Question

A person's prescription for her new bifocal glasses calls for a refractive power of -0.450 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.

Enter your answers numerically separated by a comma.

Answer 1

Far point of the eye is 22.24 m

Far point of the eye is 0.4 m

Step-by-step explanation:

`(1)/(f)=-0.045`

Object distance = u

Image distance = v

Lens equation

`(1)/(f)=(1)/(u)+(1)/(v)\\\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\\\Rightarrow (1)/(v)=-0.045-(1)/(\infty)\\\Rightarrow (1)/(v)=(1)/(-0.045)\\\Rightarrow v=-22.22\ m`

Far point

`|v|+\text{Position from eye}\\ =|-22.22|+0.02\\ =22.24\ m`

Far point of the eye is 22.24 m

Object distance = u = 0.25-0.02 = 0.23 m

`(1)/(f)=1.75`

`(1)/(f)=(1)/(u)+(1)/(v)\\\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\\\Rightarrow (1)/(v)=1.75-(1)/(0.23)\\\Rightarrow v=-0.38\ m`

Near point

`|v|+\text{Position from eye}\\= |-0.38|+0.02\\ =0.4\ m`

Far point of the eye is 0.4 m