- #1

givemeknowledge

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## 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!