00$ diopters and they sit 2.00 cm in front of the eyes. This means that the person's far point is at a distance of $f = \frac{1}{P}$, where $P$ is the power of the eyeglasses. Substituting the given values, we get $f = \frac{1}{-4.00 \, \text{diopters}} = -0.25 \,
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