Explanation Given info: The accommodation limits of eyes are 18 cm and 80 cm . Write the expression for the focal length of the eye glasses from the thin lens equation. 1 p + 1 q = 1 f (1) Here, p is the object distance. q is the image distance. f is the focal length. The image is on the same side as the object so the image distance is − 80 m and the object length for starlight going through a nearsighted person’s glasses is ∞ . Substitute − 80 cm for q and ∞ for p in the equation (1). 1 ∞ + 1 ( − 80 cm ) ( 10 − 2 m 1 cm ) = 1 f 1 f = 1 − 0.8 m = − 1.25 diopters Thus, the focal length is − 1.25 diopters . Write the expression for a nearby object which image is virtual. 1 p + 1 q = 1 f Substitute − 80 cm for q and − 1

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

8th Edition

ISBN: 9781439048382

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Chapter 36, Problem 36.61P

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The minimum distance from where the person can see object clearly.

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Chapter 36, Problem 36.60P

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The accommodation limits for Nearsighted Nick's eyes are 20.0 cm and 78.0 cm. When he wears his glasses, he is able to see faraway objects clearly. At what minimum distance is he able to see objects clearly? cm=?

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns 1mm2 is typical near the center of the eye. We shall model the eye as a sphere 2.50 cm in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0 mm in diameter. (a) What is the smallest object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes (1° = 60 min), and compare it with the typical experimental value of about 1.0 min.

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns 1mm2 is typical near the center of the eye. We shall model the eye as a sphere 2.50 cm in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0 mm in diameter. (a) What is the smallest object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes 11° = 60 min2, and compare it with the typical experimental value of about 1.0 min

Chapter 36 Solutions

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

Ch. 36 - You are standing approximately 2 m away from a...Ch. 36 - You wish to start a fire by reflecting sunlight...Ch. 36 - Consider the image in the mirror in Figure 35.14....Ch. 36 - Prob. 36.4QQ Ch. 36 - Prob. 36.5QQ Ch. 36 - What is the focal length of a pane of window...Ch. 36 - A camera can be modeled as a simple converging...Ch. 36 - Two campers wish to start a fire during the day....Ch. 36 - The faceplate of a diving mask can be ground into...Ch. 36 - Lulu looks at her image in a makeup mirror. lt is...

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The minimum distance from where the person can see object clearly.

Solution Summary: The author explains that the minimum distance from where the person can see object clearly is 23.2cm.

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