Mathematical shape for a lens with no spherical aberration

[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

 

[Antti]

Im no expert on this, but I do know that aspheric lenses are used in high-end camera objectives http://en.wikipedia.org/wiki/Aspheric_lens

But as you may know, no lens can be perfect for all wavelengths.

 

[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.

 

[jtbell]

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.

 

[Hobnob]

Im no expert on this, but I do know that aspheric lenses are used in high-end camera objectives http://en.wikipedia.org/wiki/Aspheric_lens

But as you may know, no lens can be perfect for all wavelengths.

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

 

[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.

 

[Hobnob]

There isn't a single surface that produces a perfect image, a parabola does for a collimated beam.

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.

 

[cesiumfrog]

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 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.

 

[Hobnob]

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.

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.

 

[Hobnob]

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.

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

Merry Christmas...
Hob

 

[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.

 

[Dadface]

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 each other,this shares the deviation over four lens faces thereby reducing the aberration.

 

[Redbelly98]

... 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, ...

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:
https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=3810
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.

 

[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.

 

[Dadface]

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.

 

[jambaugh]

[...] the image is a diffraction pattern [...]we can reduce it by using large apertures.

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.

 

[Redbelly98]

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.

 

[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.

 

[Redbelly98]

So far I followed most of that. Two questions:

What is ρ? I'm having trouble seeing why

s² = [s_o + ρ(r_o-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 r_o = x_o, rather than z_o?

That is if my reasoning w.r.t. Fermat's principle is correct

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

Regards,

Mark

 

[jambaugh]

So far I followed most of that. Two questions:

What is ρ? I'm having trouble seeing why

s² = [s_o + ρ(r_o-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 r_o = x_o, rather than z_o?


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

Regards,

Mark

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.

 

[Dadface]

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.

 

[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.

 

[Dadface]

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?

 

[Redbelly98]

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?

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.

 

[Dadface]

Magnification is just an example.The point is that in lens design we have to consider all of the properties of light.With several applications of lenses we have no control over the wavelength so if we wish to reduce diffraction we must increase the aperture and increasing the aperture makes spherical aberration worse[remember that the simplest way to reduce spherical aberration is to confine the rays to a narrow beam on the central portion of the lens].We cannot win ,diffraction and spherical aberration are linked together and improving one thing worsens the other thing. We cannot tell light to obey the laws of refraction but disregard the laws of diffraction.

 

[Redbelly98]

James,

Your derivation (in post #22) looks right to me. I'd like to do a ray trace of a lens following your prescription, but it will have to wait as I'm busy with Holiday stuff this week and it is a little involved to define such a surface in my ray tracing code.

Ideally all rays that intersect the surface would converge to the same point, so that would be a good check of your analysis. It would also be interesting to see what aberrations arise for off-axis source points.

Thanks for doing this,

Mark

 

[Redbelly98]

Well, something is bothering me now about these curves. I've entered the equations into Excel and reproduced the results from Post 22, but I am having the following issues:

1. For large enough angles (for example at theta 90 degrees), looking at the figures it seems such a ray cannot be refracted toward the image point.

2. When I reverse n1 and n2, so that the rays go from air into glass, I get a negative discriminant (Δ<0), and therefore no solution, for many values of theta. (Ro=100, So=10, rho=1/1.5)

I'll have to think about this some more to resolve these issues.

Mark

p.s. As A→∞ the surface appears to be an ellipse rather than a circle.

 

[jambaugh]

Well, something is bothering me now about these curves. I've entered the equations into Excel and reproduced the results from Post 22, but I am having the following issues:

1. For large enough angles (for example at theta 90 degrees), looking at the figures it seems such a ray cannot be refracted toward the image point.

2. When I reverse n1 and n2, so that the rays go from air into glass, I get a negative discriminant (Δ<0), and therefore no solution, for many values of theta. (Ro=100, So=10, rho=1/1.5)

I'll have to think about this some more to resolve these issues.

Mark

p.s. As A→∞ the surface appears to be an ellipse rather than a circle.

Yes, at the point were the line to the image point is tangent to the surface you transition to internal reflection. You could cook up an impractical 2-dim situation where you reflect while simultaneously transitioning to the the other medium. The actual lens would be defined for a restricted angle.

