Question

The cornea, a boundary between the air and the aqueous humor, has a 3.0 cm focal length when acting alone.

What is its radius of curvature?

Answer 1

Since the 1/s+1/s'=f relationship applies to image-object distances to the lens with the same medium (e.g., air) on both sides, I think you have to modify it for application to the eye, where the internal focal distance in the vitreous will be greater, by a factor close to n. If exactly by n, you'll end up with R = 1.02 cm.
By simple refraction, convergent rays become less convergent going from air to a medium of higher n, by a factor n*cos? that approaches n as incident convergence angle ? approaches 0; thus the longer focal distance. However, parallel rays remain parallel. Thus an object in air at s = f would produce internal parallel rays so s' = infinity.

Therefore another form of the equation could be 1/3 cm = (1.34-1)/R ==> R = 1.02 cm.

This all assumes f is the air-to-air focal length f(air) of the lens. If it's the internal FL f(int), then f(air) would be calculated as f(int)/n2.