A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.570 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

Explanation:

The refractive index of the lens is

d = 1.570 dioptres

Therefore, her focal length is

f = 1 / d

f = 1 / 1.570

f = 0.6369 m

f = 63.69cm

The distance between the newspapers and her eyes is 42cm

Distance from her eye to the glasses is 2cm

We want to calculate her near point?

So, the distance between the newspaper and the glass

do = 42 - 2 = 40cm

The distance between the glass and the virtual image formed is given as

di = f•do / (do —f)

di = 63.69 × 40 / (40 — 63.69)

di = 2547.6cm / -23.69

di = —107.54cm

The relation use is obtain from the lens equation and the negative shows that virtual image formed.

The near point from her eye is

107.54 + 2 = 109.54cm

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