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A converging lens of focal length \(8.000 \mathrm{cm}\) is \(20.0 \mathrm{cm}\) to the left of a diverging lens of focal length \(-6.00 \mathrm{cm} .\) A coin is placed \(12.0 \mathrm{cm}\) to the left of the converging lens. Find (a) the location and (b) the magnification of the coin's final image.

Short Answer

Expert verified
The final image is 12.0 cm to the left of the diverging lens and the magnification is 6.0.

Step by step solution

01

Find the location of the image formed by the converging lens

Using the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f = 8.0\, \text{cm} \) and \( d_o = 12.0\, \text{cm} \) (object distance), calculate the image distance \( d_i \) for the converging lens. Rearranging gives \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{8.0} - \frac{1}{12.0} \). Solving for \( d_i \), we find \( d_i = 24.0\, \text{cm} \).
02

Determine the object distance for the diverging lens

The image from the converging lens (found previously) acts as the object for the diverging lens. The distance between the lenses is \( 20.0\, \text{cm} \), so the object distance for the diverging lens \( d_{o2} = 20.0\, \text{cm} - 24.0\, \text{cm} = -4.0\, \text{cm}\) (using the convention that positive is to the right and negative to the left).
03

Find the location of the image formed by the diverging lens

Apply the lens formula again for the diverging lens: \( \frac{1}{f_2} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}} \), where \( f_2 = -6.0\, \text{cm} \) and \( d_{o2} = -4.0\, \text{cm} \). Solving \( \frac{1}{d_{i2}} = \frac{1}{-6.0} - \frac{1}{-4.0} \), we find \( d_{i2} = -12.0\, \text{cm} \). This means the final image is located \( 12.0\, \text{cm} \) to the left of the diverging lens.
04

Calculate the magnification for each lens

The magnification \( m_1 \) from the converging lens is given by \( m_1 = -\frac{d_i}{d_o} = -\frac{24.0}{12.0} = -2.0 \). The magnification \( m_2 \) from the diverging lens is \( m_2 = -\frac{d_{i2}}{d_{o2}} = -\frac{-12.0}{-4.0} = -3.0 \).
05

Find the total magnification

The total magnification \( M \) is the product of the magnifications from each lens: \( M = m_1 \times m_2 = -2.0 \times -3.0 = 6.0 \).

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Key Concepts

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

Focal Length
Focal length is a fundamental concept in optics. It is the distance between the center of a lens and its focus, where light rays converge or diverge. When dealing with lenses, knowing the focal length helps in determining how the lens will affect light traveling through it.

For a converging lens, which has a positive focal length, it converges parallel light rays to a point on the opposite side of the lens.
  • A smaller focal length means the lens is more powerful, converging rays more sharply.
  • A larger focal length indicates a weaker lens, as it takes longer for the rays to meet.


On the other hand, a diverging lens has a negative focal length. It causes parallel light rays to spread out, as if they are originating from a point on the same side as the incoming rays. This makes diverging lenses essential for procedures that involve spreading light or managing light paths for specific design and purposes.
Magnification
Magnification measures how much larger or smaller an image is compared to the object itself. It is represented by the ratio of the image height to the object height.
  • A positive magnification means the image is upright compared to the object.
  • A negative magnification indicates an inverted image.


In lens systems like the exercise example, you can calculate magnification for individual lenses using the formula: \[ m = -\frac{d_i}{d_o} \] where \( m \) is magnification, \( d_i \) is image distance, and \( d_o \) is object distance.

For compound lens systems, total magnification is the product of individual magnifications. Later on, calculations can show how lenses might, together, greatly enlarge or reduce the size of the image. Adjusting the lens properties and combinations allows for controlling image qualities in practical applications like microscopy or photography.
Image Distance
Image distance refers to the distance between the lens and the position where the image is formed. It is pivotal in determining the behavior and placement of the image.
  • For converging lenses, a positive image distance indicates a real image located on the opposite side of the lens from the object.
  • For diverging lenses, a negative image distance suggests a virtual image on the same side as the object.


The lens formula connects image distance \( d_i \), object distance \( d_o \), and focal length \( f \) as follows: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

Understanding the image distance is key for practical applications, such as adjusting cameras or designing optical instruments. By controlling this distance, users can decide where an image forms and how it aligns, enabling precise operations in fields like visual arts, astronomy, and scientific research.
Optics
Optics is the branch of physics that studies light and its properties. It is a comprehensive field that encompasses how light interacts with objects like lenses, mirrors, and filters. Optics plays a crucial role in developing technologies for communications, health, and engineering.

Key areas within optics include:
  • Geometrical Optics: Deals with rays, lenses, and the creation of images, focusing on real and virtual images.
  • Physical Optics: Discusses the wave nature of light, interference, diffraction, and polarization.


Understanding optics concepts allows for the design and use of technologies like glasses, microscopes, telescopes, and cameras effectively. Optics enable us to manipulate light, improving how we view and analyze the world.

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Most popular questions from this chapter

You have two lenses: lens 1 with a focal length of \(0.45 \mathrm{cm}\) and lens 2 with a focal length of \(1.9 \mathrm{cm}\). If you construct a microscope with these lenses, which one should you use as the objective? Explain.

With unaided vision, a physician can focus only on objects that lie at distances between \(5.0 \mathrm{m}\) and \(0.50 \mathrm{m}\). (a) Which type of lens (converging or diverging) is needed to correct her nearsightedness? Explain. (b) Which type of lens will correct her farsightedness? Explain. (c) Find the refractive power needed for each part of the bifocal contact lenses that will give the physician normal visual acuity from \(25 \mathrm{cm}\) out to infinity.

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?

The Moon is \(3476 \mathrm{km}\) in diameter and orbits the Earth at an average distance of \(384,400 \mathrm{km} .\) (a) What is the angular size of the Moon as seen from Earth? (b) A penny is \(19 \mathrm{mm}\) in diameter. How far from your eye should the penny be held to produce the same angular diameter as the Moon?

without his glasses, Isaac can see objects clearly only if they are less than \(4.5 \mathrm{m}\) from his eyes. What focal length glasses worn \(2.1 \mathrm{cm}\) from his eyes will allow lsaac to see distant objects clearly?

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