Converging Lens Magnification Problem

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SUMMARY

The discussion focuses on calculating the magnification of a virtual image formed by a converging lens with a focal length of 15.0 cm. The user initially calculated the object distance (p) as 35.5 cm and the magnification (M) as -0.733. However, the magnification was marked incorrect due to a sign convention error, as the image distance (q) for a virtual image should be negative. The correct magnification is derived from the formula M = -(q/p), with the proper sign convention applied.

PREREQUISITES
  • Understanding of lens formulas, specifically (1/p) + (1/q) = (1/f)
  • Knowledge of sign conventions for virtual images in optics
  • Familiarity with magnification calculations in optics
  • Basic principles of converging lenses
NEXT STEPS
  • Study the sign conventions for lenses in optics
  • Learn about the properties of virtual images formed by converging lenses
  • Explore advanced magnification calculations for different lens types
  • Review practical applications of lens formulas in optical systems
USEFUL FOR

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

soccerjayl
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What am I missing here?

A person looks at a gem using a converging lens with a focal length of 15.0 cm. The lens forms a virtual image 26.0 cm from the lens. Determine the magnification.

(1/p)+(1/q) = (1/f)
p=?
q=26 cm
f=15 cm

(1/15)-(1/26)= Ans^-1 = 35.5 cm (p)

Magnification= -(q/p)
M= -(26/35.5)
M= -0.733

Answer [-.733] has been marked wrong. What did I do?
 
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check your sign convention

soccerjayl said:
(1/p)+(1/q) = (1/f)
p=?
q=26 cm
f=15 cm
Your sign convention is incorrect. "q" for a virtual image is negative.

Welcome to PF, by the way.
 
whoa thanks

i apologize, i knew that...just dumb tonight

thanks for the welcome
 

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