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A person's prescription for his new bifocal eyeglasses calls for a refractive power of -0.0675 diopters in the distance-vision part and a power of 1.20 diopters in the close-vision part. Assuming the glasses rest 2.00 cm from his eyes and that the corrected near-point distance is 250 cm, determine the near and far points of this person's uncorrected vision.

Respuesta :

Answer:

Far point of the eye is 14.83 m

Far point of the eye is 1.27 m

Explanation:

[tex]\frac{1}{f}=-0.0675[/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}=-0.0675-\frac{1}{\infty}\\\Rightarrow \frac{1}{v}=\frac{1}{-0.0675}\\\Rightarrow v=-14.81\ m[/tex]

Far point

[tex]|v|+\text{Position from eye}\\ =|-14.81|+0.02\\ =14.83\ m[/tex]

Far point of the eye is 14.83 m

Object distance = u = 2.5-0.02 = 2.48 m

[tex]\frac{1}{f}=1.2[/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}=1.2-\frac{1}{2.48}\\\Rightarrow v=1.25\ m[/tex]

Near point

[tex]|v|+\text{Position from eye}\\= |1.25|+0.02\\ =1.27\ m[/tex]

Far point of the eye is 1.27 m

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