Elementary Corrective Lens Problem

[EstimatedEyes]

Homework Statement

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.655 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 39.4 cm from her eyes in order to read it. She wears the eyeglasses 1.90 cm from her eyes. How far is her near point from her eyes?

Homework Equations

1/f = 1/(do) + 1/(di)

di < 0 for virtual images (what this is)

The Attempt at a Solution

do is .394m and 1/f=1.655 diopters

so 1/(1/.394m - 1.655) = di

di = 1.13m

Near Point = 1.13m + .019m = 1.15m

Where did I go wrong? Thanks!

 

[alphysicist]

Hi EsimatedEyes,

Homework Statement

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.655 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 39.4 cm from her eyes in order to read it. She wears the eyeglasses 1.90 cm from her eyes. How far is her near point from her eyes?

Homework Equations

1/f = 1/(do) + 1/(di)

di < 0 for virtual images (what this is)

The Attempt at a Solution

do is .394m

What is the variable do? It is called the object distance, but what is it? In particular, what two points is it measured between? Once you answer that, I think you'll see what do needs to equal here.

 

[EstimatedEyes]

I just figured out that the distance was from the lens so I kept getting it wrong because I was using the eyes value that they gave.