Question

A myopic (nearsighted) person has a far point distance of 150 cm. What is the refractive power of his ideally prescribed eyeglasses if he wears his eyeglasses a distance of 3.0 cm from his eyes?

Answer 1

Answer:0.68 D

Step-by-step explanation:

Given

Person has a far point distance of 150 cm

i.e.
`d_i=-150 cm`

therefore

v=-150+3=-147 cm

From Lens Formula

`(1)/(f)=(1)/(v)+(1)/(u)`

Here the object distance is infinity

`(1)/(f)=(-1)/(147)+(1)/(\infty)`

Therefore f=-147 cm

and
`f=(1)/(P)`

`P=-(100)/(147)`

P=-0.68 D