TL;DR: What is the shape of the ideal planoconvex lens?
More precise formulation, and my work so far: A lens sits in the xy plane. The lens has index of refraction n2 and the rest of the plane has index of refraction n1. The lens is bounded below by the x axis and above by a function f(x). Let a=f(0) and let f(±b)=0. Rays parallel to the y axis enter the lens from below and are all refracted so that they pass through the point (0,c).
Consider the ray that passes through the point (t,0). Since this ray enters the lens through the bottom surface orthogonally to the surface at that point, the ray inside the lens is still a vertical line. It exits the lens at the point (t, f(t)) and, by premise, is refracted towards the point (0,c).
Let line L1 be the line that contains the refracted part of this ray, and let line L2 be the line that is orthogonal to the surface of the lens at point (t, f(t)).
Let θ be the angle of incidence of the ray upon the point (t, f(t)), and let ϕ be the angle of refraction at that point. θ is measured counterclockwise from L2 to the vertical, and ϕ is measured counterclockwise from L2 to the refracted ray.
Suppose that the indices of refraction are such that total internal reflection does not occur. Note that the ray is moving upwards, so the incident side of the interface is inside the lens and the refracted side of the interface is outside the lens.
By Snell's law, we have n1sin(θ) = n2sin(ϕ). Let N = n2/n1. Then sin(θ) = N sin(ϕ).
Line L1 contains the points (0,c) and (t, f(t)), so its slope is (f(t)-c)/t.
The angle from the x axis to L2, measured counterclockwise, is π/2 - θ. The angle from the x axis to the refracted ray, measured counterclockwise, is therefore π/2 - θ + ϕ. Therefore the slope of line L1 is tan(π/2 - θ + ϕ). Equating this with the slope just obtained for that line yields (f(t)-c)/t = tan(π/2 - θ + ϕ), which simplifies to t/(f(t)-c) = tan(θ - ϕ).
Since the angle from the x axis to Line L2 is π/2 - θ, the slope of that line is tan(π/2 - θ). Since this line is orthogonal to f(x) at the point (t, f(t)), its slope is also given by -1/f'(t). So -1/f'(t) = tan(π/2 - θ), which simplifies to f'(t) = -tan(θ).
We are then left with three simultaneous equations:
sin(θ) = N sin(ϕ)
t/(f-c) = tan(θ - ϕ)
f' = -tan(θ)
After substituting to eliminate the angles, we can use some trig identities to reduce things to a single first-order very nonlinear differential equation. Solving this seems like a nightmare, so... help? Please?
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