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 :

Answer:

2.82 D

Explanation:

Near point of the person eye is 85 cm that is [tex]u=85\ cm[/tex]

He wants to see the object at 25 cm so [tex]v=25\ cm[/tex]

For the lens we know that [tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

[tex]\frac{1}{f}=\frac{1}{25}-\frac{1}{85}[/tex]

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

[tex]f=35.41\ cm[/tex]

We know that power of the lens [tex]P=\frac{1}{f\ in\ meter}[/tex]

So [tex]P=\frac{1}{0.3541}=2.8235\ D[/tex]

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