# Focal length of Convex lens

Is it true that, twice the focal length of a convex lens is equal to the radius of curvature? Please give your opinion.

• Do you mean an equiconvex lens? There are two radii of curvature. It's possible to have two different magnitudes (and even signs) for the radii and still have a positive lens. And this is not an opinion. It's a fact. – Bill N May 27 '18 at 12:49

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 $\frac{r_1}{2}$ ?
• I've just put right a typo: $n$ is the refractive index of the $glass$. – Philip Wood May 27 '18 at 13:53
• Well done! Lenses used to be taught in quite a lot of detail at high school level (at least they were in the UK). As an application of refraction, lenses are part of a subject called $geometrical\ optics$, which has been largely squeezed out of school syllabuses by more exciting topics! – Philip Wood May 29 '18 at 6:59