Finding radius of curvature of an eyeball

1. The problem statement, all variables and given/known data

Consider a simplified model of the human eye, in which all internal elements have the
same refractive index of n = 1.40. Furthermore, assume that all refraction occurs at the
cornea, whose vertex is 2.50 cm from the retina. Calculate the radius of curvature of the
cornea such that the image of an object 40.0 cm from the vertex of the cornea is focussed
on the back of the eye (the retina).

2. Relevant equations

1/s+1/s'=1/f

1/f=(n-1)(1/R_1-1/r_2)

3. The attempt at a solution

I attempted to find the focal point with

1 / -40 + 1 / 2.5 = 0.375cm
With that I figured I should use the lensmaker equation but I've never seen a problem where you have to solve the radius of curvature and couldn't find any examples like this online.

I changed the equation to 1/f=(n-1)R and solved for R but I'm not sure if that would be the right thing to do.

Thanks

For a start
You have the wrong value for the focal length of the eye lens.

The equation is 1/40 + 1/2.5 = 1/f

In the real is positive convention, all lengths here are positive.

oh yeah. so f = 40/17

Correct.

The lensmakers formula you've used is the thin lens approximation.
Is this eye lens thin?
Try the one for a thick lens here
http://en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation

im not sure how i'd find r_1 and r_2 just given the focal length and the refractive index