an eye can be fully corrected by a -5.50 D contact lens. if the power of the eye is +63.50 D, what is the distance from the first focal point of the correcting lens to the retina?
1.333 as refractive index inside eye

So I made the primary medium the air outside the eye and the secondary medium is the inside of the eye with my refractive surface being the eye.

So I know in order to find the primary or secondary focal points, you find the point at which parallel light rays converge after passing through a refractive surface.
Using formulas F' = n'/f' and F = -n/f

well because the eye is corrected by a -5.50 lens, then the power is +58.00 and being corrected means their far point is at infinity, which means light rays entering the eye are parallel. Consequently the distance from the surface of the eye to the retina can be found with f' = 1.333/58.00 D = +0.02298 m

Now how do I find the primary focal point? is it correct to just plug into the formula? f = -1.000/58.00 D = -0.01724 m

So the total distance between them is 0.04022 m?

But for some reason, a classmate said the answer should be 0.15883 m?

e: I think he got 0.15883 because that last sentence that says primary focal point of the correcting lens, meaning the primary focal point of a -5.50 D lens, which would be -(1.000/-5.50) or +0.18181 so that both distance are to the right of the surface of the eye, meaning the distance between the primary focal point and the retina would be 0.15883

This post was edited by bulkathos_son on Oct 14 2013 06:01pm