Question

Part A Where is the near point of an eye for which a contact lens with a power of 2.80 diopters is prescribed? (Assume that near point for an average viewer is 25 cm.) s′s ′ = nothing m Request Answer Part B Where is the far point of an eye for which a contact lens with a power of -1.60 diopters is prescribed for distant vision? s′s ′ = nothing m Request Answer Provide Feedback

Answer 1

-83.33 cm

-62.5 cm

Step-by-step explanation:

u = Object distance

v = Image distance

f = Focal length

Power is given by

`P=(1)/(f)\\\Rightarrow f=(1)/(P)\\\Rightarrow f=(1)/(2.8)\\\Rightarrow f=0.35714\ m=35.714\ cm`

Lens equation

`(1)/(f)=(1)/(u)+(1)/(v)\\\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\\\Rightarrow (1)/(v)=(1)/(35.714)-(1)/(25)\\\Rightarrow v=-83.33\ cm`

The object distance is -83.33 cm

`P=(1)/(f)\\\Rightarrow f=(1)/(P)\\\Rightarrow f=(1)/(-1.6)\\\Rightarrow f=-0.625\ m=-62.5\ cm`

`(1)/(f)=(1)/(u)+(1)/(v)\\\Rightarrow(1)/(f)=(1)/(\infty)+(1)/(v)\\\Rightarrow (1)/(f)=(1)/(v)\\\Rightarrow f=v\\\Rightarrow v=-62.5\ cm`

The far point of the lens is -62.5 cm