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Q13PE

Expert-verifiedFound in: Page 956

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**The power of a physician’s eyes is 53.0 D while examining a patient. How far from her eyes is the feature being examined?**

The feature is 33.3 cm from the physician's eyes.

**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.**

Power of the physician’s eyes P = 53.0 D.

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

The power of the eyes of a normal person can be calculated as,

$P=\left(\frac{1}{{d}_{\circ}}+\frac{1}{{d}_{i}}\right)\phantom{\rule{0ex}{0ex}}53.0D=\left(\frac{1}{0.02m}\right)+\frac{1}{{d}_{\circ}}\phantom{\rule{0ex}{0ex}}\frac{1}{{d}_{\circ}}=3.0D$

Substituting data values in the above expression gives,

${d}_{\circ}=\frac{1}{3.0D}\left(\frac{1D}{1{m}^{-1}}\right)\phantom{\rule{0ex}{0ex}}=0.334m\left(\frac{100cm}{1m}\right)\phantom{\rule{0ex}{0ex}}=33.4cm$

Therefore, the feature is 33.3 cm from the physician's eyes.

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