/*! 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 54 Contact Lenses. Contact lenses a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 \(\mathrm{cm}\) from his eyes instead of the usual 25.0 \(\mathrm{cm} .(\mathrm{a})\) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

Short Answer

Expert verified
The person is farsighted and needs a converging lens with a focal length of 16.07 cm and a power of 6.22 diopters.

Step by step solution

01

Define Nearsightedness and Farsightedness

Nearsighted individuals can see close objects clearly but have difficulty seeing distant objects. Farsighted individuals can see distant objects clearly but struggle to see close ones. This person can see distant objects well but not close ones (closer than 45 cm). Thus, this person is farsighted.
02

Determine the Type of Lens Needed

Farsightedness is corrected using converging lenses, which help in focusing light on the retina for closer objects. Therefore, this person requires a converging lens for correction.
03

Calculate Focal Length Using Lens Formula

To find the required focal length, we use the lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\). Here, the desired near point \(v\) is 25.0 cm, and the actual near point \(u\) is 45.0 cm. Since contact lenses are worn directly on the eye, \(v = 25\) cm (positive for a real image), \(u = -45\) cm (negative for the virtual object). Substituting these values, \(\frac{1}{f} = \frac{1}{25} - \frac{1}{-45} = \frac{1}{25} + \frac{1}{45}\).
04

Solve for Focal Length

Let's calculate the focal length from the previous equation: \[\frac{1}{f} = \frac{45 + 25}{1125} = \frac{70}{1125}\].Hence, \(f = \frac{1125}{70} \approx 16.07\) cm.
05

Calculate Lens Power in Diopters

The power \(P\) of a lens is given by the formula \(P = \frac{1}{f}\), where \(f\) is the focal length in meters. Convert 16.07 cm to meters: \(f = 0.1607\) m.So, \(P = \frac{1}{0.1607} \approx 6.22\) diopters.

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 91Ó°ÊÓ!

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 where distant objects are seen clearly, but close objects appear blurry. It occurs when the light entering the eye is focused behind the retina rather than directly on it. This can be due to an eyeball that is too short or a lens that is not curved enough.

If you're farsighted, your eye struggles to bring near objects into focus, causing you to experience strain and fatigue when reading or focusing on anything up close. Ordinary tasks like reading a book or using a smartphone can become challenging.

Some key facts about farsightedness include:

  • It often runs in families and can be detected early in life.
  • Many children are slightly farsighted but outgrow it as their eyes grow longer and the lens can focus light more effectively.
  • Farsightedness is corrected by using convex lenses which help to converge the light rays before they enter the eye.
Understanding farsightedness is the first step towards finding the right corrective measures.
Converging Lens
A converging lens, commonly referred to as a convex lens, is thicker in the middle than at the edges. Its primary function is to bend light rays towards each other, or converge them, to focus on a single point. This is particularly useful for farsighted individuals, as these lenses help focus light rays on the retina instead of behind it, providing a clearer image of close objects.

Here’s how a converging lens works:

  • Light rays that pass through the lens are bent towards the center, making them converge.
  • The point where the rays converge is called the focal point, and the distance from the lens to this point is the focal length.
  • By adjusting the focal length, these lenses can adapt to correct various levels of farsightedness.
Converging lenses are integral to eyeglasses and contact lenses designed to rectify hyperopia, helping individuals see both near and far objects more clearly.
Lens Power in Diopters
The power of a lens is measured in diopters, a unit that quantifies the ability of the lens to bend light. A diopter describes the refractive power of the lens, calculated as the inverse of its focal length (in meters). This means that a lens with a shorter focal length has a higher diopter value and thus greater light-bending ability.

To understand lens power better, consider the formula:
\[ P = \frac{1}{f} \]
Where `P` is the lens power in diopters and `f` is the focal length in meters. For example, if a lens has a focal length of 0.1607 meters, its power would be approximately 6.22 diopters.

Here are some key points about diopters:

  • Positive diopter values indicate converging (convex) lenses, ideal for treating farsightedness.
  • Negative diopter values are used for diverging (concave) lenses, which correct nearsightedness.
  • Optometrists use diopters to prescribe the appropriate lens strength for correcting vision issues.
By understanding how diopters work, individuals can better comprehend the specifications of their corrective lenses.

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

You have a camera with a 35.0 - \(\mathrm{mm}\) -focal-length lens and 36.0 -mm- wide film. You wish to take a picture of a \(12.0\)-m-long sailboat but find that the image of the boat fills only \(\frac{1}{4}\) of the width of the film. (a) How fare you from the boat? (b) How much closer must the boat be to you for its image to fill the width of the film?

A lens forms an image of an object. The object is 16.0 \(\mathrm{cm}\) from the lens. The image is 12.0 \(\mathrm{cm}\) from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 \(\mathrm{mm}\) tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

A thin lens with a focal length of 6.00 \(\mathrm{cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 \(\mathrm{cm}\) from the eye, and that the lens is very close to the eye.

The focal length of the eyepiece of a certain microscope is 18.0 \(\mathrm{mm}\) . The focal length of the objective is 8.00 \(\mathrm{mm}\) . The distance between objective and eyepiece is 19.7 \(\mathrm{cm} .\) The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

A photographer takes a photograph of a Boeing 747 airliner (length 70.7 \(\mathrm{m} )\) when it is flying directly overhead at an altitude of 9.50 \(\mathrm{km} .\) The lens has a focal length of 5.00 \(\mathrm{m} .\) How long is the image of the airliner on the film?

See all solutions

Recommended explanations on Physics Textbooks

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.