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?
[Physics] Radius of curvature and focal length
curvaturelensesopticsrefraction
Related Solutions
A lens has two surfaces. One of these surfaces still has refracting power, due to the difference in refractive index and angle of incidence.
Refraction on one surface
Let parallel rays start in glas. Radius sign convention results $R=-200\,mm$ since incident light is on same side as center of curvature.
Radius of curvature and the two media influence the focal point. For your rays propagate from right to left (air) the FFL (front focal length, as depicted) is $$f=R\cdot \frac{n_1}{n_1-n_2}=R\cdot \frac{1}{1-n_2}$$ This is the reciprocal element of your transfer matrix. Focal length is proportional to radius of curvature. Increasing refractive index $n_2$ of glas shortens it. $$f=-200\,mm \cdot \frac{1}{-0.5}=400\,mm$$
Remark
The cited above lensmaker equation with $R_1=\infty$ will return the same result. Since on-axis parallel rays are not refracted on plane surface, as can be seen from Snell's law.
Any formula in physics comes with a set of definitions of what each variable in the equation represents, and how to interpret positive or negative values. This is particularly rue in the case of lens and mirror formulae. In each case, a different form of the equation, with a different set of definitions, will give the same correct result.
In this case, Wikipedia https://en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation adds to the above equation:
The signs of the lens' radii of curvature indicate whether the corresponding surfaces are convex or concave. The sign convention used to represent this varies, but in this article a positive R indicates a surface's center of curvature is further along in the direction of the ray travel (right, in the accompanying diagrams), while negative R means that rays reaching the surface have already passed the center of curvature. Consequently, for external lens surfaces as diagrammed above, R1 > 0 and R2 < 0 indicate convex surfaces (used to converge light in a positive lens), while R1 < 0 and R2 > 0 indicate concave surfaces. The reciprocal of the radius of curvature is called the curvature. A flat surface has zero curvature, and its radius of curvature is infinity
Thus, your "on the other hand" individual is not respecting the sign convention, and will not get the correct result...
EDIT to expand This source, http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenmak.html, with its links, define exactly how the direction of light flow, sign of radius of curvature and position of focal point are to be defined. If you fail to follow these conventions with this form of the equation, chaos results...
Best Answer
Not necessarily. The lens equation, as provided by Nordic, is
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
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.