Solving a Focal Length Puzzle: Find the Image Location and Size

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SUMMARY

The discussion centers on solving a focal length problem involving a converging lens with a focal length of 25 cm. The object, measuring 1.00 cm in height, is positioned 15.0 cm in front of the lens. The calculations yield an image location of -37.5 cm, indicating the image is formed 37.5 cm behind the lens, and the image height is calculated to be 2.50 cm. However, discrepancies arise in the ray diagram, prompting questions about sign conventions and object distance.

PREREQUISITES
  • Understanding of lens formulas, specifically the thin lens equation
  • Knowledge of magnification calculations in optics
  • Familiarity with ray diagrams and sign conventions in optics
  • Basic principles of converging lenses and focal lengths
NEXT STEPS
  • Review the thin lens equation and its applications in optics
  • Study the principles of ray diagrams for converging lenses
  • Learn about sign conventions in optics, particularly for object and image distances
  • Explore magnification calculations and their significance in image formation
USEFUL FOR

Students and educators in physics, particularly those focusing on optics, as well as anyone interested in understanding lens behavior and image formation techniques.

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A converging lens has a focal length of 25cm. A 1.00cm tall object is placed 15.0cm in front of the lens. Find the image location and size, and describe it. Draw a ray diagram.

Ok so I get 1/p + 1/q = 1/f rearranged for q=pf/p-f which equals -37.5, so the object is 37.5cm behind the lens. and then m=-q/p=2.50cm. The h'=mh(i)=2.50*1.00=2.50cm tall. This all works out

BUT the ray diagram doesn't make the image at 37.5cm behind the lens. Some one help please!
 
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Did you learned the sign conventions? what is the sign of the object distance?
 

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