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Found in: Page 956

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A severely myopic patient has a far point of 5.00 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him?

The power of his eye must be reduced by 20.0 D.

See the step by step solution

## Step 1: Definition of dioptre lens

The unit of measurement for the optical power of a lens or curved mirror is the dioptre of power of a lens, which is equal to the reciprocal of focal length.

## Step 2: Given information and Formula to be used

The lens-to-retina distance is, ${d}_{i}=2.00cm\left(\frac{1m}{100cm}\right)=0.02m$.

The far point for the myopic patient is, ${d}_{\circ }=5.00cm\left(\frac{1m}{100cm}\right)=0.05m$.

The power of a normal person is P = 50 D

The power of the severely myopic eye Pmyopic can be obtained using:

${P}_{myopic}=\frac{1}{{d}_{0}}+\frac{1}{{d}_{i}}$...............(1)

And, the number of diopters PCorrected can be obtained using,

${P}_{Corrected}={P}_{Myopic}-{P}_{Normal}$.................(2)

## Step 3: By how many diopters should the power be reduced?

Substitute values in equation (1) and it gives,

${P}_{myopic}=\left(\frac{}{0.05m}\right)+\left(\frac{1}{0.02m}\right)\dots \dots \dots \dots \dots \dots \left(3\right)\phantom{\rule{0ex}{0ex}}=70D$

Substituting data from equation (3) in equation (2), which gives,

${P}_{Corrected}=70D-50D\phantom{\rule{0ex}{0ex}}=20D$

Therefore, the power of the eye must be reduced by 20 D.