Near and far points with contact lenses

[grouper]

Homework Statement

A nearsighted person has near and far points of 11.0 and 22.0 cm , respectively. If she puts on contact lenses with power P = -4.00 D, what are her new near and far points?

Homework Equations

P=1/f

1/f=1/do+1/di where f=focal point, do=distance to object, di=distance to image

The Attempt at a Solution

The focal length of the lens is -0.25 m (and is a diverging lens, meaning the person is nearsighted). To find the new near point, the distance to the image should be at the naked-eye near point (0.11 m), so 1/-0.25=1/do+1/0.11 makes d0=-7.39 cm the new near point.

I was less sure how do approach the new far point, but I assumed that do=∞ (as it should be in a properly working naked eye) so that 1/-0.25=1/∞+1/di, which makes di=-0.25.

One or both of these answers is incorrect (they are graded together so I can't tell which one) but I have a feeling I'm way off on the far point. I think I'm on the right track with the near point but 7.39 cm seems really close, I think that's better than a normal naked eye. Any suggestions would be appreciated! Thanks.

 

[grouper]

I figured this one out; don't worry about it.

 

[Bob Dole]

I still can't figure it out!

Where is the near point of an eye for which a contact lens with a power of 2.30 diopters is prescribed? (Assume that near point for an average viewer is 25 cm.)
s' =0.588 m
Correct
Part B
Where is the far point of an eye for which a contact lens with a power of -1.20 diopters is prescribed for distant vision?
s' = m

I tried using the same equation I did for part A where Power = 1/f, so f = -0.8333333

However with 1/do + 1/di = 1/f I do not have a value of di and if I use 0.25 m I end up with new near point = -0.1923. I have no clue how to get the far point. However, I will continue working on it and maybe if someone else is stuck on this same problem this can help.

EDIT
It occurred to me that di = infinity so 1/do = 1/f!