance to an image at a finite distance. Even if you consider only coaxial point sources to point images there will, I believe, only be a perfect lens for a given object-lens distance.

This I am dredging up from a faint memory of trying to solve this very problem many years ago. Let me qualify this to say at least until the analysis is done (I'm about to give it a whirl) we shouldn't assume that there is one perfect lens for all ranges of focus.

15. Dec 20, 2008

Even if a lens could be made without defects it could not focus a point image of a point object,instead the image is a diffraction pattern with a bright central maximum surrounded by subsidiary maxima and minima.We cannot stop diffraction but we can reduce it by using large apertures.

16. Dec 20, 2008

jambaugh

And by scaling the system up relative to the wavelength. We can thus speak of the idealized case in the limit as the wavelength(s) approach 0 on the scale of the lens and focal distances.

Clearly diffraction issues are not material to the topic.

17. Dec 20, 2008

Redbelly98

Staff Emeritus
If we are talking about designing and using an actual lens, I think diffraction issues are relevant. If nothing else, it means we needn't worry about aberrations once they are reduced to below the diffraction limit.

But if instead this is more of an academic exercise, you could certainly make the case that diffraction is irrelevant since it can be made arbitrarily small in the limit λ→0.

I guess it depends on what the OP intends.

18. Dec 20, 2008

jambaugh

Here is what I think is the solution.
Firstly consider the most simple case where you have two foci, one inside medium with index of refraction$\nu_1$ and the other in medium with index of refraction $\nu_2$. Assume these points lie on an axis of symmetry for the boundary between the two mediums where refraction occurs.

(Once this case is solved we can construct a half planar lens with this shape and slightly shifted first focus.)

Now think about Fermat's principle. It states that the path of light will locally be such that the time of travel is minimized. But this must be a well defined minimum i.e. the first order differential of the travel time must be zero along the tangent to the surface. Since this must be true throughout the focusing[\u] surface the travel time must be constant throughout.

Now if we impose this condition there will be a limit at which the refraction becomes an internal reflection so imposing this condition we get two pieces one of which is the shape of the focusing mirror and the other of a finite lens.

Now picking a point on the surface of this boundary between mediums and defining variables$r$ for the distance to the first focus and $s$ as the distance to the second we get:

$$\nu_1 r + \nu_2 s = t = \nu_1 r_0 + \nu_2 s_0$$

where $r_0$ and $s_0$ are the distances for the point on the boundary chosen on the axial line of symmetry.

We can express the shape in polar coordinates using the first focus as the origin by using the law of cosines:
$$s^2 = r^2 + (s_0+r_0)^2 -2r(s_0+r_0)\cos(\theta) = [s_0 + \rho (r_0-r)]^2$$

Now if you want a rectangular (cylindrical coordinates) form let $x = r \cos(\theta)$ and [eqn ***] $r^2 = x^2 + z^2$ (and also $r_0 = z_0$ ).

$$x^2 +z^2 + (s_0+z_0)^2 -2x(s_0+z_0) = [s_0 + \rho (z_0-r)]^2$$

Mow multiply out the r.h.s. substitute eqn. ***, solve for the linear term in $r$, square this and substitute *** one more time. You get a quartic equation for the surface.

That is if my reasoning w.r.t. Fermat's principle is correct. I'll try to work up a polar plot for various values and post it.

Last edited: Dec 20, 2008
19. Dec 20, 2008

Redbelly98

Staff Emeritus
So far I followed most of that. Two questions:

What is ρ? I'm having trouble seeing why
s2 = [so + ρ(ro-r)]2

Is x parallel to the optic axis (line joining the two focal points), and z is perpendicular to the optic axis? If so, is it preferable to let ro = xo, rather than zo?

Yes, I'm pretty sure that's right. A nice starting point for setting this up.

Regards,

Mark

20. Dec 21, 2008

jambaugh

Pardon me,
$$\rho = \frac{\nu_1}{\nu_1}$$
is the ratio of indexes of refraction and...

I was thinking of z as the optical axis but I see I got things mixed up. Yes as it stands x is the optical axis and z corresponds to the distance from it since theta is the angle off the axis. To be more conventional replace z with R and x with z.

