Question

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

Calculate the power of the eye when viewing an object 3.00 maway.

Short answer

The power of the eye when viewing an object 3.00 m away is, P = +50.34 D.

Step-by-step solution

Concept Introduction

The thin lens equations can be used to examine the image formation by the eye quantitatively. The power of the lens is equal to the reciprocal of the focal length,

$\mathit{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\left(\frac{1}{d_{\circ }}+\frac{1}{d_i}\right)$

Whereis 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,$d_{\circ }=3.0m$

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

Calculation for power

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

$$\begin{gathered} P=\left(\frac{1}{d_{\circ }}+\frac{1}{d_i}\right) \\ =\left(\frac{1}{3m}+\frac{1}{0.02m}\right) \\ =+50.34D \end{gathered}$$

Therefore, the value of power is obtained as $P=+50.34D$.