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A farsighted person has a near point that is 67.0 cm from her eyes. She wears eyeglasses that are designed to enable her to read a newspaper held at a distance of 25.0 cm from her eyes. Find the focal length of the eyeglasses, assuming that they are worn (a) 2.2 cm from the eyes and (b) 3.3 cm from the eyes.

Short Answer

Expert verified
For (a), the focal length is approximately 16.7 cm; for (b), it's approximately 16.1 cm.

Step by step solution

01

Understand the Problem

A farsighted person needs glasses to read at a distance closer than their natural near point, which is 67.0 cm. The glasses adjust their near point to 25.0 cm. We need to calculate the focal length of the glasses depending on their position relative to the eyes.
02

Convert Distances to Lens Position

Consider the glasses' position in relation to the eyes. The object distance for the glasses is different from the distance from the eyes because the glasses are positioned away from the eyes. So, we calculate: \(d_o = 25.0\,\text{cm} - \text{glasses position}\).
03

Use the Lens Formula

Apply the lens formula, \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(d_o\) is the object distance calculated in the previous step, and \(d_i = 67.0\,\text{cm} - \text{glasses position}\) is the image distance for the glasses.
04

Calculate Focal Length for Part (a)

First, if the glasses are 2.2 cm from the eyes, compute \(d_o = 25.0\,\text{cm} - 2.2\,\text{cm} = 22.8\,\text{cm}\) and \(d_i = 67.0\,\text{cm} - 2.2\,\text{cm} = 64.8\,\text{cm}\). Then, use the lens formula: \[\frac{1}{f} = \frac{1}{22.8} + \frac{1}{64.8}\]This gives \(f \approx 16.7\,\text{cm}\).
05

Calculate Focal Length for Part (b)

For glasses worn 3.3 cm from the eyes, determine \(d_o = 25.0\,\text{cm} - 3.3\,\text{cm} = 21.7\,\text{cm}\) and \(d_i = 67.0\,\text{cm} - 3.3\,\text{cm} = 63.7\,\text{cm}\). Then apply the lens formula: \[\frac{1}{f} = \frac{1}{21.7} + \frac{1}{63.7}\]Calculating this gives \(f \approx 16.1\,\text{cm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Farsightedness
Farsightedness, also known as hyperopia, is a common vision condition wherein distant objects may be seen more clearly than nearby objects. This is due to the light being focused behind the retina rather than directly on it. In farsighted individuals, the eyeball is either too short or the corneal curvature is too flat, and they often experience difficulty when trying to focus on close-up tasks, such as reading or writing.

To address this, corrective lenses such as eyeglasses or contact lenses might be prescribed to help refocus the image on the retina. These lenses are typically convex, helping to reduce the strain on the eyes when viewing objects closer than the eye's natural near point distance.
Lens Formula
The lens formula is a fundamental equation used in optics to relate the different distances associated with lens-based systems. It provides a mathematical relationship between the focal length of the lens (\( f \)), the object distance (\( d_o \)), and the image distance (\( d_i \)). The lens formula is expressed as:
  • \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]
This formula is crucial for determining how a lens will focus light. By knowing any two of these quantities, the third one can be calculated. It is especially helpful in designing eyeglasses, cameras, microscopes, and other optical devices to achieve the desired focal properties.
Focal Length
Focal length is an important parameter in optics that measures the distance between the lens and the focus (where light rays converge). It is denoted by the symbol (\( f \)). A lens with a short focal length bends light rays more sharply than one with a long focal length.

Convex lenses, which are used in correcting farsightedness, help to converge light rays before they enter the eye. This adjustment helps position the image directly on the retina. Calculating the focal length for specific lens positioning is essential to ensure that glasses meet their corrective purpose. You can find this by using the lens formula, and it shows how well the lens can assist in refocusing images closer to the person’s eyes.
Object Distance
Object distance (\( d_o \)) refers to the distance from the object being viewed to the lens. In optics problems, this measurement is crucial because it helps define how light from the object will get focused through the lens. For eyeglasses correcting farsightedness, the object distance is the distance from the newspaper (or the object being read) to the lens of the eyeglasses.

To compute this in practical terms, one needs to consider any physical distance between the lens and the eyes. For example, if the newspaper is held 25.0 cm away from the eyes and the glasses are 2.2 cm away from the eyes, the effective object distance for the lens is adjusted to 22.8 cm (\( 25.0 - 2.2 \)).
This adjustment ensures the lens formulates the correct focus to improve the wearer's vision.
Image Distance
Image distance (\( d_i \)) is the distance from the lens to the image formed by the lens. For eyeglasses, the image needs to be focused onto the retina for clear vision. In farsighted individuals, without corrective lenses, this distance would naturally be longer because their lens system focuses images behind the retina.

When calculating image distance, it's important to factor in the distance between the lens and the eye itself. As in our example, if a person's natural near-point (where images are focused clearly on the retina without lenses) is 67.0 cm, and the glasses are 3.3 cm away from the eyes, the effective image distance would be adjusted to 63.7 cm (\( 67.0 - 3.3 \)).
This calculated image distance then aids in using the lens formula effectively to ensure the glasses correct the farsightedness. The goal is to alter image distances such that the person can comfortably see both near and far objects.

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Most popular questions from this chapter

A farsighted person can read printing as close as 25.0 cm when she wears contacts that have a focal length of 45.4 cm. One day, she forgets her contacts and uses a magnifying glass, as in Figure 26.39b. Its maximum angular magnification is 7.50 for a young person with a normal near point of 25.0 cm. What is the maximum angular magnification that the magnifying glass can provide for her?

A woman can read the large print in a newspaper only when it is at a distance of 65 cm or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn 2.0 cm from the eyes), so she can read the newspaper at a distance of 25 cm from the eyes?

The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscope is to provide an angular magnification of 83 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece?

Two systems are formed from a converging lens and a diverging lens, as shown in parts a and b of the drawing. (The point labeled \(" F_{\text { comverging }} "\) is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 at a distance of 10.00 cm to the left of lens 1. The focal lengths of the converging and diverging lenses are 15.00 and 20.0 cm, respectively. The distance between the lenses is 50.0 cm. Determine the final image distance for each system, measured with respect to lens 2.

A person working on the transmission of a car accidentally drops a bolt into a tray of oil. The oil is 5.00 cm deep. The bolt appears to be 3.40 cm beneath the surface of the oil, when viewed from directly above. What is the index of refraction of the oil?

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