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 ’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:
\[ \frac{1}{V} – \frac{1}{u} = \frac{1}{f}\]
\[ \frac { 1 } { \infty } – \frac {1}{u} = \frac{1}{0.5}\]
\[ u = -0.5 m \]
Rachel’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:
\[ \frac{ 1 }{ V } – \frac { 1 }{ u } = \frac{ 1 }{ f } \]
\[\frac { 1 }{\infty} – \frac {1}{u} = \frac {1}{0.33} \]
\[u = -0.33 m \]
Adam’s far point is $0.33 m$.
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