# Consider 2 bi-convex lenses, one of which is thicker than the other. Which one has greater focal length? Is there a relation between radius of curvature and focal length of a convex lens?
Assuming they are the same diameter and glass and both of the two surfaces are smooth, the thicker one has shorter focal length, f. Not much direct relationship* to radii of curvature, one can even be large and negative as is common in positive diopter eye glasses. Also if you relax the "smooth surfaces" requirement, all parts of both surfaces can be planar (Fresnel lens) and then either can have the smaller f. ------------------------------------ *With good approximation and some additonal condsiderations (and sign conventions, if lens is not "bi-convex"): (1/f) ~ (1/R1) + (1/R2) I recall, but I forget where to put "n" the index of refraction, in this to make an equation. I think (1/nf) = .... is correct as if right hand side is fixed and n is increased, then I know focal length f is reduced. I.e. product nf remains unchanged, (I think) if the two radii, R1 & R2, remain unchanged.
Thanks James. I did. My prior post's n should be n-1 assuming that the index of air is 1, which it almost is. (I almost never take the time to Google etc but with your clue, I quickly found how to correct my guess / memory. Frankly, I was surprised I remembered as well as I did after all these years.)
