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Chapter 26: Q1PE (page 956)

Unless otherwise stated, the lens-to-retina distance is 2.00cm.

What is the power of the eye when viewing an object 50.0cmaway?

Short Answer

Expert verified

The power of the eye when viewing an object 50.0cm away is, P=+52.0D.

Step by step solution

01

Concept Introduction

Equation of the thin lens can be used to examine the image formation by the eye quantitatively.

The power of the lens can be estimated from the reciprocal of the focal length

P=1f=1d鈭�+1di

Wherelocalid="1654081131156" d鈭�anddiis the distance between the object and eye as well as the distance from the lens to the retina.

02

Information Provided

The distance between the object and the eye is,

d鈭�=50.0cm(1m100cm)=0.5mThelens-to-retinadistanceis:d鈭�=2.00cm(1m100cm)=0.02m

03

Calculation for the power

The expression for the power of the eye is given as,

P=(1d鈭�+1di)=(10.5m+10.02m)=+52D

Therefore, the value of power is obtained asP=+52D

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Most popular questions from this chapter

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

If the cornea is to be reshaped (this can be done surgically or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain. Also explain how hyperopia can be corrected.

It has become common to replace the cataract-clouded lens of the eye with an internal lens. This intraocular lens can be chosen so that the person has perfect distant vision. Will the person be able to read without glasses? If the person was near-sighted, is the power of the intraocular lens greater or less than the removed lens?

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