Finding radius of curvature of an eyeball

[navm1]

Homework Statement

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).

Homework Equations

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

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

The Attempt at a Solution

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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

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[rude man]

I attempted to find the focal point with

1 / -40 + 1 / 2.5 = 0.375cm

Why is "40" prefixed by a minus sign?

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.

Make it 1/f = (n-1)/R and you got a deal. It assumes the eye lens is plano-convex, i.e. only one surface is curved. That seems to be the intent of the problem (it mentions only one radius of curvature). In reality the cornea is convex-convex, though.