Is it true that, twice the focal length of a convex lens is equal to the radius of curvature? Please give your opinion.
[Physics] Focal length of Convex lens
lensesoptics
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.
The formula $$f=\frac{c}{2}$$ where $c$ is the radius of curvature is for mirrors, not lenses.
The reason you are getting a wrong result is that you are applying a formula designed for mirrors to a lens.
Incidentally, the formula $f=\frac{c}{2}$ is still valid for mirrors underwater.
Best Answer
No. It is not true for a lens (except, possibly, rarely by numerical accident*). It is, though, true for a concave mirror (and no doubt for a convex mirror). For a (thin) converging lens in air or a vacuum, the relationship is this: $$\frac{1}{f}=(n-1)\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$$ in which $r_1$ and $r_2$ are the radii of curvature of the two lens surfaces, counted positive if convex as seen from the outside, and n is the refractive index of the glass.
*For example, for an equiconvex lens ($r_1 = r_2$) what would n have to be in order for the focal length to be equal to $\frac{r_1}{2}$ ?