# Problem: A severely myopic patient has a far point of 5.75 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him? (Assume a lens-to-retina distance of 2.00 cm.)

###### FREE Expert Solution

Power of the lens:

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

$\begin{array}{rcc}\mathbf{P}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{5}\mathbf{.}\mathbf{75}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}\end{array}$

###### Problem Details

A severely myopic patient has a far point of 5.75 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him? (Assume a lens-to-retina distance of 2.00 cm.)