How the eye lens changes with distance

The retina is $2.5\ \text{cm}$ behind the eye lens, so a far object needs $40\ \text{D}$. A nearer object at $d$ needs more: $P = 40\ \text{D} + \dfrac{100}{d}$, with $d$ in centimetres. You will make the program find the power for five distances, the extra power the nearest one needs, and the near point of an eye whose lens can reach at most $42.5\ \text{D}$. Then it draws the power against the distance.

Your tasks.

1. Inside the loop, work out P, the power needed for the distance d, in dioptres.
2. Find extra, how much more power the last (nearest) distance needs than the first (farthest).
3. Find near, the near point in centimetres of an eye that can reach at most $42.5\ \text{D}$.