/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Suppose that the lens system (co... [FREE SOLUTION] | 91影视

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Chapter 24: Problem 23

Suppose that the lens system (cornea + lens) in a particular eye has a focal length that can vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm},\) but the distance from the lens system to the retina is only \(1.90 \mathrm{cm} .\) (a) Is this eye nearsighted or farsighted? Explain. (b) What range of distances can the eye see clearly without corrective lenses?

Short Answer

Expert verified
Based on the given variable focal length range of the eye's lens system and the distance between the lens system and the retina, we have concluded that the eye is farsighted. Furthermore, we have determined the clear vision range without corrective lenses to be between 19 cm and 37 cm.

Step by step solution

01

Determine whether the eye is nearsighted or farsighted

To understand whether the eye is nearsighted or farsighted, we'll first examine the focal length range of the eye's lens system and compare it with the distance between the lens system and the retina. A nearsighted eye is unable to focus on distant objects, meaning the focal length of the lens system should be shorter than the distance to the retina. In this problem, Variable focal length range: \(1.85\,\mathrm{cm}\) to \(2.00\,\mathrm{cm}\) Distance between the lens system and retina: \(1.90\,\mathrm{cm}\) As the range of focal lengths are larger than the distance to the retina, we can conclude that the eye is farsighted.
02

Determine clear vision range

As the eye is farsighted, we'll use the thin lens formula to determine the range of object distances where images will be formed on the retina without the need for corrective lenses. The thin lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) Here, \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance (distance between the lens and the retina). To find the clear vision range, we'll calculate object distance ranges for the maximum and minimum focal lengths of the lens system.
03

Calculate object distance range for minimum focal length

Minimum focal length: \(f_{min} = 1.85\,\mathrm{cm}\) Distance between lens and retina: \(d_i = 1.90\,\mathrm{cm}\) Using the thin lens formula, we can solve for the minimum object distance: \(\frac{1}{1.85} = \frac{1}{d_{o_{min}}} + \frac{1}{1.9}\) \(d_{o_{min}} = \frac{1.85 \times 1.90}{1.90 - 1.85} = 37\,\mathrm{cm}\)
04

Calculate object distance range for maximum focal length

Maximum focal length: \(f_{max} = 2.00\,\mathrm{cm}\) Using the thin lens formula, we can solve for the maximum object distance: \(\frac{1}{2.00} = \frac{1}{d_{o_{max}}} + \frac{1}{1.9}\) \(d_{o_{max}} = \frac{2.00 \times 1.90}{1.90 - 2.00} = 19\,\mathrm{cm}\)
05

Conclude clear vision range

The clear vision range without corrective lenses for the eye is from \(19\,\mathrm{cm}\) to \(37\,\mathrm{cm}\).

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Most popular questions from this chapter

Cameras You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length \(3.70 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high, what is the image height?
A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.
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