Example 25.1 Prescribing a Corrective Lens for a Farsighted Patient Goal Apply g

Question

Example 25.1 Prescribing a Corrective Lens for a Farsighted Patient Goal Apply geometric optics to correct farsightedness. Problem The near point of a patients eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 22.9 cm away? Neglect the eye-lens distance? (b) What is the power of this lens? (c) Repeat the problem, taking into account the fact that, for typical eyeglasses, the corrective lens is 2.00 cm in front of the eye Strategy This problem requires substitution into the thin-lens equation (Eq 23.11) and then using the definition of lens power in terms of diopters. The object is at 22.9 cm, but the lens must form an image at the patient's near point, 50.0 cm, the closest point at which the patient's eye can see clearly. In part (c), 2.00 cm must be subtracted from both the object distance and the image distance to account for the position of the lens Solution (a) Find the focal length of the corrective lens, neglecting the distance from the eye. Apply the thin-lens equation. Substitute p 22.9 cm and q 50.0 cm (the latter is negative because the image must be virtual) on 22.9 cm 50.0 cm f the same side of the lens as the object. Solve for The focal length is positive, Cm corresponding to a converging lens. Your response differs from the correct answer by more than 10%. Double check your calculations. (b) What is the power of this lens? The power is the reciprocal of the focal length in meters

Explanation / Answer

a.

focal length: f = [1/p + 1/q]-1 = [(1/22.9) + (1/-50)]-1 = 42.25 cm

b.

power of the lens: p = 1/f = 1/42.25 cm = 1/42.25x10-2 = 2.3668 m-1 = 2.3668 D

C.

focal length: f = [1/p + 1/q]-1 = [(1/20.9) + (1/-48)]-1 = 37 cm

power = 1/f = 1/37 cm = 1/0.37 m = 2.7027 D

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focal length = f = 1/p = 1/2.79 = 0.3584 m = 35.84 cm

1/(-49.9) + 1/d = 1/35.84

d= 20.85937 cm

= 20.86 cm

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