GOAL Apply geometric optics to correct farsightedness. PROBLEM The near point of
ID: 1459032 • Letter: G
Question
GOAL Apply geometric optics to correct farsightedness.
PROBLEM The near point of a patient's eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 25.0 cm away? Neglect the eye-lens distance. (b) What is the power of this lens? (c) Repeat the problem, taking into account 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 and then using the definition of lens power in terms of diopters. The object is at 25.0 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 its distance from the eye.
Apply the thin-lens equation.
Substitute p = 25.0 cm and q = -50.0 cm (the latter is negative because the image must be virtual) on the same side of the lens as the object
Solve for f. The focal length is positive, corresponding to a converging lens.
f = 50.0 cm
(B) What is the power of this lens?
The power is the reciprocal of the focal length in meters.
(C) Repeat the problem, noting that the corrective lens is actually 2.00 cm in front of the eye.
Substitute the corrected values of pand q into the thin-lens equation.
f = 44.2 cm
Compute the power.
LEARN MORE
REMARKS Notice that the calculation in part (c), which doesn't neglect the eye-lens distance, results in a difference of 0.26 diopter.
QUESTION If the distance of a farsighted patient's eye to its near point is greater, the power of the required corrective lens is:
PRACTICE IT
Use the worked example above to help you solve this problem. The near point of a patients eye is 50.0 cm.
EXERCISEHINTS: GETTING STARTED | I'M STUCK!
Suppose a lens is placed in a device that determines its power as 3.08 diopters.
1 + 1 = 1 p q fExplanation / Answer
practice it:
a)f=50.8064cm
1/f=1/25.2-1/50
b)power=1/f(m)
=1/0.508064=1.96825D
c)p=23.2cm
q=-48cm
c)power=2.227D
exercise hint:
A)f=32.467cm
b)image distance is patients near point=q=-45.9cm
f=32.467cm
1/f=1/p+1/q
p=object distance=19.016107cm
glasses help to correct the refractive errors
if power of lens is negative ,it needs concave lens
power of lens is positive it ineeds convex lens
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