How Do You Calculate Image Size and Location with a Diverging Lens?

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SUMMARY

The calculation of image size and location for a diverging lens involves using the lens formula and magnification equations. For an object 3 cm tall placed 16 cm in front of a diverging lens with a focal length of -24 cm, the image distance (di) is calculated as -48 cm, indicating the image is virtual and located on the same side as the object. The image height (hi) is determined to be -9 cm, confirming that the image is smaller than the object and inverted. This aligns with the properties of diverging lenses, which always produce virtual, upright, and reduced images.

PREREQUISITES
  • Understanding of the lens formula: 1/f = 1/di + 1/do
  • Knowledge of magnification formula: hi/ho = di/do
  • Familiarity with the characteristics of diverging lenses
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the properties of diverging lenses in optics
  • Learn about real vs. virtual images in lens systems
  • Explore the impact of varying object distances on image formation
  • Practice solving lens problems with different focal lengths and object heights
USEFUL FOR

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

Robin_oto
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okay the problem reads
-an object 3 cm tall is placed 16 cm in front of a diverging lens with a focal length of 24cm. Find the location and size of the image. What kind of image is it?



1/f = 1/di+1/do

hi/ho = di/do


I'm not to sure how to solve this i used the equation; 1/f = 1/di+1/do and solved for di, getting -48cm. then pluged it into the equation hi/ho = di/do, and got -9cm. It doesn't seem right to me because when i draw out the problem it shows the image as smaller then the actual size, not bigger. please help!
 
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The focal length of a diverging lens should be taken negative: f=-24 cm

ehild
 

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