Answer
a) R=$6.48*10^{-3}$ meters
b) No, the image would focus behind the retina.
c) $s^{'}=0.0193$ meters, the image would focus in front of the retina.
Work Step by Step
a) Use the formula $(n_{a}/s)+(n_{b}/s^{'})=(n_{b}-n_{a})/R$ and solve for R. Because the object is at infinity, the term $n_{a}/s$ becomes zero. This leaves us with $R=(n_{a}-n_{b})*(s^{'}/n_{b})$. The given values are $n_{b}=1.35$, $n_{a}=1$, and $s^{'}=0.025$ meters. Plugging those values in gives us a final answer of $R=6.48*10^{-3}$ meters.
b) Because we are looking at something closer than infinity the rays that hit the lens are not parallel like they would be if the object was at infinity. This means that the image will be formed past the focal point, therefore the image will be formed behind the retina.
c) We are now given a value of $R=0.005$ meters and are asked to solve for $s^{'}$. We use the same equation as in part a, this time we just solve for something different. Rewrite the equation as $s^{'}=n_{b}*R/(n_{b}-n_{a})$. Plug in the values given earlier and solve, $s^{'}=0.0193$ meters. This is less than the diameter of the eyeball so the image is formed in front of the retina.
