Question

The near point for a person's eye is 85.0 cm. To see objects held as closely as 25.0 cm from the eye, what power contact lenses should be worn?

Answer 1

Answer:

2.82 D

Explanation:

Near point of the person eye is 85 cm that is $u=85\ cm$

He wants to see the object at 25 cm so $v=25\ cm$

For the lens we know that $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}=\frac{1}{25}-\frac{1}{85}$

\frac{1}{f}=0.04-0.01170=0.02823

$f=35.41\ cm$

We know that power of the lens $P=\frac{1}{f\ in\ meter}$

So $P=\frac{1}{0.3541}=2.8235\ D$