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How can the focal point of a ball lens be found?

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I've tried to make the formulas readable, but the math is messed up. What is $1 / 2^{0.5} / 1.3$ supposed to mean? – TMM Mar 29 '13 at 0:20
I believe the 1.3 refers to the refractive index mentioned earlier - so it would be $(1/2^0.5)/1.3$ – mardat Mar 29 '13 at 0:25
This is not a question about derived functors. – Piotr Pstrągowski Mar 29 '13 at 1:06
Note that your example that uses $45^{\circ}$ is far beyond the region of validity of the paraxial approximation which the author inherently assumes. See my solution below. – Ron Gordon Mar 29 '13 at 10:47
up vote 5 down vote accepted

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., $\theta = \arcsin{(h/R)}$):

Ball lens plot

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.

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@DarkLightA: Thanks - I used Microsoft Visio. I wish I could have said Mathematica, but I didn't feel like fussing with placing the symbols there. – Ron Gordon Mar 29 '13 at 12:06

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