But a more practical application of this part of the surface may be that you interpret the foci as virtual foci in a compound lens system. I'll have to think about this some more.

Another note. One should also be able to work with virtual foci as well and build two half lenses which focus from point to point, or from infinity to a virtual focus with one half and then from that virtual focus to a true focal point with the second half.

Final note. I worked this out several times seldom without errors. Although I'm pretty confident I have it right I can't be 100% certain until I take more time to type-set the whole derivation and check it carefully by hand. With the holidays upcoming...

 

[jambaugh]

[...]
2. When I reverse n1 and n2, so that the rays go from air into glass, I get a negative discriminant (Δ<0), and therefore no solution, for many values of theta. (Ro=100, So=10, rho=1/1.5)
[...]
p.s. As A→∞ the surface appears to be an ellipse rather than a circle.

Re. 2.) This is the same as swapping foci so you should get the same curve about the other focus with then the angle limited to a range of values. You thus better not have positive discriminant for all angles. You will in fact get a better graph this way since it will only trace the part of the curve where you do indeed get actual refraction.

As far as the A→∞ limit, it is hard to visually distinguish ellipse and circle but I can't see how the eccentricity of this ellipse could be anything but zero in the limit. The form I have is not the best to look at this case. I'll have to work out a parallel derivation.

 

[jambaugh]

Oops! You were right. In the A→∞ limit you get an ellipse. I was thinking there were no parameters to specify the eccentricity but I forgot about the indices of refraction.

I got via quick calculation semi-radii of:

$$a = \frac{\rho}{\rho+1}, b=\sqrt{\frac{\rho-1}{\rho+1}}=\frac{\sqrt{\rho^2-1}}{\rho+1}$$
(where $r_0 = 1$).

Thus eccentricity:

$$e = \frac{\sqrt{\rho^2 -\rho+1}}{\rho}$$

That's a first draft calc. so don't bet the farm on it.

 

[jambaugh]

Yet another note... In the quadratic formula the case $\pm\to -$ relied on $\rho > 1$. If you're working with the $\rho < 1$ case you must change the sign of the radical of the discriminant.

 

[Redbelly98]

Hi James,

Yes, I realized after my earlier post that for the air→glass case, rays at a large enough angle would simply miss the lens, so the negative discriminant (no solution) makes sense to me now.

Yet another note... In the quadratic formula the case $\pm\to -$ relied on $\rho > 1$. If you're working with the $\rho < 1$ case you must change the sign of the radical of the discriminant.

Yup, I noticed this in my spreadsheet.

Interestingly, as A→∞ for air→glass, the surface resembles a hyperbola. I've not confirmed whether it is an actual hyperbola or not, but the asymptotes are clearly evident and make sense physically too. I will post a sample graph shortly.

 

[Redbelly98]

For this plot:

n1 = 1
n2 = 1.5
ρ = n1/n2 = 2/3
Other info as given in graph

 

[jambaugh]

Hi James,
[...] Interestingly, as A→∞ for air→glass, the surface resembles a hyperbola. I've not confirmed whether it is an actual hyperbola or not, but the asymptotes are clearly evident and make sense physically too. I will post a sample graph shortly.

It appears locally to be a hyperbola but note that there is a symmetry in reversing the two foci and refraction indices. You get the same curves as with $\rho > 1$. The analytic curves continue to enclose the far focus. [See attached plot.]

If you graph both solutions to the quadratic then I think you'll get the other half of the curves.

 

[Redbelly98]

I'm finding a problem with this solution. It seems that Snell's Law is violated, if I have done things correctly.

When I have time I can show my derivation, but for now I'll just outline what I did.

We have, for the lens surface points (z,x):

z = A r cos(θ)
x = A r sin(θ)​

where we use the function r(θ) as shown in Post #22.

Then

dz/dθ = A ( dr/dθ cos(θ) - r sin(θ) )
dx/dθ = A ( dr/dθ sin(θ) + r cos(θ) )​

We can calculate the angle of the surface normal from

θ_norm = atan(-dz/dx)

= atan( -(dz/dr) / (dx/dr) )
​

From the normal angle, we can caculate the angles of incidence and transmission, and use the refractive indices to check whether Snell's Law holds.

I'm seeing a violation of Snell's Law. I'll try to post the derivation later, and perhaps somebody can either confirm it or point out an error.

Mark