Anaclastic lenses

It all began as a daydream. I was bored in class and for some reason got to thinking of melting things (in particular this eraser I was playing with). I devised a contraption composed of a large parabolic reflector and a set of small lenses or other mirrors that would focus the light from the reflector into a death beam. But I realized that to get a high quality beam, I’d need a lens that would have no geometric aberrations (i.e. a convergent lens which focuses all rays parallel to its focal axis to the same point). There are two (simple) geometric cases I could consider here but I’ll only do one (something magic will happen to take care of the other).

Positive Meniscus Lens

You might be asking yourself (figuratively and literally) "if this is the lens, why does it only have one side?" Well, I neglect the other side for simplicity. However, I can construct the other side to be a sphere centered at the focal point. That means that if the leading face refracts all the rays towards the focal point, then the trailing edge will not affect their path. So, from this diagram, we can start writing some equations. Ultimately what we want is a differential equation involving the function g, the focal length f, and the vertical position y. Let’s start with the obvious, Snell’s law. Without loss of generality I will assume the region left of the leading face has a refractive index of 1 (you can always normalize all refractive indices by this one).

\\sin\\theta_i = n \\sin\\theta_r

Note that it is not necessary for n to be greater than 1 in this equation. Now, for my next trick…err…equation, we’ll relate the slope of the lens face to the incident angle.

g\'(y) = \\tan\\theta_i

So far we haven’t written any equations involving the refracted angle, so the last equation we’ll need is on which relates that angle to all the others. Here it is:

\\tan(\\theta_i - \\theta_r) = \\frac{y}{f - g(y)}

This equation will hold true also if our lens face bends to the left instead of the right. If that happens, that means that g and g’ are both negative instead of positive. If we use the same equations as above, this makes the angles of incidence and refraction both negative as well and we get the same set of equations. I’ll explain more about why this is important later. Now, let’s make this last equation a little more useful by splitting apart the tangent.

\\tan(\\theta_i - \\theta_r) = \\frac{\\tan\\theta_i - \\tan\\theta_r}{1 + \\tan\\theta_i\\tan\\theta_r} = \\frac{y}{f - g(y)}

Now, we already know how to write the tangent of the incident angle in terms of g, but that still leaves us with the refraction angle lying around. So, let’s get rid of that. We already know the sine of that angle from Snell’s law above, so let’s transform that into something involving the tangent of that angle.

\tan\theta_r = \frac{\sin\theta_r}{\cos\theta_r} = \frac{\sin\theta_r}{\sqrt{1 - \sin^2\theta_r}}

Now we’ve got a pesky sine in there. But we know the tangent of the angle of incidence already, so let’s use that to get rid of it.

\begin{align*} \sin^2\theta_i &= \tan^2\theta_i \cos^2\theta_i = [g'(y)]^2 (1 - \sin^2\theta_i) \Rightarrow \\ \sin^2\theta_i &= \frac{[g'(y)]^2}{1+[g'(y)]^2} \end{align*}

Now, putting everything back together we find this pretty horrendous looking differential equation:

\frac{y}{f - g(y)} = \frac{g'(y) - \frac{1}{\sqrt{n^2\frac{1+[g'(y)]^2}{[g'(y)]^2} - 1}}}{1 + \frac{g'(y)}{\sqrt{n^2\frac{1+[g'(y)]^2}{[g'(y)]^2} - 1}}}

Simplifying a bit:

\frac{y}{f - g(y)} = \frac{g'(y)\sqrt{n^2\frac{1+[g'(y)]^2}{[g'(y)]^2} - 1} - 1}{\sqrt{n^2\frac{1+[g'(y)]^2}{[g'(y)]^2} - 1} + g'(y)}

This is still a non-linear, non-homogeneous first-order differential equation. How do we solve this beast? Well, the solution is less than obvious, especially from this point. However, you can move some things around and get g’ by itself. What you’re left with after the algebraic acrobatics is this:

g'(y) = \frac{n y}{n(f - g(y)) \pm \sqrt{(f - g(y))^2 + y^2}}

Now comes a trick. We’re going to do a change of variables:

\begin{align*} r(\theta) \cos\theta &= f - g(y)\\ r(\theta) \sin\theta &= y \end{align*} g'(y) = \frac{dg}{dy} = \frac{\frac{dg}{d\theta}}{\frac{dy}{d\theta}} \begin{align*} g(y) = f - r(\theta) \cos\theta &\Rightarrow \frac{dg}{d\theta} =  r(\theta) \sin\theta - r'(\theta) \cos\theta\\ y = r(\theta) \sin\theta &\Rightarrow \frac{dy}{d\theta} = r(\theta) \cos\theta + r'(\theta) \sin\theta \end{align*}

Putting it all together:

\frac{r(\theta) \sin\theta - r'(\theta) \cos\theta}{r(\theta) \cos\theta + r'(\theta) \sin\theta} = \frac{n r(\theta) \sin\theta}{n r(\theta)\cos\theta \pm r(\theta)} = \frac{n \sin\theta}{n \cos\theta \pm 1}

Doing some more algebra (only cross-multiplying and canceling out some terms) lands us here:

\frac{r'(\theta)}{r(\theta)} = \frac{\pm\sin\theta}{n \pm \cos\theta}

But wait a minute, we can actually integrate this!

