This question aims to find the near point and far point of Rachel when she wears +2.0 D reading glasses. Rachel has a good distant vision but she has a touch of presbyopia. Her near point is 0.60 m.

The **maximum distance** at which the eyes can see things properly is called the **far point** of the eye. It is the farthest point at which an image is formed on the retina within the eye. The normal eye has a far point equal to infinity.

The **minimum distance** where an eye can focus and creates the image on the retina is called the **near point** of an eye. The range of an eye at which it can see a closely-placed object is the near point of an eye. The distance of a normal human eye is 25 cm.

**Presbyopia** is an eye condition in which the eye focus gets blurred. Blurry images are formed by the retina. It is most commonly present in **adults** and this condition becomes worse after the â€™40s.

The **power of the lens** is the ability of the lens to bend the light falling on it. If the light entering the lens has a **shorter wavelength**, then it means the lens will have more power.

## Expert Answer

According to the given data:

Power = $ +2D $

The near point without glasses is $ 0.6 m $:

\[ ( P ) = \frac { 1 } { f } = + 2D , V = – 0.6 m \]

Where $P$ is the power of the lens, $f$ is the **focal length** of the lens, $u$ is the **object-distance** for the first lens, and $v$ is the object-distance for the second lens.

By using the equation for lens, we get:

\[\frac{1} {V} – \frac {1}{u} = \frac{1}{f}\]

By putting values in the equation:

\[\frac {-1}{0.6} – \frac {1}{u} = 2 \]

\[ u = – 0.27 m \]

**The near point of Rachel is $-0.27 m$.**

To find the far point, $V$ = $\infty$ :

\[P = \frac {1}{f} \]

\[2 = \frac {1}{f} \]

\[f = \frac {1}{2} \]

\[ f = 0.5 m \]

## Numerical Solution

By using the lens equation, we get:

\[ \frac{1}{V} – \frac{1}{u} = \frac{1}{f}\]

\[ \frac { 1 } { \infty } – \frac {1}{u} = \frac{1}{0.5}\]

\[ u = -0.5 m \]

**Rachelâ€™s far point is $0.5 m$.**

## Example

Find the far point if Adam wears reading glasses of $+3.0 D$.

To find the far point, $V$ = $\infty$ :

\[ P = \frac {1}{f}\]

\[ 3 = \frac{1}{f}\]

\[ f = 0.33 m \]

By using the lens equation, we get:

\[ \frac{ 1 }{ V } – \frac { 1 }{ u } = \frac{ 1 }{ f } \]

\[\frac { 1 }{\infty} – \frac {1}{u} = \frac {1}{0.33} \]

\[u = -0.33 m \]

**Adamâ€™s far point is $0.33 m$.**

*Image/Mathematical drawings are created in Geogebra. *