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Chapter 27: Problem 53

To engrave wishes of good luck on a watch, an engraver uses a magnifier whose focal length is \(8.65 \mathrm{cm}\). If the image formed by the magnifier is at the engraver's near point of \(25.6 \mathrm{cm}\) find (a) the distance between the watch and the magnifier and

Short Answer

Expert verified
The distance between the watch and the magnifier is approximately \(6.47 \text{ cm}\).

Step by step solution

01

Understand the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance (the distance from the watch to the magnifier). In this problem, \( f = 8.65 \text{ cm} \) and \( v = -25.6 \text{ cm} \) (negative because the image is virtual and on the same side as the object).
02

Substitute the Given Values

Substitute the given values into the lens formula: \( \frac{1}{8.65} = \frac{1}{-25.6} - \frac{1}{u} \).
03

Rearrange the Equation

Rearrange the equation to solve for \( u \): \( \frac{1}{u} = \frac{1}{-25.6} - \frac{1}{8.65} \).
04

Calculate \( \frac{1}{u} \)

Calculate \( \frac{1}{u} \) by computing \( \frac{1}{-25.6} \approx -0.03906 \) and \( \frac{1}{8.65} \approx 0.11561 \). Therefore, \( \frac{1}{u} = -0.03906 - 0.11561 \approx -0.15467 \).
05

Calculate \( u \)

Now calculate \( u \) by taking the reciprocal of \( \frac{1}{u} \): \( u \approx \frac{1}{-0.15467} \approx -6.47 \text{ cm} \). The negative sign indicates the object is on the same side as the light source, which is expected for a magnifier.

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Key Concepts

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

Focal Length
The focal length is a fundamental concept in optics that indicates how strongly a lens converges or diverges light.
The focal length is denoted by the symbol \( f \).
For a magnifying lens, like the one used by our engraver, it determines the power of magnification.
  • A positive focal length implies a converging lens, which brings light rays together.
  • A negative focal length (not applicable here) corresponds to a diverging lens.
The focal length, measured in centimeters (cm) or meters (m), is crucial for calculating the object and image distances in a lens system.
In our exercise, the lens has a focal length of \( 8.65 \text{ cm} \), which influences the positions where the object (watch) and image (engraved text view) are located.
To use the lens formula effectively, understanding the role of the focal length helps to predict how the system behaves.
Image Distance
Image distance, denoted by \( v \), is another vital concept in understanding lens functioning.
It represents the distance from the lens to the image formed.
  • A positive image distance means the image is formed on the opposite side of the lens from the object, making it a real image.
  • A negative image distance suggests a virtual image, meaning the image forms on the same side as the object.
For our engraver's magnifying lens, the image is at the engraver's near point of \( 25.6 \text{ cm} \).
This given image distance is negative (\( v = -25.6 \text{ cm} \)) as it indicates a virtual image created by the magnifier.
Virtual images are typical in magnifying lenses because they provide an enlarged view of the object without flipping it upside down.
Object Distance
Object distance, represented by \( u \), is the distance between the object and the lens.
It's essential for understanding where an object (like a watch) needs to be placed to achieve a certain image through the lens.
  • If \( u \) is positive, the object is on the opposite side of the lens relative to the incoming light.
  • If \( u \) is negative, as in the case with the engraver, the object lies on the same side as the light source or observer.
By employing the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), we can determine \( u \).
Substituting the focal length and image distance, we find \( u \approx -6.47 \text{ cm} \).
The negative sign reflects the placement of the watch and confirms that the lens behaves as a magnifier, rendering a virtual image.

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Most popular questions from this chapter

Two concave lenses, cach with \(f=-12 \mathrm{cm},\) are separated by \(6.0 \mathrm{cm}\). An object is placed \(24 \mathrm{cm}\) in front of one of the lenses. Find (a) the location and (b) the magnification of the final image produced by this lens combination.

Find the focal length of contact lenses that would allow a farsighted person with a near-point distance of \(176 \mathrm{cm}\) to read a book at a distance of \(10.1 \mathrm{cm}\)

Pricey Stamp A rare 1918 "Jenny" stamp, depicting a misprinted, upside-down Curtiss JN-4 "Jenny" airplane, sold at auction for \(\$ 525,000 . A collector uses a simple magnifying glass to examine the "Jenny," obtaining a linear magnification of 2.5 when the stamp is held \)2.76 \mathrm{cm} from the lens. What is the focal length of the magnifying glass?

Four camera lenses have the following focal lengths and \(f\) -numbers: $$ \begin{array}{lcc} \hline \text { Lens } & \text { Focal length }(\mathrm{mm}) & \text { f-number } \\ \hline \mathrm{A} & 150 & f / 1.2 \\ \mathrm{B} & 150 & f / 5.6 \\ \mathrm{C} & 35 & f / 1.2 \\ \mathrm{D} & 35 & f / 5.6 \\ \hline \end{array} $$ Rank these lenses in order of increasing aperture diameter. Indicate ties where appropriate.

The "tube length" of a microscope is defined to be the difference between the (objective) image distance and objective focal length: \(L=d_{i}-f_{\text {objective }}\). Many microscopes are standardized to a tube length of \(L=160 \mathrm{mm}\). Consider such a microscope whose objective lens has a focal length \(f_{\text {objective }}=7.50 \mathrm{mm}\)
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