Is the radius of curvature of a convex or concave lens longer than the focal length of the lens? Does the center or curvature affect the focal point in a lens?

  • $\begingroup$ Depends on the index of refraction of the material the lens is made of. The radius of curvature is probably way longer than the focal length. $\endgroup$ – ahemmetter Mar 6 '15 at 13:24
  • $\begingroup$ I think you should consult optics book for this question. Such as Hecht $\endgroup$ – hsinghal Jun 16 '16 at 3:31

Not necessarily. The lens equation, as provided by Nordic, is

enter image description here

You can play around with the numbers, and a few things should become obvious.

1) If you increase the radius of one surface to a very large number, the lens becomes either a plano-convex or a plano-concave lens. For the moment, think only about a plano-convex lens, with R2 essentially infinite. Then

                          1/F = (n-1)(1/R1)

                     or     F = R1 / (n-1)

where n is the refractive index.

2) So, for n < 2, the focal length of a plano-convex thin lens will be greater than the radius of curvature, and for n > 2, the focal length will be less.

Most optical materials have an index of refraction in the range of 1.3 to 1.7 for visible light, so for most lenses, the focal length will be greater than the radius of curvature. Diamond, though, has an index of refraction of about 2.4, so such a lens will reverse the usual order. And although it is reflective in the visible, germanium is transparent in the mid to far infrared, and has an index of refraction of about 4 at these wavelengths.

  1. Yes, radius of curvature is longer than focal length for both types of lenses (O1 and O2 are the centers of curvature, F is focal point): enter image description here
  2. Yes, curvatures of lens surfaces (R1 and R2) along with refractive index of lens material (n) affect its focal length (F) , hence the position of focal point(s):

enter image description here



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