# Homework Help: New Near Point with Corrective Lens?

1. Nov 21, 2016

### givemeknowledge

1. The problem statement, all variables and given/known data
So I've calculated the person's corrective lens to need the power of -2.00D now I must find the new near point with the corrective lens on.

The initial near point is 15cm and I converted the -2.00D in to cm, which is -50cm (I hope).

2. Relevant equations
I've used this formula -
1/f = 1/v + 1/u

v = image distance

u = object distance

3. The attempt at a solution
My attempt so far looks like this... I have an answer but need someone more experienced to tell me if it seems right:

1/f = 1/0.019m + 1/-0.35m = 49.7m

So would the new near point be 49.7m?

Any help would be really appreciated, thanks!

2. Nov 21, 2016

The nearpoint without glasses is 15 cm. With the glasses on (assuming they near the eye), you need to find the object distance that gives a virtual image (right side up) from this corrective lens at m=-15 cm. You are correct in using $1/f=1/b+1/m$ with $f=-50$ cm. Your "b" will be positive. (editing this=originally, I thought the focal length of the corrective lens was positive.) It appears the person is nearsighted and needs glasses to help him see distant objects. In the process, it will move his nearpoint farther out. (Note: I don't know where your .019 and -.35 came from.)