Power of a lens:

$\overline{){\mathbf{D}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{f}}}$

Lens maker equation:

$\overline{)\frac{\mathbf{1}}{{\mathit{s}}_{\mathit{o}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathit{s}}_{\mathit{i}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathit{f}}}$

Consider a person whose eyes have a fully accommodated power of 56.5 D. Assume the distance from the lens to the retina is a fixed 2.00 cm

What is the near point of that person in cm?

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