Question

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.52 meters and 5.3 meters.

1) What type of lens is needed to correct his nearsightness?

(1) converging

(2) diverging

2 CORRECT

2) What type of lens is needed to correct his farsightness?

(1) converging

(2) diverging

1 CORRECT

3) Assume that the physics instructor would like to have normal visual acuity from 23 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

1/f = ? diopters  

4) What is the refractive power of the portion of the lense that will correct the instructors farsightedness?

1/f = ? diopters

Answer 1

1] The person here has a far point of 5.3 m

this means that the correction lens should be such that a distant object must appear at a distance 5.3 m from the eyes.

taking the distance between the eye and the lens to be approximately 2.0 cm, this suggests that the image and the object are both on the same side. Therefore, a diverging lens is needed for correction.

2] The near point is 0.52 m and so the object distance must be closer than the focal length to create a corrected image. Therefore, a converging lens is needed for correction.

3] A distant object must produce an image 23 cm from the eye.

Since the distance of the lens from the eye is 2cm, the image distance here is: v = 23-2 = 21 cm

object distance = u = infinity

use the lens formula

-1/f = 1/v + 1/u

- 1/f = 1/23 + 0

=> f = - 23 cm = - 0.23 m

or 1/f = - 4.348 D

4] Here, 1/f = + 4.348 D.