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### College Physics (Urone)

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.

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

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

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

## Step 3: How far from her eyes is the feature being examined

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.