Question

Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye. Part A If you hold the book 28.0 cmcm from your eye, what is your nearpoint distance

Answer 1

The near point is
`n =44.8 \ cm`

Step-by-step explanation:

From the question we are told that

The power is
`P = 1.50`

The distance from the eye is
`k = 1.8 \ cm`

The distance of the book from the eye is
`z = -28 \ cm`

Generally the focal length of the glasses is

`f = (1)/(P)`

=>
`f = (1)/(1.50 )`

=>
`f = 0.667 \ m`

=>
`f = 66.7 \ cm`

The object distance is evaluated as

`u = z + k`

=>
`u = -28 + 1.8`

=>
`u = -26.2 \ cm`

The image distance is evaluated from lens formula as

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

=>
`(1)/(v) = (1)/(66.7) + (1)/(-26.2)`

=>
`v=- (1)/(0.0232)`

=>
`v=- 43 \ cm`

The near point is evaluated as

`n = -v + k`

=>
`n =-(-43) + 1.8`

=>
`n =44.8 \ cm`