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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.

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Answer:

Far point of the eye is 22.24 m

Far point of the eye is 0.4 m

Explanation:

[tex]\frac{1}{f}=-0.045[/tex]

Object distance = u

Image distance = v

Lens equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=-0.045-\frac{1}{\infty}\\\Rightarrow \frac{1}{v}=\frac{1}{-0.045}\\\Rightarrow v=-22.22\ m[/tex]

Far point

[tex]|v|+\text{Position from eye}\\ =|-22.22|+0.02\\ =22.24\ m[/tex]

Far point of the eye is 22.24 m

Object distance = u = 0.25-0.02 = 0.23 m

[tex]\frac{1}{f}=1.75[/tex]

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=1.75-\frac{1}{0.23}\\\Rightarrow v=-0.38\ m[/tex]

Near point

[tex]|v|+\text{Position from eye}\\= |-0.38|+0.02\\ =0.4\ m[/tex]

Far point of the eye is 0.4 m

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