New Near Point with Corrective Lens?

[givemeknowledge]

Homework Statement

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 into cm, which is -50cm (I hope).

Homework Equations

I've used this formula -
1/f = 1/v + 1/u

v = image distance

u = object distance

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!

 

[Charles Link]

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.)