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An engraver uses a magnifying glass \((f=9.50 \mathrm{cm})\) to examine some work, as in Figure \(26.39 b .\) 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) 15.31 cm; (b) 3.63.

Step by step solution

01

Understand the Problem

We are given a focal length, the distance to the near point, and we need to find the distance between the object and the lens, as well as the angular magnification.
02

Identify Given Values

The focal length of the magnifying glass is given as \( f = 9.50 \, \mathrm{cm} \). The image distance from the magnifying glass to the eye, equal to the near point, is \( v = 25.0 \, \mathrm{cm} \).
03

Calculate the Object Distance

Use the lens formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) to find the object distance \( u \). Rearranging for \( u \), we have:\[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]Substituting the given values:\[ \frac{1}{u} = \frac{1}{9.50} - \frac{1}{25.0} \]\[ \frac{1}{u} = 0.1053 - 0.0400 = 0.0653 \]Therefore:\[ u = \frac{1}{0.0653} \approx 15.31 \, \mathrm{cm} \]
04

Calculate Angular Magnification

The angular magnification \( M \) of a magnifying glass can be calculated using the formula:\[ M = 1 + \frac{D}{f} \]where \( D = 25.0 \, \mathrm{cm} \) is the near point of a person with normal eyesight. Substituting:\[ M = 1 + \frac{25.0}{9.50} \approx 1 + 2.63 = 3.63 \]
05

Conclusion

The distance between the work and the magnifying glass is approximately \( 15.31 \, \mathrm{cm} \), and the magnification is approximately \( 3.63 \).

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Key Concepts

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

Focal Length
In optics, the focal length is a fundamental concept when it comes to lenses and vision enhancement tools like magnifying glasses. Simply put, the focal length of a lens is the distance over which initially parallel rays of light are brought to a point, or focus. It determines how strongly the lens converges (or diverges) light. The shorter the focal length, the greater the lens's ability to bend light beams. A lens with a positive focal length, as in the case of a simple magnifying glass, focuses incoming light to a point beyond the lens. In the given problem, we know that the focal length of the lens used by the engraver is 9.50 cm. This means the lens is fairly strong at converging light within a relatively short distance, allowing the engraver to examine details closely. It's crucial to remember that focal length forms a relationship with other properties of light in systems, specifically through equations like the lens formula, influencing both image and object distance.
Lens Formula
The lens formula is one of the essential equations in optics, capturing the relationship between the object distance (the distance from the object being viewed to the lens), the focal length of the lens, and the image distance (from the lens to the image location perceived by the viewer).The formula is expressed as:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]Here:- \( f \) is the focal length,- \( v \) is the image distance, and- \( u \) is the object distance.This equation helps solve the distances involved in the optics system. For instance, in the engraver's scenario, the known values (\( f = 9.50 \text{ cm} \) and \( v = 25.0 \text{ cm} \)) were used to find \( u \). By rearranging the formula to solve for \( u \), we find:\[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]Plugging the values, this becomes:\[ \frac{1}{u} = 0.1053 - 0.0400 = 0.0653 \]which, when inverted, gives \( u \approx 15.31 \text{ cm} \). This shows that the object (engraving) is positioned 15.31 cm from the magnifying glass for the image to be focused correctly on the engraver's near point.
Angular Magnification
Angular magnification is a measure of how much larger a lens can make an object appear to the human eye compared to the object's actual size. It's a critical measure when using tools like magnifying glasses, especially in detail-oriented tasks, such as engraving or reading small print.The formula for angular magnification \( M \) of a simple magnifying glass is:\[ M = 1 + \frac{D}{f} \]where \( D \) is the near point for a typical human, usually 25.0 cm, and \( f \) is the lens's focal length.Applying the formula for the engraver's magnifying glass with \( f = 9.50 \text{ cm} \), we calculate:\[ M = 1 + \frac{25.0}{9.50} \approx 1 + 2.63 = 3.63 \]This means the magnifying glass makes the engraving appear 3.63 times larger than its actual size. Knowing the magnification aids in selecting the correct tools for tasks requiring enhanced precision, ensuring that the magnified view aligns well with human visual capabilities.

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Most popular questions from this chapter

When light strikes the surface between two materials from above, the Brewster angle is \(65.0^{\circ} .\) What is the Brewster angle when the light encounters the same surface from below?

The near point of a naked eye is \(25 \mathrm{cm} .\) When placed at the near point and viewed by the naked eye, a tiny object would have an angular size of \(5.2 \times 10^{-5}\) rad. When viewed through a compound microscope, however, it has an angular size of \(-8.8 \times 10^{-3}\) rad. (The minus sign indicates that the image produced by the microscope is inverted.) The objective of the microscope has a focal length of \(2.6 \mathrm{cm},\) and the distance between the objective and the eyepiece is \(16 \mathrm{cm} .\) Find the focal length of the eyepiece.

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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, measured 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 layer of oil \((n=1.45)\) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid. The angle of incidence is \(64.0^{\circ},\) and the angle of refraction is \(53.0^{\circ} .\) What is the index of refraction of the unknown liquid?

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