Is the radius of curvature of a lens correspond the the radius of the sphere in which the lens rises from?

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    Yes, imagine you'd complete the sphere that the lens is cut from; that will define the radius of curvature. – ahemmetter Mar 6 '15 at 13:22
  • @andynitrox: The polished side of a mirror is cut from one sphere. While, each surface of a lens is cut from a sphere. Hence, it takes two spheres to define a lens. Therefore, each surface will have it's own radius of curvature which may or may not be equal. – Nick Mar 6 '15 at 14:36
  • Yes of course, I missed that point. When I wrote the comment I had a plano-convex lens in mind. – ahemmetter Mar 6 '15 at 14:39

From this answer, I wish to make it abundantly clear to you that while one sphere can define two mirrors (concave and convex), it takes two spheres to define a lens.

In the case of a perfect concave or convex mirror, you can complete the sphere and by the definition of radius of curvature, the radius of the sphere is the same as that of the mirror. See figure below: mirror

Now, in the case of lenses. Let us consider a common biconvex lense. The lense has two surfaces unlike a mirror which has only one. Each of these surfaces can be thought of as being a segment of a sphere as shown in the below figure:

enter image description here

The figure shows that the lens surfaces are part of distinct spheres and hence each surface has unique radius. Similarly, you can imagine a biconcave lens being part of two spheres.

Now, not all lenses are symmetrical like the above. the radii of lenses can be different from one another. By different variations of the radii, the following lenses can be made: lenses

PS: It is interesting to note that from the lensmaker's formula, it found that for an asymmetrical thin lens, the focal length is proportional to the ratio between the product and sum of the radii: $$f\propto \frac{R_1\cdot R_2}{R_1 + R_2}$$

  • @Floris: Yes, I made a mistake there, but I think that you are also making a mistake. According to the formula, $$\frac{1}{f} \approx (n-1)\left( \frac{1}{R_1} - \frac{1}{R_2}\right) \implies f\propto \frac{R_1 \cdot R_2}{R_2 - R_1}$$ – Nick Mar 6 '15 at 15:33
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    Your words say "sum" but your formula shows difference. Doesn't that bother you? I assume that R is positive for a convex surface - then the + sign is right. Otherwise a symmetrical lens would have infinite focal length... – Floris Mar 6 '15 at 17:15
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    @Floris: I understand your point. In my defence, equations in optics are beyond common sense because of these sign conventions. Glad I'm not going into this feild. – Nick Mar 7 '15 at 7:31

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