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Chapter 14: Problem 28

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

She can see clearly up close but not beyond 75.0 cm. This means she is nearsighted (myopic). In myopia, distant objects are not clear.
02

Determine Lens Type

Nearsightedness is corrected with diverging (concave) lenses. These lenses help focus distant objects on the retina by diverging the light before it enters the eye.
03

Convert Distance to Meters

The farthest distance the person can see clearly is given. Convert this from centimeters to meters: \(75.0\, \text{cm} = 0.75\, \text{m}\).
04

Use Lens Formula

Apply the lens formula to find the focal length \( f \) of the contact lens.\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]For corrective lenses, \(d_o\) (object distance) is infinity (a hypothetical very far object) \((\infty)\) and \(d_i\) (image distance) is -0.75\, \text{m} (the image should appear at the far point).
05

Calculate Focal Length

Based on the above, substitute \( \infty \) for \( d_o \) and \(-0.75\, \text{m}\) for \( d_i \):\[ \frac{1}{f} = \frac{1}{\infty} + \left (-\frac{1}{0.75} \right ) \]\[ \frac{1}{f} = 0 - \frac{1}{0.75} \approx -1.33 \]\[ f = -0.75 \text{ m} \]
06

Calculate Lens Power

Power \( P \) of a lens in diopters is defined as the reciprocal of the focal length in meters:\[ P = \frac{1}{f} \]\[ P = -1.33 \text{ D} \]

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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 condition where individuals have difficulty seeing distant objects clearly, but can see nearby objects without any issues. This happens because the eye is unable to focus light directly on the retina, causing distant images to be focused in front of it. As a result:
  • Distant objects appear blurry.
  • Close objects can be seen clearly.
  • Usually, it begins during childhood.
Most people who are nearsighted can correct their vision using glasses or contact lenses. These optical aids adjust the focus of light as it enters the eye, ensuring that distant images are properly focused on the retina.
Lens Formula
The lens formula is a fundamental equation used to determine the relationship between the focal length of a lens, the object distance, and the image distance. It is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:
  • \( f \) represents the focal length of the lens,
  • \( d_o \) is the distance of the object from the lens,
  • \( d_i \) is the distance of the image from the lens.
This formula is crucial in solving problems related to lens corrections, like determining the correct contact lens specifications for a nearsighted person. By using this formula, we can calculate how a lens will alter light paths to ensure proper vision.
Lens Power Calculation
After determining the focal length using the lens formula, the next step is to calculate the lens power. The power of a lens, measured in diopters (D), is a measure of its ability to converge or diverge light. The formula for lens power is:\[ P = \frac{1}{f} \]Here, \( P \) is the power in diopters, and \( f \) is the focal length in meters.
  • If \( f \) is negative, the lens is diverging.
  • If \( f \) is positive, the lens is converging.
For a nearsighted person, a diverging lens is typically used, which will have a negative focal length, leading to a negative power in diopters. Calculating this ensures the lens provides the right correction to improve vision.
Vision Correction
Vision correction involving nearsightedness focuses on correcting the eye's inability to properly focus light on the retina. The goal is to adjust the light path so that it ends directly on the retina, providing a clearer image:
  • Corrective lenses, such as glasses or contacts, are common solutions.
  • Surgery, like LASIK, is another option for some individuals.
  • Orthokeratology involves wearing special contact lenses that reshape the cornea temporarily.
Corrective measures specifically for myopia use lenses that diverge light, helping reposition the focus onto the retina. This adjustment is essential for those who struggle to see faraway objects.
Diverging Lenses
Diverging lenses, often referred to as concave lenses, are specifically designed to address nearsightedness. They work by spreading out light rays that enter them before the light reaches the eye:
  • This causes the light to appear as if it is originating from a point further away than it actually is.
  • They have a thinner center compared to their edges.
  • Diverging lenses have a negative focal length.
For nearsighted individuals, these lenses are advantageous because they help focus distant images onto the retina, compensating for the eye's natural focus that occurs in front of it. Choosing the correct diverging lens strength ensures blurry distant objects become clear, significantly improving vision.

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Most popular questions from this chapter

An object to the left of a lens is imaged by the lens on a screen \(30.0 \mathrm{~cm}\) to the right of the lens. When the lens is moved \(4.00 \mathrm{~cm}\) to the right, the sereen must be moved \(4.00 \mathrm{~cm}\) to the left to refocus the image. Determine the focal length of the lens.

(a) For a lens with focal length \(f\), find the smallest distance possible between the object and its real image. (b) Ciraph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?

The refracting angle of a prism is \(90^{\circ}\). If \(\gamma\) is the angle of minimum deviation and \(\beta\) is the deviation of ray which enters at grazing incidence, prove that \(\sin \gamma=\sin ^{2} \beta\) and \(\cos \gamma=\mu \cos \beta, \mu\) is refractive index of material of the prism.

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.

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.
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