How Does Lens Focal Length Affect Image Size?

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SUMMARY

The discussion focuses on the calculation of focal lengths and image sizes using lens formulas. The first part involves determining the focal length of a lens that produces a real image twice the height of the object, with an object distance of 26.0 cm. The second part addresses a diverging lens with the same focal length, where the object height is 5.00 cm, and the task is to find the height of the image formed. Key formulas used include magnification (m = -i/o) and the lens formula (1/o + 1/i = 1/f).

PREREQUISITES
  • Understanding of lens formulas and sign conventions in optics
  • Familiarity with the concepts of real and virtual images
  • Knowledge of magnification calculations
  • Basic principles of diverging and converging lenses
NEXT STEPS
  • Study the derivation of the lens formula (1/o + 1/i = 1/f)
  • Learn about the properties of diverging lenses and their image formation
  • Explore practical applications of magnification in photography
  • Investigate the effects of varying object distances on image characteristics
USEFUL FOR

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

thurrucane
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Homework Statement



An object is located 26.0cm from a certain lens. The lens forms a real image that is twice as high as the object.

(i) What is the focal length of this lens?

(ii) Now replace the lens used in Part 1 with another lens. The new lens is a diverging lens whose focal points are at the same distance from the lens as the focal points of the first lens. If the object is 5.00cm high, what is the height of the image formed by the new lens? The object is still located 26.0cm from the lens.

Homework Equations



Since the image is real and 2X as large, the magnification m = -2. The formula
m = -i/o
1/o + 1/i = 1/f

The Attempt at a Solution



After attempting part 1 i got 5.77cm which is incorrect.
 
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You got to take care of the sign convention.
 

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