Question

Unless otherwise stated, the lens-to-retina distance is $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$.

What is the power of the eye when viewing an object $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$away?

Short answer

The power of the eye when viewing an object $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ away is, $\mathbf{P}\mathbf{=}\mathbf{+}\mathbf{52}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{D}$.

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

$$\begin{gathered} \mathbf{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{\circ }}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{i}}} \end{gathered}$$

Wherelocalid="1654081131156" ${\mathbf{d}}_{\mathbf{\circ }}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{d}}_{\mathbf{i}}$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,

$$\begin{gathered} {\mathbf{d}}_{\mathbf{\circ }}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}\mathbf{\left(}\frac{\mathbf{1}\mathbf{}\mathbf{m}}{\mathbf{100}\mathbf{}\mathbf{cm}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{m}\mathbf{} \\ \mathrm{The}\mathrm{lens}-\mathrm{to}-\mathrm{retina}\mathrm{distance}\mathrm{is}: \\ {\mathbf{d}}_{\mathbf{\circ }}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}\mathbf{}\mathbf{\left(}\frac{\mathbf{1}\mathbf{}\mathbf{m}}{\mathbf{100}\mathbf{}\mathbf{cm}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{m}\mathbf{} \end{gathered}$$

Calculation for the power

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

$$\begin{gathered} \mathbf{P}\mathbf{=}\left(\frac{1}{d_{\circ }}+\frac{1}{d_i}\right) \\ \mathbf{}\mathbf{}\mathbf{=}\left(\frac{1}{0.5m}+\frac{1}{0.02m}\right) \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{+}\mathbf{52}\mathbf{}\mathbf{D} \end{gathered}$$

Therefore, the value of power is obtained as$\mathbf{P}\mathbf{=}\mathbf{+}\mathbf{52}\mathbf{}\mathbf{D}$