Walter Clyburn

2022-01-03

Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. When she wears +2.0 D reading glasses, what is her near point? Her far point?

Bertha Jordan

Beginner2022-01-04Added 37 answers

Given

Power =+2D

Initial near point =0.6m

a) power$\left(P\right)=\frac{1}{f}=+2D,\text{}V=-0.6\text{}m$

Here, using the lens equation as following below,

$\frac{1}{V}-\frac{1}{u}=\frac{1}{f}$

$-\frac{1}{0.6}-\frac{1}{u}=2$

$u=-0.27\text{}m$

Her near point is 0.27 m

b) Now, her new near point$V=\mathrm{\infty}$

$P=\frac{1}{f}$

$2=\frac{1}{f}$

$f=\frac{1}{2}m$

$f=0.5\text{}m$

Now, using the lens equation,

$\frac{1}{V}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{\mathrm{\infty}}-\frac{1}{u}=\frac{1}{0.5}$

$u=-0.5m$

Raches far point is 0.5 m

Power =+2D

Initial near point =0.6m

a) power

Here, using the lens equation as following below,

Her near point is 0.27 m

b) Now, her new near point

Now, using the lens equation,

Raches far point is 0.5 m

22+64

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