Focus of a ball lens

We had a question involving how a simple equation in geometrical optics is derived. Although very simple for experienced Optics people like myself, it is very difficult for those not schooled in the geometry of refraction to put together.

In *Optical Design Fundamentals for Infrared Systems 2nd ed.*, Mr. Riedl writes:

A sphere or ball performs surprisingly well as a lens. At closer scrutiny, one fmds that such an element can be broken down into two plano-convex lenses, separated by a plane-parallel plate. The positive lenses have under-corrected spherical aberration and the plane-parallel plate is over-corrected. Therefore, there is a compensating effect. A ball lens finds its main application as a coupling element for optical fibers. It deserves an analysis, especially since there are certain limitations that need to be understood.

Figure $5.13$ indicates that the focal length is measured from the center of the sphere, where the extensions of the entering and exiting rays meet. This means that the back focal distance bfl is merely the difference between the focal length and the radius $R$ of the sphere. One can see immediately that the focus falls on the rear surface of the sphere if $f= R$.

ball lens 2

ball lens 1

My solution is as follows.

The formula assumes the angles are small, as I will show below.

The focal length may be shown by the geometry to be

$$f=R \left [ \cos{(2 \theta’-\theta)}+ \frac{\sin{(2 \theta’-\theta)}}{\sin{(2 \theta – 2 \theta’)}} \cos{(2 \theta-2 \theta’)} \right]$$

where $\theta$ is the angle of incidence from the air, and $\theta’$ is the angle of refraction in the glass. We now assume the so-called paraxial approximation in which the sines are replaced by their arguments and the cosine is replaced by $1$. Using the paraxial form of Snell’s Law:

$$N \theta’ = \theta$$

we find that

$$f \approx R \left [ 1 + \frac{(2-N) \theta’}{2 (N-1) \theta’} \right ] = \frac{N}{2(N-1)} R $$

I will elaborate on how I got that exact formula for the focal length. Here’s a diagram:

ball lens

Note that the solid lines represents the ray path and the dotted lines are for measurement. The spherical geometry is expressed in the fact that the trangle in the circle is isosceles. Therefore the angle $\Delta$ in the picture is

$$\Delta = \pi – [\theta + (\pi – 2 \theta’)] = 2 \theta’-\theta$$

Then $x=R \cos{(2 \theta’-\theta)}$. The other leg of that right triangle of which $x$ is a leg is $z=R \sin{(2 \theta’-\theta)}$. The length $y$ is determined from the angle of refraction out of the glass, which is $\theta$ by Snell’s Law; the angle of the right triangle of which $y$ is a leg is $\theta-\Delta$ (alternate interior angles). Thus, $\tan{(\theta-\Delta)} = \tan{(2 \theta – 2 \theta’)} = z/y$, and

$$y = R \cos{(2 \theta-2 \theta’)} \frac{\sin{(2 \theta’-\theta)}}{\sin{(2 \theta-2\theta’)}}$$

The focal length $f=x+y$, and the result follows.

You should see how well the paraxial approximation fares for this ball lens. Here I take your example value of $N=1.3$ and $R=1$, and present a plot of focal length vs. initial ray height $h$ off the optical axis (i.e., $\theta = \arcsin{(h/R)}$):


Two things to note: 1) the exact focal length is less than the paraxial focal length, and 2) the paraxial approximation only works for a pencil of rays that are less than about $0.15$ from the axis.

One Comment

  • Venkat Balagurusamy wrote:

    Thanks for this nice derivation. The formula also tells us about the size of the front aperture one might use to keep mostly paraxial rays.

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