/*! 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 56 A person can see clearly up clos... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 56

A person can see clearly up close but cannot focus on objects beyond 75.0 \(\mathrm{cm} .\) She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?

Short Answer

Expert verified
(a) Nearsighted, (b) Diverging lens, (c) Focal length: -0.75 m, Power: -1.33 D.

Step by step solution

01

Identify Vision Problem

The person can only see objects clearly that are close and cannot focus on objects beyond 75.0 cm. This indicates that the person is nearsighted, or myopic, as they can see close objects clearly but not far ones.
02

Determine Lens Type

For correcting nearsightedness, diverging lenses are used. These lenses help in adjusting the focal point of the light so that distant objects can be focused correctly on the retina.
03

Calculate Focal Length

For nearsighted vision correction, the focal length of the lens should allow distant objects to come into focus at the near point of 75.0 cm, or 0.75 m. Using the lens formula for a lens correcting at infinity, which is \\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \where \(v\) is the image distance (set to be the near point, 75 cm or -0.75 m, because virtual images are formed on the same side of the lens as the object for diverging lenses) and \(u\) is at infinity (i.e., \( \frac{1}{u} = 0 \)). The formula simplifies to \\[ \frac{1}{f} = -\frac{1}{v} = -\frac{1}{0.75} \Rightarrow f = -0.75 \text{ m} \] \So, the focal length needed is -0.75 meters.
04

Calculate Power of Lens

The power \(P\) of a lens in diopters is given by the formula \\[ P = \frac{1}{f(\text{in meters})} \] \Thus, using the calculated focal length, \\[ P = \frac{1}{-0.75} = -1.33 \text{ diopters} \]. \The negative sign indicates a diverging lens.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Nearsightedness
Nearsightedness, also known as myopia, is a common vision issue where individuals can see objects close to them clearly but struggle to focus on distant objects. This occurs because the light entering the eye is focused in front of the retina instead of directly on it. This can make things like road signs or boards difficult to read unless they get closer. The eye shape, either being too long or the cornea being too curved, usually causes this condition. To correct nearsightedness, wearing lenses is one solution that can refocus the light precisely on the retina, providing clear vision at both near and far distances.
Diverging Lens
Diverging lenses, also known as concave lenses, are used to correct nearsightedness. These lenses have a distinct shape that is thinner at the center and thicker at the edges. When the light passes through a diverging lens, it spreads out or "diverges," creating a virtual image that the eye can focus on correctly. This adjustment helps direct light rays onto the retina rather than in front of it, which is necessary for clear distant vision.
  • Diverging lenses are identified by their characteristic concave shape.
  • They help correct the focal point of light so it aligns properly with the retina.
  • The negative focal length or negative diopter indicates diverging lens power.
Understanding diverging lenses is critical for effective vision correction in people with myopia. These lenses need to be chosen carefully to ensure they match the specific needs of the individual.
Focal Length
The focal length of a lens is an essential factor in correcting vision. In the case of diverging lenses for myopia, the focal length is negative. This negative measure reflects the lens鈥檚 ability to spread out light rays before they reach the eye.

In this scenario, where a person cannot see beyond 75.0 centimeters, adjustments in the focal length through corrective lenses bring distant objects into the range of clear vision. The lens formula, \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), is used here. Since the goal is to correct vision for objects at infinity, the formula simplifies, permitting us to find the negative focal length needed: \( f = -0.75 \) m. This calculation guides the choice of lens necessary to focus the light properly on the retina.
Diopters
Diopters measure the refractive power of lenses and play a significant role in eye prescriptions. Specifically, in the context of nearsightedness correction, diopters express how much the lenses must diverge light to focus it on the retina.
  • The calculation for diopter strength involves taking the inverse of the focal length in meters: \( P = \frac{1}{f} \).
  • For our example, a focal length of \(-0.75\) meters results in a lens power of \(-1.33\) diopters.
  • The negative sign in diopters indicates the use of a diverging lens.
Simply put, diopters help optometrists understand what kind of lens and what strength will correct the vision impairment, ensuring clearer distant vision for those with myopia.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Lens of the Eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 \(\mathrm{mm}\) , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 \(\mathrm{cm}\) tall is placed 30.0 \(\mathrm{cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)

A tank whose bottom is a mirror is filled with water to a depth of 20.0 \(\mathrm{cm} .\) A small fish floats motionless 7.0 \(\mathrm{cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

Combination of Lenses I. A 1.20 -cm-tall object is 50.0 \(\mathrm{cm}\) to the left of a converging lens of focal length 40.0 \(\mathrm{cm} . \mathrm{A}\) second converging lens, this one having a focal length of \(60.0 \mathrm{cm},\) is located 300.0 \(\mathrm{cm}\) to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it \(I_{1} )\) formed by the lens with a focal length of 40.0 \(\mathrm{cm} .\) (b) \(I_{1}\) is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

You hold a spherical salad bowl 90 \(\mathrm{cm}\) in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a \(35-\) cm radius of curvature. (a) Where is the image of your 2.0 -cm- tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1},\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right|\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1} .\) Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q,\) parallel rays would still be brought to a focus at \(I^{\prime}\) . Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{cm}\) \(f_{2}=-18.0 \mathrm{cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{cm},\) find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{cm} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Electricity and Magnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Radiation

Read Explanation

Space Physics

Read Explanation

Engineering Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.