Suppose a lens is placed in a device that determines its power as 2.86 diopters. (a) Find the focal length of the lens. f = cm (b) Find the minimum distance at which a patient will be able to focus on an object if the patient's near point is 32.0 cm. Neglect the eye-lens distance. d min cm

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Suppose a lens is placed in a device that determines its power as 2.86 diopters.
(a) Find the focal length of the lens.
f=
cm
(b) Find the minimum distance at which a patient will be able to focus on an object if the patient's
near point is 32.0 cm. Neglect the eye-lens distance.
d min
cm
Transcribed Image Text:Suppose a lens is placed in a device that determines its power as 2.86 diopters. (a) Find the focal length of the lens. f= cm (b) Find the minimum distance at which a patient will be able to focus on an object if the patient's near point is 32.0 cm. Neglect the eye-lens distance. d min cm
SOLUTION
(A) Find the focal length of the corrective lens, neglecting its distance from the eye.
Apply the thin-lens equation.
1 1
р
q f
1
1
+
25.0 cm -50.0 cm f
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
f = 50.0 cm
Solve for f. The focal length is positive,
corresponding to a converging lens.
(B) What is the power of this lens?
1
1
The power is the reciprocal of the focal
length in meters.
===
+2.00 diopters
f 0.500 m
(C) Repeat the problem, noting that the corrective lens is actually 2.00 cm in front of the eye.
Substitute the corrected values of p
and q into the thin-lens equation.
1 +1. 1
1
→
р 9 23.0 cm (-48.0 cm) f
f = 44.2 cm
Compute the power.
1
1
P
+2.26 diopters
f 0.442 m
P
+
+
=
Transcribed Image Text:SOLUTION (A) Find the focal length of the corrective lens, neglecting its distance from the eye. Apply the thin-lens equation. 1 1 р q f 1 1 + 25.0 cm -50.0 cm f 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 f = 50.0 cm Solve for f. The focal length is positive, corresponding to a converging lens. (B) What is the power of this lens? 1 1 The power is the reciprocal of the focal length in meters. === +2.00 diopters f 0.500 m (C) Repeat the problem, noting that the corrective lens is actually 2.00 cm in front of the eye. Substitute the corrected values of p and q into the thin-lens equation. 1 +1. 1 1 → р 9 23.0 cm (-48.0 cm) f f = 44.2 cm Compute the power. 1 1 P +2.26 diopters f 0.442 m P + + =
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