# Problem: Ellen wears eyeglasses with the prescription -1.0D. What is her far point without the glasses?

###### FREE Expert Solution

Power of a lens:

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

We'll substitute for 1/f in the lens equation with D.

Power of the naked eye:

$\overline{)\frac{\mathbf{1}}{{\mathbf{s}}_{\mathbf{i}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{s}}_{\mathbf{o}}}{\mathbf{=}}{{\mathbit{D}}}_{{\mathbf{0}}}}$

Power of the eye when a correction lens is used:

$\overline{)\frac{\mathbf{1}}{{\mathbf{s}}_{\mathbf{i}}}{\mathbf{=}}{{\mathbit{D}}}_{{\mathbf{0}}}{\mathbf{+}}{\mathbit{D}}}$

The remote point so goes to infinity, thus 1/so becomes zero.

###### Problem Details

Ellen wears eyeglasses with the prescription -1.0D. What is her far point without the glasses?