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(a) Where is the near point of an eye for which a contact lens with a power of \(+2.75\) diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of \(-1.30\) diopters is prescribed for distant vision?

Short Answer

Expert verified
Near point is 36.4 cm, and far point is 76.9 cm (virtual).

Step by step solution

01

Understand the Formula

The formula to find the near or far point using the lens formula is \( P = \frac{1}{f} \), where \( P \) is the power in diopters and \( f \) is the focal length in meters. The near point or far point when a lens is used can be calculated by rearranging this formula to find \( f \).
02

Calculate Near Point for +2.75 Diopters

Given the power \( P = +2.75 \) diopters, we rearrange the lens formula to find the focal length \( f \): \( f = \frac{1}{P} = \frac{1}{2.75} \approx 0.364 \, \text{meters} \). The near point is at approximately 36.4 cm from the eye.
03

Calculate Far Point for -1.30 Diopters

Given the power \( P = -1.30 \) diopters, we rearrange the lens formula to find the focal length \( f \): \( f = \frac{1}{P} = \frac{1}{-1.30} \approx -0.769 \, \text{meters} \). This indicates that the far point is at approximately 76.9 cm (negative value suggesting the virtual or behind the eye positioning).
04

Interpretation of Results

For lenses with positive power (near point), the calculated distance shows where a person can focus on nearby objects. For lenses with negative power (far point), the calculated distance shows the virtual far point for distant vision correction.

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Key Concepts

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

Understanding Focal Length
Focal length is a fundamental concept in optics and relates to how lenses converge or diverge light. It is the distance between the center of a lens and its focus. Focal length is typically measured in meters.
When a lens has a positive focal length, it means that it can bring parallel rays of light to a point. This type of lens is known as a converging lens, and it is useful for people whose near point is too far away.
In contrast, a negative focal length signifies a diverging lens. Such lenses help those with distant vision problems by moving the focal point closer or behind the lens, creating a virtual image. Understanding focal length is crucial for calculating where an object will be in focus when using lenses.
Diopters and Their Role in Vision Correction
Diopters measure the refractive power of lenses, indicating how much they can bend light. The formula used is \[ P = \frac{1}{f} \] where \( P \) is the power in diopters and \( f \) is the focal length in meters.
Positive diopter values indicate lenses that aid in seeing closer objects more clearly, which is useful for hyperopia or farsightedness. This means the lenses are converging, reducing the near point distance.
Negative diopter values are used for nearsightedness or myopia. These lenses diverge light and place the focus behind the eye, helping individuals see distant objects more clearly. Diopters provide a straightforward way to determine the corrective power needed in a lens.
Exploring the Lens Formula
The lens formula is central to finding out the focal length of lenses based on their power, calculated using \[ P = \frac{1}{f} \] This relationship helps in determining both near and far points for different vision correction needs.
To find a lens's focal length, simply rearrange the formula to get \[ f = \frac{1}{P} \] This makes it easy to see how changing the power of a lens affects its ability to bring objects into focus. For example, a lens with +2.75 diopters has a focal length of roughly 36.4 cm, indicating its near point. However, a lens with -1.30 diopters has a virtual focal point of about 76.9 cm, referring to its corrected far point.
What is the Near Point?
The near point is the closest distance at which the eye can focus on an object clearly. For people with healthy vision, this distance is typically about 25 cm or closer.
However, for those with farsightedness, their natural near point might be too far, making it difficult to see close objects clearly without the aid of lenses. A lens with positive diopters can bring this near point closer to the standard comfortable distance, allowing clearer focus.
  • For instance, in the problem, a lens with +2.75 diopters puts the near point approximately at 36.4 cm.
  • This adjustment enhances the ability to focus on nearby objects, making daily activities easier.
The Concept of the Far Point
The far point is the maximum distance at which an object can be seen clearly by the eyes. For those with myopia, or nearsightedness, this point is closer than it ideally should be.
Negative diopter lenses help push the focus to the back of the eye, effectively correcting the focal distance.
  • The far point for a -1.30 diopter lens can adjust to approximately 76.9 cm, indicating the point where far objects can be clearly seen.
  • This negative adjustment is necessary for those who have trouble seeing faraway objects.
Understanding the far point helps in designing corrective lenses that improve distant 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 screen must be moved 4.00 \(\mathrm{cm}\) to the left to refocus the image. Determine the focal length of the lens.

Pinhole Camera. A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image without a lens.(a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (Hint: Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is 25 \(\mathrm{cm}\) square and 20.0 \(\mathrm{cm}\) deep, with the hole in the middle of one of the 25 \(\mathrm{cm} \times 25 \mathrm{cm}\) faces. If this camera is used to photograph a fierce chicken that is 18 \(\mathrm{cm}\) high and 1.5 \(\mathrm{m}\) in front of the camera, how large is the image of this bird on the film? What is the magnification of this camera?

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the 95.0 -cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

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?

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 \(\mathrm{m}\) to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

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