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Chapter 26: Problem 121

At age forty, a man requires contact lenses \((f=65.0 \mathrm{cm})\) to read a book held \(25.0 \mathrm{cm}\) from his eyes. At age forty-five, while wearing these contacts he must now hold a book \(29.0 \mathrm{cm}\) from his eyes. (a) By what distance has his near point changed? (b) What focal-length lenses does he require at age forty-five to read a book at \(25.0 \mathrm{cm} ?\)

Short Answer

Expert verified
(a) 13.62 cm; (b) 71.56 cm.

Step by step solution

01

Calculate Initial Near Point (Age 40)

At age 40, the contact lenses have a focal length, \( f = 65.0 \text{ cm} \). The object distance (distance from the book to the eyes) is \( d_o = 25.0 \text{ cm} \). We use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substituting the known values:\[\frac{1}{65} = \frac{1}{25} + \frac{1}{d_i}\]Solving for \( d_i \), the image distance (effective near point) is calculated as approximately \( d_i \approx 81.25 \text{ cm} \).
02

Calculate New Near Point (Age 45)

At age 45, the object distance changes to \( d_o = 29.0 \text{ cm} \). We use the same lens formula:\[\frac{1}{65} = \frac{1}{29} + \frac{1}{d_i'}\]Solving for the new image distance \( d_i' \) yields approximately \( d_i' \approx 94.87 \text{ cm} \).
03

Determine Change in Near Point

The change in the near point is the difference between the new and initial near points. Calculate: \[\Delta d = d_i' - d_i = 94.87 \text{ cm} - 81.25 \text{ cm} = 13.62 \text{ cm}\]
04

Calculate Required Focal Length at Age 45

To find the new focal length required for reading at \( 25.0 \text{ cm} \) using lenses: \( d_o = 25 \text{ cm} \) and the desired near point \( d_i' = 94.87 \text{ cm} \) from the earlier step. Using the lens formula:\[\frac{1}{f_{new}} = \frac{1}{25} + \frac{1}{94.87}\]Solving gives the new focal length \( f_{new} \approx 71.56 \text{ cm} \).

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Key Concepts

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

Lens Formula
The lens formula is a crucial equation in optics that relates the focal length of a lens to the object distance and image distance. It is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:
  • \( f \) is the focal length of the lens
  • \( d_o \) is the object distance (distance from the object to the lens)
  • \( d_i \) is the image distance (distance from the lens to the image formed)
The lens formula allows us to calculate one of these variables if the other two are known. This is essential when determining how a lens needs to be adjusted in optical devices, such as glasses or contact lenses, to correct vision.
The lens formula helps us understand why changes in object distance or eye condition can affect vision and what adjustments must be made to retain clear eyesight.
In our exercise, this formula helped us determine the effective near point of vision for different scenarios and calculate the lens adjustments needed over time.
Focal Length
Focal length is a measure of how strongly a lens converges or diverges light. A lens with a shorter focal length bends light rays more sharply, bringing them to a focus at a shorter distance from the lens. Conversely, a longer focal length means a gentler bending of light.
In the context of vision correction, the focal length of a lens prescribes how it will adjust the path of incoming light to help focus an image directly onto the retina. This is particularly important for people whose eyes can't naturally focus close-up objects clearly. A prescribed focal length in glasses or contact lenses compensates for the deficits in natural eye focusing ability.
In the provided exercise, the focal length of the lenses at age 40 was 65.0 cm, and it needed adjustment to 71.56 cm by age 45. These changes helped maintain the ability to focus on objects held at a consistent distance.
Near Point
The near point is the closest distance at which the eye can focus on an object. For a person with normal vision, this is typically around 25 cm. However, as people age, the near point tends to move farther away because of the eye's decreasing ability to accommodate or focus.
In dealing with vision problems, understanding the near point is crucial because it defines the closest point of clear vision. When a person's near point has shifted outward, corrective lenses can help push it back to a manageable distance.
According to the problem, at age 45, the man's near point had shifted from approximately 81.25 cm to 94.87 cm. This change indicates a decrease in the eye's accommodation ability, prompting a need for lenses with a different focal length to allow him to read comfortably at his preferred reading distance.

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Most popular questions from this chapter

A woman can read the large print in a newspaper only when it is at a distance of \(65 \mathrm{cm}\) 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 \(2.0 \mathrm{cm}\) from the eyes), so she can read the newspaper at a distance of \(25 \mathrm{cm}\) from her eyes?

A narrow beam of light from a laser travels through air \((n=1.00)\) and strikes point \(A\) on the surface of the water \((n=1.33)\) in a lake. The angle of incidence is \(55^{\circ} .\) The depth of the lake is \(3.0 \mathrm{m} .\) On the flat lake-bottom is point \(\mathrm{B},\) directly below point \(\mathrm{A}\). (a) If refraction did not occur, how far away from point \(B\) would the laser beam strike the lake-bottom? (b) Considering refraction, how far away from point \(B\) would the laser beam strike the lake-bottom?

A glass block \((n=1.56)\) is immersed in a liquid. A ray of light within the glass hits a glass-liquid surface at a \(75.0^{\circ}\) angle of incidence. Some of the light enters the liquid. What is the smallest possible refractive index for the liquid?

A plate glass window \((n=1.5)\) has a thickness of \(4.0 \times 10^{-3} \mathrm{m} .\) How long does it take light to pass perpendicularly through the plate?

A slide projector has a converging lens whose focal length is \(105.00 \mathrm{mm}\) (a) How far (in meters) from the lens must the screen be \(10 \mathrm{c}-\) ated if a slide is placed \(108.00 \mathrm{mm}\) from the lens? (b) If the slide measures \(24.0 \mathrm{mm} \times 36.0 \mathrm{mm},\) what are the dimensions (in \(\mathrm{mm}\) ) of its image?
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