| Brad
Rodgers |
Is it possible from 'first principles' (whatever those might be) to prove there is a focus for a given lens? Everything I've read seems to just assume that all light will automatically converge at one point, but I don't see a priori why it should. Brad |
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| Hauke
Worpel |
I can't remember off the top of my head how, but it's possible to prove using Snell's law. |
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| Liz F |
From work we've done in physics, I don't think there is only a particular focus for light of each particular wavelength. If you study chromatic abberation you discover that light of different colours (and therefore wavelengths) is focused at different points. Do you want a proof that light of a certain wavelength is focused at a point? It would be interesting, but I don't really see how it could be done absolutely from first principles. Similarly, does anyone have a proof (maybe using the actual definition of a parabola) that a light shone on a parabolic mirror will always focus to a point. I can kind of believe that the definition of a parabola does make this true, but I don't see the actual proof! |
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| Matthew
Smith |
The result that a parabolic mirror focuses parallel light to a point is fairly straightforward to prove, if you can differentiate. Using a standard form for the equation of a parabola, y2=4a x (where a is a constant) consider a beam coming in at a distance h from the axis (i.e. along the line y=h). Then you can work out the slope of the tangent to the parabola at y=h (by differentiation), and hence the slope of the normal. Once you have this, you can get an expression for tan(j), where j is the angle between the incident ray and the normal to the mirror. Because the angle of incidence equals the angle of reflexion, the light is reflected at an angle 2j below the horizontal. By using a trigonometric identity for tan(2j), and simple trigonometry, it should be possible to work out where the reflected ray meets the axis y=0. If your answer is independent of h, then you've proved that the rays all meet at a focus. If you need any more help with the proof, post again and I'll try to explain further. I think there are more elegant ways of proving this result (after all, it was known to the ancient Greeks, who worked out the position of tangents by purely geometrical means), but the way I've outlined above is fairly straightforward if you've done co-ordinate geometry. As for the lens, it is possible to 'prove' the existence of a focus from first principles (i.e. Snell's law, or Fermat's principle), but only by making approximations. This is because, even with monochromatic light, a circular lens doesn't actually focus quite to a point. If you're interested in the approximate proof, I can find or work it out for you. |
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| Brad
Rodgers |
Yes, I'd be interested in the "proof" you mention. I'm familiar with Snell's law and Fermat's principle, as well as the derivation of the former from the latter. Thanks, Brad |
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| Arun
Iyer |
Its a very interesting question indeed.I must admit that i thought the existence of focus came from the definition of convex lens.Now that i think over it,i might have not been right after all. love arun P.S-> i will try to find out some material over it.I would be interested in Hauke's proof as well. |
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| Matthew
Smith |
The standard derivation of the focal length of a lens uses Snell's law, lots of angles and lots of approximations. Instead of that, I've tried to put together a 'proof' using Fermat's principle, which still uses lots of approximations, but is a little shorter. I hope it makes sense... Consider a convex lens with two spherical surfaces, and assume, for simplicity, equal radii of curveture. Take a point source a distance u along the axis on one side, and the image a distance v away on the other. Obviously a ray along the axis will pass through both points without being deflected. We need, therefore, to prove that all rays from u to v have the same optical path length as this central ray: then, by Fermat's principle, they are all real rays, and light from u really is focused at v. First work out the optical path length along the axis: this is just u + v + t0 (n-1), where t0 is the thickness of the lens at the middle, and n is the refractive index of the glass/air interface. Now work out the optical path length of a hypothetical ray passing through the lens a height h above the axis. This is approximately sqrt(u2 + h2 ) + sqrt (v2 + h2 ) + t(n-1), where t is the thickness of the lens at height h. Using the binomial expansion, the path length is approximately u + v + (1/u + 1/v)h2 /2 + t(n-1). Thus the difference in path lengths is about (t0 -t)(n-1) - (1/u + 1/v)h2 /2. Now we need to work out t0 - t in terms of h. To do this, draw the circles of which the lenses are part, with radii R. Call the length of the common chord, which is the diameter of the lens, d. Then by simple geometry, again making several approximations, we have 2R(t0 /2) = d2 , 2R(t/2) = (d+h)(d-h) = d2 - h2 , and, combining, t0 -t = h2 /R. Substituting this into the equation above, we get a path difference of (n-1)h2 /R - (1/u + 1/v)h2 /2. This is zero whenever 1/u + 1/v = 2(n-1)/R, regardless of h. This proves (!) that, provided the lens formula condition (which should look familiar) is met, the light from u is brought to a focus at v. As a bonus, it provides an expression for the focal length in terms of R and n. Incidentally, a similar sort of argument from Fermat's principle can be used to prove the focal property of a parabolic mirror very easily. You just need to use the focus/directrix property of the parabola, and should be able to show it in a few lines. |
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| Matthew
Smith |
A quick footnote to the above: the Snell's law version of the above isn't as bad as I remembered, since you only need to get a formula for one curved glass/air boundary. Then, by judicious changes of sign, you can apply it again to the other boundary. You can also do the proof using Fermat's principle that way, and it's arguably more illuminating (if you'll excuse the pun), as you can think of each boundary adding to the 'curveture of the ray' (and also deal with asymmetrical lenses easily). If anyone's interested in more details, I'll post them; alternatively, it's worth looking up the treatment in the first volume of the 'Feynmann lectures in Physics', if you can get hold of a copy. |