Hyperopia, Far Sightedness With hyperopia, the focal point is behind the retina, shouldn't this mean that the image is flipped on the retina itself from what is usual? 
I must be drawing my ray diagrams wrong.
A little something different from: Optics, lenses and our eyes
 A: Well the image is flipped on the retina (your brain fixes that), but that doesn't change if you go slightly in front or behind the focus. 
The rays that you draw in paraxial approximation ( parallel becomes focal, focal becomes parallel , center stays center)  are simply not all in one point if you move your detection plane slightly away from the perfect image plane, so there is a larger 'circle of confusion' or blur.
A: Here are ray diagrams that show what is going on.  In the top case, a weak (thin) lens doesn't have the power to pull the rays together tight enough.  An object farther away than the tree would make rays converge on the retina.  This is farsightedness.
Remember the fundamental formula for thin lenses (using some appropriate sign convention):
$$
{1\over f}={1\over D_1}+{1\over D_2}
$$
If $D_1$ increases, then $D_2$ must decrease.  Thus rays of farther away objects converge in a shorter distance after the lens, hopefully on the retina.
In the second diagram, the lens is of proper thickness, and each part of the tree and nearby paraphernalia focus on the retina.
Just for comparison, the bottom diagram shows a lens too strong (thick) causing rays to converge too quick, before reaching the retina.  To meet at the retina, a slightly longer distance, the outside world object must decrease its distance - this is nearsightedness.
Note that in all cases, the tree image on the retina is upside down. The only difference is if it's sharp or blurry.

