Locating Object & Image with a Converging Lens

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SUMMARY

A converging lens with a focal length of 7 cm produces an erect image measuring 1.10 cm tall from a real object that is positioned to the left of the lens. The object distance can be calculated using the lens equation 1/f = 1/o + 1/i. Additionally, a second converging lens with a focal length of 12 cm creates a virtual image 5 mm tall, located 19 cm to the right of the lens, requiring the object distance to be determined with the correct sign using the same lens equation and the magnification formula m = -i/o.

PREREQUISITES
  • Understanding of lens equations, specifically 1/f = 1/o + 1/i
  • Knowledge of linear magnification, defined as m = -i/o
  • Familiarity with the concepts of real and virtual images
  • Basic skills in solving algebraic equations
NEXT STEPS
  • Explore the application of the lens equation in various optical systems
  • Learn about the characteristics of real versus virtual images in optics
  • Investigate simulation tools for visualizing lens behavior, such as ray tracing software
  • Study the effects of different focal lengths on image formation
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the principles of lens behavior and image formation.

eku_girl83
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1) Aconverging lens with a focal length of 7 cm forms an image of a 4 mm tall real object that is to the left of the lens. The image 1.10 cm tall and erect.
Where is the object located?
Where is the image located?
2) A converging lens with a focal length of 12 cm forms a virtual image 5 mm tall, 19 cm to the right of the lens.
Determine the position of the object. (Give the object distance with the correct sign).
Determine the size of the object.
 
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You'll need the lens equation: 1/f = 1/o + 1/i, and the definition of linear magnification: m = -i/o.

Look up how these equations are used and give the problems a try.
 
eku_girl83 said:
1) Aconverging lens with a focal length of 7 cm forms an image of a 4 mm tall real object that is to the left of the lens. The image 1.10 cm tall and erect.
Where is the object located?
Where is the image located?
2) A converging lens with a focal length of 12 cm forms a virtual image 5 mm tall, 19 cm to the right of the lens.
Determine the position of the object. (Give the object distance with the correct sign).
Determine the size of the object.
There is a good simulation program picture what about it
http://www.geocities.com/mrfarag
 

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