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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!