Question

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

What is the near point of a person whose eyes have an accommodated power of 53.5 D?

Short answer

The near point of a person whose eyes have an accommodated power of 53.5 D is, d₀= 0.286 m.

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,

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

Where d₀and d_i is the distance between the object and eye as well as the distance from the lens to the retina.

Information Provided

Power of the relaxed eyes, P = 53.5 D.

The lens-to-retina distance is, d₀= 0.286 m.

Calculation for distance

Rearrange equation (1) and solve for the distance between the far point and the person's eyes,

$\frac{1}{d_{\circ }}=P-\frac{1}{d_i}$

Substitute values in the above expression,

$$\begin{gathered} \frac{1}{d_{\circ }}=53.5D-\frac{1}{0.02m} \\ =53.5D-50.0D \\ =3.5D\left(\frac{1m^{-1}}{1D}\right) \\ =3.5m^{-1} \end{gathered}$$

This gives, d₀ = 0.286 m from the above expression.

Therefore, the distance of the object is d₀ = 0.286 m.