\begin{align*} \ln(r(\theta)) &= -\ln(n \pm \cos\theta) + C \\ r(\theta) &= \frac{e^C}{n \pm \cos\theta} \end{align*}

And a remarkable result appears. The solution to that nasty differential equation is actually a conic section. But you might wonder where the focal length parameter has gone. Well, it’s been pulled up into the boundary conditions. From our original picture, we drew g(0) at the origin, so g(0) = 0. Using this fact, we find:

\left.\begin{matrix} r(\theta_0) \cos\theta_0 = f - g(0) = f \\ r(\theta_0) \sin\theta_0 = 0 \end{matrix}\right\} \Rightarrow \theta_0 = 0, r(0) = f

The focal point is actually one focus of the conic section solution. Notice also that there are two solutions. I haven’t yet tested the validity of both solutions. Hopefully some kind person will tell me about this. I should also note that the only restriction to this solution is that n must be greater than zero and not one. This makes sense, though. A refractive index of 1 would leave us with a lens that does nothing and a negative refractive index has no meaning. So what about if 0 < n <1? What does this correspond to? This is actually the case where the lens has one flat side (which the parallel rays hit first and are unaffected). In this case, n is equal to 1/n‘ where n‘ is the refractive index of the lens material. Now, to bring this full circle (or ellipse, hyperbola, or parabola) let’s get this back in terms of g and y:

\begin{align*} g(y) &= \frac{n}{n^2-1}\left((n + 1)f \pm \sqrt{(n + 1)^2 f^2 - (n^2-1) y^2}\right) \\ g(y) &= \frac{n}{n^2-1}\left((n - 1)f \pm \sqrt{(n - 1)^2 f^2 - (n^2-1) y^2}\right) \end{align*}

Pretty neat, right? Is there anything conic sections can’t do?

11 Responses to “Anaclastic lenses”

  1. kc lc says:

    Yes, very neat. But I might have tackled it differently right from the start. By inspection, theta = thetaI – thetaR. And tan(theta) = y/(f-g(y)).

    And since tan(thetaI) = g’(y) you can quickly write down a differential equation for g(y) like in your equation 8. But the 1/sqrt() term is replaced by tan(thetaR). Trouble is… removing that left over tan(thetaR) term…

    hmm… I guess I can’t see any shortcut by staying in polar/parametric form right from the start.

  2. Yourself says:

    I noticed that as well, but I didn’t actually put much thought into inspecting that. It just didn’t seem to me to be entirely worthwhile to do it that way.

  3. kc lc says:

    I guess you’re right. Seems like it just leads to the same place once you re-write that tan(thetaR) term.

    Here’s something sort of related… to conic sections anyway: shadows from a sphere. Place a sphere on a plane, then place lights in various locations to cast shadows. Depending on the light locations, the shadow will be a circle, ellipse, parabola, or one branch of a hyperbola:

    http://www.flyingbanjo.com/temp/conic_shadows.png

    I left out the circle (too obvious). The light location to get an elliptical shadow is pretty obvious too. But less obvious with the parabola and hyperbola. I know the positions needed for them, but I haven’t tried to prove it.

  4. Yourself says:

    Well, with a perfect point light, the shadow volume will be a cone frustum (the vertex of the cone would be at the light source). The actual shadow is where this volume intersects the plane. It should be pretty obvious from here why you get conic sections. You’re cutting a cone with a plane and that’s where all conic sections come from.

    The type of conic section depends on the source’s position with respect to a plane tangent to the top of the sphere (or, in general, tangent to the sphere and opposite from and parallel to the shadow receiving plane). If the light source is on the far side of this plane from the shadow receiver, you get an ellipse. If it’s on the plane, you get a parabola. If it’s below the plane (but not below the sphere), you get a hyperbola.

  5. kc lc says:

    Nicely explained. I was thinking about this too hard — it’s actually easy to see with your “shadow volume cut by a plane” idea.

    What about the reverse problem? I’ve read before that all conics can cast a circular shadow. Easy to see this for an ellipse. What about the others?

  6. Yourself says:

    You mean a flat cut-out of a conic section casting a shadow? In that case, construct the cone that the particular conic belongs to. Place the point light at the vertex, and the shadow receiving plane perpendicular to the axis of the cone.

    The unbounded conics could never practically cast a fully circular shadow; at least not as far as I can imagine. At the very least the cast shadow will be a portion of a circle. What might be slightly more interesting is placing a pinhole camera at the vertex of the cone, pointed along the cone’s axis. In this case, a parabola would form a complete circle (well, with the exception of a single point). Hyperbolas still wouldn’t form complete circles.

  7. xot says:

    Very clever illustration, KC.

  8. Makerofthegames says:

    I’m too dumb to understand any of this..Perhaps in college..

  9. Alpha Man says:

    KC LC! I didn’t know you were fond of the higher maths!

  10. PF says:

    I may not have understood the whole thing, but it was certainly intriguing. Good luck on your death ray. :)

    -PF

  11. Mgamerz says:

    You might be asking yourself (figuratively and literally) “if this is the lens, why does it only have one side?”
    I’m asking you WHY ARE YOU THINKING THIS!!!!! I guess your just really smart like that. And like death rays.