What is the power of the lens in diopters?

AI Thread Summary
To determine the power of a lens in diopters, the problem involves magnifying reading material by a factor of 3.5× with the book placed 9.5 cm behind the lens. The relationship between object distance (do), image distance (di), and magnification (M) is used to derive di as -33.25 cm. The formula 1/f = 1/di + 1/do is applied to find the focal length, leading to a calculation of 1/f = 0.075. Clarification is needed regarding the sign conventions for di and the units of focal length in diopters, which should be in meters.
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Homework Statement


It is desired to magnify reading material by a factor of 3.5× when a book is placed 9.5 cm behind a lens.
What is the power of the lens in diopters?

Homework Equations


1/f = 1/di + 1/do
M = -di/do
P=1/f

The Attempt at a Solution


I set equal the magnification to -di/do and solved for di
3.5=-di/9.5
di = -33.25 cm
Then I plugged into equation to find 1/f which = P
1/f = 1/di + 1/do
1/f = 1/-33.25 + 1/9.5
1/f = 0.075
I don't understand why this isn't right??
 
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What are the units of f in 1/f when you want 1/f to be in diopters??
 
(If 1/f is in diopters, f is in meters).

The magnification is negative; di is positive. If you do that, it should work out.
 
Go with Gene N. but also pay attention to what kuruman said. Together they have the answer!
 
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