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:

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

Power of the eye when a correction lens is used:

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

The remote point s_{o} goes to infinity, thus 1/s_{o} becomes zero.

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

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Thin Lens And Lens Maker Equations concept. You can view video lessons to learn Thin Lens And Lens Maker Equations. Or if you need more Thin Lens And Lens Maker Equations practice, you can also practice Thin Lens And Lens Maker Equations practice problems.