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## 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.