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

Answer:

-83.33 cm

-62.5 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length

Power is given by

$P=\frac{1}{f}\\\Rightarrow f=\frac{1}{P}\\\Rightarrow f=\frac{1}{2.8}\\\Rightarrow f=0.35714\ m=35.714\ cm$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35.714}-\frac{1}{25}\\\Rightarrow v=-83.33\ cm$

The object distance is -83.33 cm

$P=\frac{1}{f}\\\Rightarrow f=\frac{1}{P}\\\Rightarrow f=\frac{1}{-1.6}\\\Rightarrow f=-0.625\ m=-62.5\ cm$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow\frac{1}{f}=\frac{1}{\infty}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{v}\\\Rightarrow f=v\\\Rightarrow v=-62.5\ cm$

The far point of the lens is -62.5 cm