 # A severely myopic patient has a far point of 10.00 keche0b 2021-12-14 Answered
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?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it 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 $=60-50=10D$
###### Not exactly what you’re looking for? 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
${p}_{c\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}rected}=60D-50D$
$=10D$