How Does Lens Power Correction Work for Nearsightedness?

[sixpence]

I'm having trouble deciding what to use as the object, p, and the image, q. For example here's a question:

An individual is nearsighted; his near point is 13cm and far point is 50cm. What lens power is needed to correct his nearsightedness? When lenses are in use, what is this person's near point?

so i got 1/infinity + 1/-.5 = 1.f

1/-.5 = 1/-.13 +1/q

these are the right answers but they don't make sense to me. How would the focal length from the far point have anything to do with wearing glasses. I also don't know when to make things negative or keep it positive, but that's mainly b/c I don't know what's the image and what's the object. I don't know if the questions are just worded poorly but I can't seem to get them. Is the near point always the object or can it be object and image? This is frustrating meeeeeee

 

[G01]

The near point is the closest something can be so that the person can still see it clearly. The far point is the farthest away something can be so that the person can still see it.

Since the person can't see any farther away than the far point, we want to make glasses from lens that will take objects which are really far away (object distance =infinity) and put the image at the far point.

So, the object distance in this problem is infinity. The image point will be the near point. Does this clear things up?