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An object has an angular size of 0.0150 rad when placed at the near point \((21.0 \mathrm{~cm})\) of an eye. When the eye views this object using a magnifying glass, the largest possible angular size of the image is 0.0380 rad. What is the focal length of the magnifying glass?

Short Answer

Expert verified
The focal length of the magnifying glass is approximately 13.69 cm.

Step by step solution

01

Understand the Problem

We need to find the focal length of a magnifying glass. When an object is viewed with the magnifying glass at the near point of an eye, the largest angular size is given. The angular size without the magnifying glass is also known. The formula for angular magnification can help relate these parameters to the focal length.
02

Angular Magnification Formula

The angular magnification \( M \) of a magnifying glass, when the object is placed at the focal point, is given by \( M = \frac{0.0380}{0.0150} \). Calculate \( M \) by dividing the angular size of the image by the angular size of the object.
03

Calculate Angular Magnification

Compute the angular magnification \( M = \frac{0.0380}{0.0150} = 2.5333 \). This value represents how much larger the image appears compared to the object.
04

Focal Length Relation

The angular magnification \( M \) is also related to the focal length \( f \) of the magnifying glass and the near point distance \( D \) by the formula \( M = 1 + \frac{D}{f} \). Here, \( D = 21.0 \text{ cm} \), which needs to be converted to meters \(D = 0.21 \text{ m}\).
05

Rearrange Formula to Solve for Focal Length

Use the magnification formula: \( M = 1 + \frac{D}{f} \) to express the focal length \( f \): \( \frac{D}{f} = M - 1 \) thus \( f = \frac{D}{M - 1} \).
06

Substitute Known Values and Solve

Substitute \( D = 0.21 \text{ m} \) and \( M = 2.5333 \) into the equation to find \( f \): \( f = \frac{0.21}{2.5333 - 1} \). Calculate the focal length.
07

Calculate Result

Perform the division: \( f = \frac{0.21}{1.5333} \approx 0.1369 \text{ m} \). Convert the focal length into centimeters: \( f = 13.69 \text{ cm} \).

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Key Concepts

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

Angular Magnification
Angular magnification is a key concept when dealing with optical instruments like magnifying glasses. It describes how much larger an object appears when viewed through a magnifying glass compared to viewing it with the naked eye. The magnification factor is determined by the formula:\[ M = \frac{\theta'}{\theta} \]where \( \theta' \) is the angular size of the image seen through the magnifying glass, and \( \theta \) is the angular size of the object without the magnifying glass.
This ratio gives a numerical value indicating how many times the image is enlarged. In our example, the angular magnification is calculated as \( M = \frac{0.0380}{0.0150} \), which equals \( 2.5333 \).
This means the object appears more than twice its normal size when viewed with the magnifying glass.
  • Understanding magnification helps in determining the effectiveness of a magnifying glass.
  • It aids in applications requiring precise visual enlargement, such as detailed inspection or reading fine text.
Near Point Distance
The near point distance is an important parameter in optics, particularly when discussing vision and optical aids. It represents the closest distance at which the human eye can focus comfortably and clearly. For a typical human eye, this distance is about 25 centimeters, but it can vary.
In this exercise, the near point distance is given as 21 centimeters, indicating the specific condition for the eye viewing the object.
This distance is crucial when using a magnifying glass because it acts as a baseline for calculating angular magnification and determining viewing ease.
By placing an object at the near point with a magnifying glass, the angular size is effectively increased:
  • The improvement in angular size facilitates easier reading or observation.
  • Understanding your own near point can help choose the correct optical aids.
Optics
Optics is the branch of physics focusing on the behavior and properties of light, including its interactions with matter and the construction of instruments to use or detect it. Essential concepts in optics include reflection, refraction, diffraction, and the formation of images.
When you use a magnifying glass, you exploit these principles to enlarge an image. The key optical element here is the converging lens of the magnifying glass.
Some core ideas in optics that relate to magnification include:
  • Refraction: The bending of light as it passes through materials like glass.
  • Focal Length: The distance from the lens where light converges to a point.
  • Image Formation: The process by which lenses form an image by bending light rays.
Understanding optics allows you to appreciate how light and lenses work together to make small objects appear larger.
Magnifying Glass
The magnifying glass is a simple yet powerful optical tool used to enlarge the appearance of objects. It consists of a convex lens, which bends light rays outward, thereby making the image appear larger.
The effectiveness of a magnifying glass is largely determined by its focal length, which is inversely related to its angular magnification. The shorter the focal length, the greater the potential magnification, which can enhance the detail observed in small objects.
The magnifying glass is commonly used in:
  • Reading small print or detailed documents.
  • Inspecting objects closely, such as in hobbies like stamp collecting.
  • Scientific investigations requiring magnified observation.
By understanding the role of the focal length in magnifying glasses, users can select the right tool for their needs, tailoring the viewing experience to their specific requirements.

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Most popular questions from this chapter

An amateur astronomer decides to build a telescope from a discarded pair of eyeglasses. One of the lenses has a refractive power of 11 diopters, and the other has a refractive power of 1.3 diopters. (a) Which lens should be the objective? (b) How far apart should the lenses be separated? (c) What is the angular magnification of the telescope?

Red light \((n=1.520)\) and violet light \((n=1.538)\) traveling in air are incident on a slab of crown glass. Both colors enter the glass at the same angle of refraction. The red light has an angle of incidence of \(30.00^{\circ} .\) What is the angle of incidence of the violet light?

(a) For a diverging lens \((f=-20.0 \mathrm{~cm}),\) construct a ray diagram to scale and find the image distance for an object that is \(20.0 \mathrm{~cm}\) from the lens. (b) Determine the magnification of the lens from the diagram.

Two converging lenses \(\left(f_{1}=9.00 \mathrm{~cm}\right.\) and \(\left.f_{2}=6.00 \mathrm{~cm}\right)\) are separated by \(18.0 \mathrm{~cm} .\) The lens on the left has the longer focal length. An object stands \(12.0 \mathrm{~cm}\) to the left of the lefthand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, is the final image (d) upright or inverted and is it (e) larger or smaller?

The contacts wom by a farsighted person allow her to see objects clearly that are as close as \(25.0 \mathrm{~cm}\), even though her uncorrected near point is \(79.0 \mathrm{~cm}\) from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of \(217 \mathrm{~cm}\) from her eyes. (a) How far away is the poster actually located? (b) If the poster is \(0.350 \mathrm{~m}\) tall, how tall is the image formed by the contacts?

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