Question

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.660 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes. How far is her near point from her eyes?

• A. 75 cm
• B. 125 cm
• C. 225 cm
• D. 121.05 cm

Answer 1

Answer: (D)

121.05 cm

The refractive power of the lens is $1.660$ dipters. So the focal length of the lens is $f=\frac{1}{1.660}=0.6024m=60.24cm$

The distance between the newpaper and her eyes is $42.00cm$

The distance between her eyes and glasses is $2.00cm$

So, the distance between the lglasses and newspaper is $(42.00-2.00)cm=40.00cm.$ That is, $d_0=40.00cm$

The distance $\left(d_i\right)$ between the glasses and the virtual image formed by the lens given by,

$d_i=\frac{f{d}_0}{d_0-f}=\frac{60.24-40.00}{40.00-60.24}=-119.05cm$

This relation is obtained from thin lens equation and the negative sign implies the image is virtual.

This position is her near point.

So the near point from her eyes is, $119.05cm+2.00cm=121.05cm$