Question

A farsighted person breaks her current eyeglasses and is using an old pair whose refractive power is 1.550 diopters. Since these eyeglasses do NOT completely correct her vision, she must hold her phone 44.0 cm from her eyes in order to read it. She wears the eyeglasses 2.0 cm from her eyes. How far is her near point from her eyes?

Answer 1

122.4 cm

Step-by-step explanation:

`d_(p)` = distance of phone from eye = 44 cm

`d_(e)` = distance of eyeglasses from eye = 2.0 cm

`d_(o)` = Object distance =
`d_(p)` -
`d_(e)` = 44 - 2 = 42 cm

P = Power of the eyeglasses = 1.55 diopter

focal length of eyeglass is given as

`f = (1)/(P)`

`f = (100)/(1.55)`

f = 64.5 cm

`d_(i)` = image distance

using the lens equation

`(1)/(d_(o)) + (1)/(d_(i)) = (1)/(f)`

`(1)/(42) + (1)/(d_(i)) = (1)/(64.5)`

`d_(i)` = - 120.4 cm

`d_(n)` = distance of near-point

distance of near-point is given as

`d_(n)` = |
`d_(i)`| +
`d_(e)`

`d_(n)` = 120.4 + 2

`d_(n)` = 122.4 cm