# A severely myopic patient has a far point of 10.00

A severely myopic patient has a far point of 10.00 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him?
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Linda Birchfield
Step 1
Given:
Far point $=10cm$
Step 2
The power of the normal eye is given by
$P=\left(\frac{1}{f}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{v}\right)$
$u=\text{object distance}=\text{far point of a normal human eye is infinity}$
$v=\text{image distance}=\text{the image is formed on the retina which is about 2 cm from the lens}$
$⇒P=\left(\frac{1}{\mathrm{\infty }}\right)+\left(\frac{1}{0.02}\right)=50D$
The power of the myopic eye is given by
$P=\left(\frac{1}{f}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{v}\right)$
$u=\text{object distance}=\text{far point of the myopic eye}=10cm$
$v=\text{image distance}=\text{the image is formed on the retina which is about 2 cm from the lens}$
$⇒P=\left(\frac{1}{0.1}\right)+\left(\frac{1}{0.02}\right)=60D$
Reduction in power $=60-50=10D$
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temnimam2
Step 1
The power of severely myopic eye ${p}_{myo\pi c}$ can be obtained,
${P}_{myo\pi c}=\frac{1}{{d}_{0}}+\frac{1}{{d}_{i}}$
Step 2
The quantity of dioptres ${p}_{c\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}rected}$ is,
${P}_{c\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}rected}={p}_{myo\pi c}-{p}_{‖a‖l}$
Step 3
Thus, ${P}_{myo\pi c}=\frac{1}{10.00×{10}^{-2}m}+\frac{1}{2.00×{10}^{-2}m}=60.0D$
Step 4
The power for the normal is 50D thus the corrected power is
${p}_{c\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}rected}=60D-50D$
$=10D$