Question

A person's prescription for her new bifocal glasses calls for a refractive power of -0.500 diopters in the distance-vision part, and a power of 1.75 diopters in the close-vision part. What are the near and far points of this person's uncorrected vision? Assume the glasses are 2.00 cm from the person's eyes, and that the person's near-point distance is 25.0 cm when wearing the glasses.

Answer 1

Far point = 202 cm

Near point = 38.5 cm

Step-by-step explanation:

For distance vision we know that

`P = -0.500 diopter`

we know that

`P = (1)/(F)`

`F = (1)/(-0.500)`

`F = -2 m`

now by lens formula we know that

`(1)/(d_i) + (1)/(d_o) = (1)/(F)`

so we have an object positioned at infinite we have

`(1)/(d_i) + 0 = (1)/(2)`

`d_i = 200 cm`

now distance from eye is given as

`d = 202 cm`

Now similarly for near point given at 25 cm from eye

the distance of the image should be 25 - 2 = 23 cm

power for near point is given as

`P = (1)/(F)`

`F = (1)/(1.75)`

`F = 0.57 m`

now again by lens formula

`(1)/(d_i) + (1)/(d_o) = (1)/(F)`

`(1)/(d_i) + (1)/(23) = (1)/(57)`

`d_i = 38.5 cm`