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A magnifying glass produces an angular magnification of 4.5 when used by a young person with a near point of 18 cm. What is the maximum angular magnification obtained by an older person with a near point of \(45 \mathrm{cm} ?\)

Short Answer

Expert verified
The maximum angular magnification obtained by an older person with a near point of \(45\,cm\) is \(11.25\).

Step by step solution

01

Understand the formula for angular magnification

The formula for angular magnification (M) is given by the ratio of the near points of the older person to the younger person: \[M = \frac{Near Point\:of\:Older\:Person}{Near Point\:of\:Younger\:Person}\]
02

Find the angular magnification for the older person

Using the given values in the problem, we can plug them into the formula for angular magnification: \[M = \frac{45\, cm}{18\, cm}\]
03

Calculate the angular magnification

Divide 45 by 18 to find the angular magnification for the older person: \[M = \frac{45}{18} = 2.5\]
04

Find the maximum angular magnification

Finally, multiply the magnification of the younger person by the angular magnification we calculated in step 3: \[Maximum\:Angular\:Magnification = 4.5 * 2.5 = 11.25\]
05

Write down the maximum angular magnification

The maximum angular magnification obtained by an older person with a near point of 45 cm is 11.25.

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Key Concepts

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

Optics
Optics is the branch of physics that deals with the properties and behavior of light. It involves understanding how light interacts with different materials and how it can be manipulated to enhance or alter our perception.

In the study of optics, we explore concepts such as reflection, refraction, and diffraction, which describe how light waves change direction when encountering surfaces or transitioning between mediums. These concepts are crucial when designing instruments like microscopes and telescopes, which harness these principles to enhance vision.

Another important component of optics is the understanding of lenses and mirrors, which are used to direct and focus light. Lenses, such as those in magnifying glasses and glasses for vision correction, bend light rays to adjust how they converge or diverge, thereby affecting how images are viewed.
  • Reflection and refraction help analyze light paths.
  • Lenses are essential for modifying images we perceive.
  • Optics combine physics and technology for problem-solving.
Near Point
The near point of vision refers to the closest distance at which the eye can focus on an object clearly. For young, healthy eyes, this distance is typically about 25 cm. This ability decreases with age, and the near point moves further away, a condition known as presbyopia.

When using a magnifying glass or other optical aids, the near point is critical in determining how much magnification one can achieve comfortably. As the near point changes, so does the effectiveness and usage of magnifying techniques.

Understanding the near point can help tailor optical devices to individual needs, ensuring optimal usage and comfort. In the context of the exercise, the young person's near point of 18 cm allows them different magnification potential compared to an older person whose near point may be, for example, 45 cm.
  • The near point varies with age and conditions.
  • It influences the use of magnifying tools.
  • Adjustments are necessary for personalized optical aid.
Vision Correction
Vision correction is essential for individuals whose eyes do not naturally produce clear images due to various conditions like myopia or hyperopia. This involves using lenses in glasses or contact lenses to address these issues.

As people age, presbyopia often necessitates additional vision correction. This condition causes difficulty in seeing nearby objects, and is corrected using converging lenses or progressive lenses in spectacles. These corrective measures work by adjusting the focal length to bring images into clear focus at the near point, which has shifted.

In our exercise, understanding vision correction can explain how different near points affect the choice of optical aids. The magnification needed changes depending on how far the near point is from the eyes, and appropriate corrective lenses can modify this magnification effectively.
  • Glasses and contacts correct focus issues.
  • Presbyopia demands specific lens adjustment.
  • Optical aids are tailored to the user's needs.
Magnifying Glass
A magnifying glass is a convex lens used to produce a magnified image of an object. It works by bending light rays to enlarge the image, making it appear larger than the actual size. This is particularly useful for examining small details and is widely used in various fields for detailed inspection.

The magnifying glass can provide angular magnification, enhancing the details perceived by the eye. For instance, in our exercise, a younger individual using a given magnifying glass achieved an angular magnification of 4.5. Depending on one's near point, the effective magnification can vary significantly.

The formula for angular magnification, which is the near point ratio, helps determine the potential magnification one might achieve, connecting directly with the concepts covered. Magnifying glasses are not just tools for seeing small things but bridges between basic optics and practical vision correction.
  • Uses convex lenses for magnification.
  • Enhances the apparent size of objects.
  • Effectiveness varies with near point distance.

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Most popular questions from this chapter

An object of height 3 cm is placed at a distance of 25 \(\mathrm{cm}\) in front of a converging lens of focal length \(20 \mathrm{cm},\) to be referred to as the first lens. Behind the lens there is another converging lens of focal length \(20 \mathrm{cm}\) placed \(10 \mathrm{cm}\) from the first lens. There is a concave mirror of focal length 15 \(\mathrm{cm}\) placed \(50 \mathrm{cm}\) from the second lens. Find the location, orientation, and size of the final image.

A very myopic man has a far point of \(20.0 \mathrm{cm}\). What power contact lens (when on the eye) will correct his distant vision?

Unreasonable Results Your friends show you an image through a microscope. They tell you that the microscope has an objective with a 0.500 -cm focal length and an eyepiece with a \(5.00-\mathrm{cm}\) focal length. The resulting overall magnification is 250,000 . Are these viable values for a microscope? Unless otherwise stated, the lens-to-retina distance is 2.00 \(\mathrm{cm}\)

An object \(1.50 \mathrm{cm}\) high is held \(3.00 \mathrm{cm}\) from a person's cornea, and its reflected image is measured to be 0.167 cm high. (a) What is the magnification? (b) Where is the image? (c) Find the radius of curvature of the convex mirror formed by the comea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer.)

Ray tracing for a flat mirror shows that the image is located a distance behind the mirror equal to the distance of the object from the mirror. This is stated as \(d_{i}=-d_{\mathrm{o}}\) since this is a negative image distance (it is a virtual image). What is the focal length of a flat mirror?

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