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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.

2 Answers

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Final answer:

The person's uncorrected far point is approximately 14.86 meters away, while the uncorrected near point is about 82 centimeters away, calculated using the lens formula and taking into account the refractive powers for the bifocal eyeglasses and the vertex distance.

Step-by-step explanation:

To determine the near and far points of the person's uncorrected vision, we make use of the lens formula and the concept of refractive power.

The refractive power P is given in diopters (D), and it is related to the focal length f of a lens in meters by P = 1/f. The negative sign in the distance vision part indicates myopia (nearsightedness), which means the person can see nearby objects clearly but not distant ones. The positive sign in the close-vision part indicates hypermetropia (farsightedness) for close-up tasks such as reading.

The distance between the glasses and the eyes (vertex distance) also affects the effective power of the lenses when worn. Hence, we use the formula P' = P / (1 - dP), in which P' is the effective power when the glasses are worn, P is the prescribed power, and d is the vertex distance.

We'll first find the far point for the myopia correction:
P' = -0.0675 D / (1 - 0.02 m * -0.0675 D)
=> P' = -0.0673 D

Therefore, the focal length for the far point is
f = 1/P'
=> f = 1/(-0.0673 D)
=> f ≈ -14.86 m

Thus, the person's far point without glasses is approximately 14.86 m away.

Next, for the near point with the hypermetropia correction:
P' = 1.20 D / (1 - 0.02 m * 1.20 D)
=> P' = 1.22 D

Since the corrected near-point distance is 250 cm:
f = 1/P = 1/1.22 D
=> f ≈ 0.82 m

This implies the person's near point without glasses is about 0.82 m, or 82 cm away.

User Adam Beck
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3 votes

Answer:

Far point of the eye is 14.83 m

Far point of the eye is 1.27 m

Step-by-step explanation:


(1)/(f)=-0.0675

Object distance = u

Image distance = v

Lens equation


(1)/(f)=(1)/(u)+(1)/(v)\\\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\\\Rightarrow (1)/(v)=-0.0675-(1)/(\infty)\\\Rightarrow (1)/(v)=(1)/(-0.0675)\\\Rightarrow v=-14.81\ m

Far point


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

Far point of the eye is 14.83 m

Object distance = u = 2.5-0.02 = 2.48 m


(1)/(f)=1.2


(1)/(f)=(1)/(u)+(1)/(v)\\\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\\\Rightarrow (1)/(v)=1.2-(1)/(2.48)\\\Rightarrow v=1.25\ m

Near point


|v|+\text{Position from eye}\\= |1.25|+0.02\\ =1.27\ m

Far point of the eye is 1.27 m

User Christophor
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