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An engraver uses a magnifying glass \((f=9.50 \mathrm{~cm})\) to examine some work, as in Figure \(26-40 b\). The image he sees is located \(25.0 \mathrm{~cm}\) from his eye, which is his near point. (a) What is the distance between the work and the magnifying glass? (b) What is the angular magnification of the magnifying glass?

Short Answer

Expert verified
(a) The object distance is 79.37 cm. (b) The angular magnification is 3.63.

Step by step solution

01

Identify the Given Information

The focal length of the magnifying glass (\(f\)) is \(9.50\text{ cm}\). The image distance is \(25.0\text{ cm}\). We are asked to find the object distance (\(d_o\)) and the angular magnification (\(M\)).
02

Apply the Lens Formula

The lens formula is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). We need to find \(d_o\), and we know \(f = 9.50\text{ cm}\) and \(d_i = 25.0\text{ cm}\). Rearrange the formula to solve for \(d_o: \frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}\).
03

Calculate the Object Distance

Substitute the given values into the rearranged lens formula: \(\frac{1}{d_o} = \frac{1}{9.50} - \frac{1}{25.0}\). Calculate \(\frac{1}{d_o}\) which equals approximately \(0.0526 - 0.04 = 0.0126\). Thus, \(d_o = \frac{1}{0.0126} \approx 79.37\text{ cm}\).
04

Calculate the Angular Magnification

The angular magnification is given by the formula \(M = 1 + \frac{D}{f}\), where \(D \)is the near point distance \(= 25.0\text{ cm}\). Substitute the known values:\[M = 1 + \frac{25.0}{9.50} \approx 1 + 2.63 = 3.63\].

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Key Concepts

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

Lens Formula
Optical lenses are fascinating tools used to form images, and the lens formula is central in understanding how they work. This formula connects three key values: the focal length \(f\), the object distance \(d_o\), and the image distance \(d_i\). The lens formula is given by:
\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]
In practical terms, this means you can determine any of these three values if the other two are known. This formula applies to both converging lenses (like the magnifying glass in the exercise) and diverging lenses, although the sign conventions may vary.When we want to find the object distance, as in our exercise, we rearrange this formula to:
\[\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}\]
By substituting known values, it's possible to solve for the unknown distance, which is essential for practical applications like focusing a magnifying glass.
Angular Magnification
Angular magnification is a measure of how much larger or smaller an object appears when viewed through an optical instrument compared to the naked eye. It is especially useful in tools like telescopes and magnifying glasses, allowing for detailed examination of small objects.The angular magnification \(M\) of a magnifying glass is calculated using:
\[M = 1 + \frac{D}{f}\]
Here, \(D\) represents the near point distance, or the closest point at which the eye can comfortably see an object. For most people, this is around 25 cm.This formula makes it straightforward to calculate how much a magnifying glass enlarges an object. In the given exercise, the magnification factor of 3.63 means that the object appears 3.63 times larger than its actual size when viewed at a normal reading distance. This is crucial for tasks like engraving or circuit board examination, where seeing fine detail with clarity is necessary.
Focal Length
The focal length \(f\) is one of the most important parameters of a lens. It is defined as the distance from the lens to the point where the light rays converge to a focus. This concept is central in designing optical systems, such as cameras, glasses, telescopes, and more.A shorter focal length results in a stronger lens, creating a larger magnification with a shorter distance. Conversely, a longer focal length offers less magnification but can capture a broader view. In our exercise, the focal length of the magnifying lens is given as 9.50 cm. This shorter focal length indicates a powerful magnification capability, useful for examining fine details.Understanding focal lengths is crucial when choosing or using optical equipment, as it directly impacts the size and clarity of the image produced. A proper grasp of focal length helps in selecting the right lens for a specific need, whether for reading, hobby work, or advanced scientific applications.

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Most popular questions from this chapter

A compound microscope has a barrel whose length is \(16.0 \mathrm{~cm}\) and an eyepiece whose focal length is \(1.4 \mathrm{~cm}\). The viewer has a near point located \(25 \mathrm{~cm}\) from his eyes. What focal length must the objective have so that the angular magnification of the microscope will be \(-320 ?\)

An object is placed \(20.0 \mathrm{~cm}\) to the left of a diverging lens \((f=-8.00 \mathrm{~cm}) .\) A concave mirror \((f=12.0 \mathrm{~cm})\) is placed \(30.0 \mathrm{~cm}\) to the right of the lens. (a) Find the final image distance, mea sured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?

A macroscopic (or macro) lens for a camera is usually a converging lens of normal focal length built into a lens barrel that can be adjusted to provide the additional lens-tofilm distance needed when focusing at very close range. Suppose that a macro lens \((f=\) \(50.0 \mathrm{~mm}\) ) has a maximum lens-to- film distance of \(275 \mathrm{~mm}\). How close can the object be located in front of the lens?

Consult Interactive Solution 26.51 at to review the concepts on which this problem depends. A camera is supplied with two interchangeable lenses, whose focal lengths are 35.0 and \(150.0 \mathrm{~mm}\). A woman whose height is \(1.60 \mathrm{~m}\) stands \(9.00 \mathrm{~m}\) in front of the camera. What is the height (including sign) of her image on the film, as produced by (a) the 35.0 -mm lens and (b) the 150.0 -mm lens?

The back wall of a home aquarium is a mirror that is a distance \(L\) away from the front wall. The walls of the tank are negligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. (a) Does the fish appear to be at a distance greater than, less than, or equal to \(\frac{1}{2} L\) from the front wall? (b) The mirror forms an image of the fish. How far from the front wall is this image located? Express your answer in terms of \(L\). (c) Assume that your answer to Question (b) is a distance \(D\). Does the image of the fish appear to be at a distance greater than, less than, or equal to \(D\) from the front wall? (d) Could the image of the fish appear to be in front of the mirror if the index of refraction of water were different than it actually is, and, if so, would the index of refraction have to be greater than or less than its actual value? Explain each of your answers. The distance between the back and front walls of the aquarium is \(40.0 \mathrm{~cm} .\) (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall. Verify that your answers are consistent with your answers to the Concept Questions.

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