# Modeling a lense using the intersection of two spheres

1. Aug 21, 2009

### Unit

Hi everybody!

I am trying to model a lense shape using two circles of radii R and r, with one at the origin and the other offset upwards vertically by distance D. D must be less than R + r, but it must be greater than the larger of R or r.

Thus, I have two equations, and the region of their intersection is a lense-shape!

$$x^2 + y^2 = R^2$$ called the "origin sphere"

$$x^2 + (y - D)^2 = r^2$$ called the "offset sphere"

Anyway, for the numerical analysis I'm doing, I have R = r = 5 (the size of my protractor) and D = 8, which works out nicely because the two circles intersect at (-3, 4) and (3, 4).

What I do:
-start off with incident light, represented by vertical rays, like x = 2
-find the point at which it collides with the offset sphere (xray, yray)
-draw a tangent (using the derivative)
-find the necessary angles (using the arctangent of slope)
-use Snell's law once ($n_{air} \sin{\theta_1} = n_{glass} \sin{\theta_2}$)
-determine an equation of a line y that has this equivalent angle and passes through the collision point: y = m(x - xray) + yray
-find there this new line collides with the origin sphere (the top curve of the lense)
-draw a tangent
-Snell's law again
-determine an equation of the final line, the twice-refracted line
-the y-intercept of that line is the "focal" point (judging by symmetry)

This is a lengthy process, each of these steps involving not-very-simplifyable expressions, so a "master equation" that does all this for me would be unwieldy.

What I have found, though (to my great disappointment), is that the "focus" is actually a region of intersections and not simply one point. So, I'm trying to think, would adding a second "lense" remove all spherical aberration? Is it even possible to remove all spherical aberration? By "lense", I mean any extra, defineable shape that will correct the aberration.

Thanks!

-Unit

Last edited: Aug 21, 2009