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?

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Linda Birchfield

Answered 2021-12-15
Author has **39** answers

Step 1

Given:

Far point$=10cm$

Step 2

The power of the normal eye is given by

$P=\left(\frac{1}{f}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{v}\right)$

$u=\text{object distance}=\text{far point of a normal human eye is infinity}$

$v=\text{image distance}=\text{the image is formed on the retina which is about 2 cm from the lens}$

$\Rightarrow P=\left(\frac{1}{\mathrm{\infty}}\right)+\left(\frac{1}{0.02}\right)=50D$

The power of the myopic eye is given by

$P=\left(\frac{1}{f}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{v}\right)$

$u=\text{object distance}=\text{far point of the myopic eye}=10cm$

$v=\text{image distance}=\text{the image is formed on the retina which is about 2 cm from the lens}$

$\Rightarrow P=\left(\frac{1}{0.1}\right)+\left(\frac{1}{0.02}\right)=60D$

Reduction in power$=60-50=10D$

Given:

Far point

Step 2

The power of the normal eye is given by

The power of the myopic eye is given by

Reduction in power

temnimam2

Answered 2021-12-16
Author has **36** answers

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$p}_{c{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}rected$ is,

$P}_{c{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}rected}={p}_{myo\pi c}-{p}_{\Vert a\Vert l$

Step 3

Thus,${P}_{myo\pi c}=\frac{1}{10.00\times {10}^{-2}m}+\frac{1}{2.00\times {10}^{-2}m}=60.0D$

Step 4

The power for the normal is 50D thus the corrected power is

${p}_{c{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}rected}=60D-50D$

$=10D$

The power of severely myopic eye

Step 2

The quantity of dioptres

Step 3

Thus,

Step 4

The power for the normal is 50D thus the corrected power is

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