Chapter 26: Q1PE (page 956)

Unless otherwise stated, the lens-to-retina distance is $2.00 cm$ .
What is the power of the eye when viewing an object $50.0 cm$ away?

Short Answer

Expert verified

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

Step by step solution

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

$P = \frac{1}{f} = \frac{1}{d_{\circ}} + \frac{1}{d_{i}}$

Wherelocalid="1654081131156" $d_{\circ} and d_{i}$ is the distance between the object and eye as well as the distance from the lens to the retina.

Information Provided

The distance between the object and the eye is,

$d_{\circ} = 50.0 cm (\frac{1 m}{100 cm}) = 0.5 m The lens - to - retina distance is : d_{\circ} = 2.00 cm (\frac{1 m}{100 cm}) = 0.02 m$

Calculation for the power

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

$P = (\frac{1}{d_{\circ}} + \frac{1}{d_{i}}) = (\frac{1}{0.5 m} + \frac{1}{0.02 m}) = + 52 D$

Therefore, the value of power is obtained as $P = + 52 D$

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Unless otherwise stated, the lens-to-retina distance is $2.00 cm$ .
What is the power of the eye when viewing an object $50.0 cm$ away?

Short Answer

Step by step solution

Concept Introduction

Information Provided

Calculation for the power

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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

Step by step solution

Concept Introduction

Information Provided

Calculation for the power

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Force

Solid State Physics

Work Energy and Power

Physical Quantities And Units

Famous Physicists

Nuclear Physics

Study anywhere. Anytime. Across all devices.

Unless otherwise stated, the lens-to-retina distance is $2.00 cm$ .
What is the power of the eye when viewing an object $50.0 cm$ away?