21. Dec 21, 2008

Nevertheless diffraction exists and is one of the major factors to be taken into account with lens design particularly for telescopes and microscopes.There is no point in getting large magnification if we cannot see detail because of diffraction blurring.To eliminate diffraction completely the aperture[not focal length]to wavelength ratio must be infinite which clearly is impossible to achieve.To reduce diffraction we can increase aperture sizes but the costs start to become prohibitive the lens starts to distort under its own weight and its defects,particularly spherical aberration,are exacerbated.We do have some control over the wavelengths used in microscopes hence the large magnifications we achieve with electron microscopes.Diffraction must be an issue to this topic otherwise we are ignoring the nature of light itself the very thing that our lens is dealing with.Best wishes from the UK.

22. Dec 21, 2008

jambaugh

Here's some further analysis on the problem:

I'm changing the variables I use around a bit.
Now let $R$ be the distance to the boundary from the first focus and $S$ be the distance to the other focus. With initial values chosen on the optical axis.
Let $\rho=\frac{\nu_1}{\nu_2}$ be the ratios of indices of refraction and we can take it to be the index of refraction for the first medium taking the second medium as the vacuum.

We then define the values:
$$A = R_0+S_0$$ (total distance between foci)
and
$$B = \rho R_0+S_0$$ (optical distance = time of travel in c=1 units)

We have via equal time condition:

$$S+\rho R = B \to \quad S = B-\rho R$$
and via law of cosine
$$S^2 = R^2 + A^2 -2AR\cos(\theta)$$

Thence with some work we get the quadratic equation in R:
$$(\rho^2 - 1)R^2 - 2(\rho B - A\cos(\theta)) R + B^2-A^2 = 0$$
You can check that with $\theta = 0$ you get a solution $R=R_0$.

We can choose to use A as our scale factor (or pick A=1).
Define:
$$r = \frac{R}{A}, \beta = \frac{B}{A}$$
thence
$$(\rho^2 - 1)r^2 - 2(\rho \beta - \cos(\theta)) r + \beta^2-1 = 0$$

Now consider the discriminant ( divided by 4):

$$\Delta = (\rho\beta - \cos(\theta) )^2 - (\rho^2-1)(\beta^2-1)$$

$$= \rho^2\beta^2 - 2\rho\beta\cos(\theta) +\cos^2(\theta) -\rho^2\beta^2 + \rho^2+\beta^2 - 1$$

$$= - 2\rho\beta\cos(\theta) +\cos^2(\theta) + \rho^2+\beta^2 - 1$$

or
$$\Delta = \rho^2+\beta^2 - 2\rho\beta\cos(\theta) -\sin^2(\theta)$$

$$r = \frac{\rho\beta - \cos(\theta)\pm\sqrt{\Delta}}{\rho^2 - 1}$$

$$\beta = (\rho-1)r_0 + 1$$

Note that at $\theta=0$ we get $r=r_0$ only if we choose $\pm \to -$.

Thus we have:
$$r = \frac{\rho\beta - \cos(\theta)-\sqrt{\Delta}}{\rho^2 - 1}$$

[I've not triple checked this algebra but I think it's right.]

Taking typical glass, $\rho=1.5$ gives $\beta = 0.5 r_0 + 1$,
$$\Delta = 2.25+(0.5r_0+1)^2 - 4.5(0.5 r_0 + 1)\cos(\theta) -\sin^2(\theta)$$

$$r = \frac{1.5(0.5r_0+1) - \cos(\theta)-\sqrt{\Delta}}{1.25}$$

Attached are two sets of graphs. The second shows that as $A\to\infty$ we get nearly a spherical lens.

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23. Dec 21, 2008

Diffraction being a property of light is relevant to the topic,in fact it is one of the most important considerations when designing lenses for telescopes and microscopes.What point is there in getting large magnification if detail is lost because of diffraction blurring?

24. Dec 21, 2008

Redbelly98

Staff Emeritus
The point here is not to get large magnification, but to eliminate aberration effects. These are every bit as important as diffraction.

Once aberration is eliminated or at least minimized, we can still calculate what blurring due to diffraction would be. But manipulating the surface prescription will not change the diffraction, that simply is what it is given diameter, focal length, and wavelength.

25. Dec 21, 2008