Suggested languages for you:

Americas

Europe

Q11PE

Expert-verifiedFound in: Page 956

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.

**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.**

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 P_{myopic} can be obtained using:

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

And, the number of diopters P_{Corrected} can be obtained using,

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

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.

94% of StudySmarter users get better grades.

Sign up for free