- #1
Unit
- 182
- 0
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!
[tex]x^2 + y^2 = R^2[/tex] called the "origin sphere"
[tex]x^2 + (y - D)^2 = r^2[/tex] 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 ([itex]n_{air} \sin{\theta_1} = n_{glass} \sin{\theta_2}[/itex])
-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
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!
[tex]x^2 + y^2 = R^2[/tex] called the "origin sphere"
[tex]x^2 + (y - D)^2 = r^2[/tex] 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 ([itex]n_{air} \sin{\theta_1} = n_{glass} \sin{\theta_2}[/itex])
-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
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