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Everywhere I read online, it is said that biconvex lenses have two spherical surfaces, focusing parallel light onto a single focal point.

Now, consider a solid sphere, made out of glass. It must be a biconvex lens, with two symmetrical spherical surfaces. Yet, it does not have such property of converging parallel light onto a single point (as far as I understand?).

I have just made a simulation of a perfect biconvex lens in GeoGebra using Snell's law. Yet, the lens does not have a focal point. I might have done something wrong, but I could not find any mistakes.

Is it just massive simplifications online? Could you recommend any proper introductory reads regarding lenses without straight lies? Could it be I have misunderstood or misinterpreted something?

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  • $\begingroup$ TLDR: spherical lenses don't work. They only kinda-sorta work, and only if the radius of the sphere is large compared to the aperture. Compound lenses with multiple, spherical "elements" come pretty close to working. Aspheric lenses work better still. The reason we even have spherical lenses is, if you're going to make optical elements by grinding and polishing glass, there's really only two shapes that are practical: Spherical, and planar. (And, one could argue that a planar surface really is just a spherical surface of infinite radius.) $\endgroup$ Dec 12, 2023 at 21:22
  • $\begingroup$ Spheres are a pretty good approximation of an ideal lens. A large snowglobe in direct sunlight can start a fire by focusing the sun's rays. gazettenet.com/Snow-globe-starts-fire-in-Easthampton-49374809 $\endgroup$
    – Mark H
    Dec 13, 2023 at 2:44

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It is just an approximation that a biconvex lenses have two spherical surface. The ideal surface is in fact not a sphere, for more information, you can look here.

Usually the lenses are manufactured with a spherical because it's much cheaper this way, this deviation from the ideal surface, which introduces optical aberrations.

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  • $\begingroup$ Thank you! As far as I understand, though, this is more looking at it from the real-world manufacture perspective. What would be the ideal, mathematical shape? I have just checked this post – is it hyperbola or ellipsoid? It still feels a bit confusing, but I think I got the idea, that it is complicated. $\endgroup$
    – drevoksi
    Dec 12, 2023 at 21:18
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I would urge you to edit your question with a screenshot of your GeoGebra Simulation. What do you mean by the lens not having a focal point? Maybe if you added a picture of what you are referring to, it would be much easier to provide an answer for your question.

Additionally, could you elaborate on why you feel that a sphere cannot converge light to a single point?

Now, consider a solid sphere, made out of glass. It must be a biconvex lens, with two symmetrical spherical surfaces. Yet, it does not have such property of converging parallel light onto a single point (as far as I understand?).

For example, if you have a glass sphere with a radius of 2 centimetres and a refractive index of 1.5. When a parallel beam of light from infinity from air is incident on it, we can calculate a point where it would converge as follows.

$$\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$$ $$\frac{1.5}{v'}-\frac{1}{-∞}=\frac{1.5-1}{2}$$ $$\frac{1.5}{v'}=\frac{0.5}{2}$$ $$v'=\frac{1.5×2}{0.5}=6$$

The image on the first surface is formed 6cm from the centre of this surface. For the second, this real image would act as an object positioned at $6-4=2$ centimetres from the second surface's centre.

$$\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$$ $$\frac{1}{v}-\frac{1.5}{2}=\frac{1-1.5}{-2}$$ $$\frac{1}{v}=\frac{1}{4}+\frac{3}{4}$$ $$v=\frac{1}{1}=1$$

A point image is formed 1cm in front of the sphere by these calculations. Hence proved, a glass sphere can converge a beam of light just like a lens.

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  • $\begingroup$ What I meant is that the lens in my simulation does not have such point, through which all the initially parallel rays of light will pass after the refractions. The GeoGebra project, though it may contain errors. And the closer the lens' shape is to the circle, the greater the variance in the supposed focal point – intersection between the straight line through its middle – x-axis – and a refracted arbitrary ray, initially parallel to it. $\endgroup$
    – drevoksi
    Dec 12, 2023 at 21:42
  • $\begingroup$ I actually did think at some point that it might depend on other properties, like the refractive index or specific radiuses, but so far other answers here suggest that the spherical–surface lenses do not actually, really work, and are an approximation. $\endgroup$
    – drevoksi
    Dec 12, 2023 at 21:46

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