Is the Image from a 12cm Focal Length Lens Virtual or Real?

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SUMMARY

The discussion centers on the optical properties of a positive lens with a focal length of 12 cm and an object placed 3 cm from the lens. Using the lens formula, 1/o + 1/i = 1/f, the calculated image distance (i) is -4 cm, indicating that the image is virtual and located on the same side as the object. The conclusion confirms that since the object is within the focal length, the resulting image is erect and virtual. The importance of drawing a ray diagram before calculations is emphasized for clarity.

PREREQUISITES
  • Understanding of the lens formula: 1/o + 1/i = 1/f
  • Knowledge of focal length and object distance in optics
  • Familiarity with ray diagrams for lenses
  • Basic concepts of real and virtual images
NEXT STEPS
  • Study the principles of ray diagrams for different lens types
  • Learn about the characteristics of real vs. virtual images
  • Explore the effects of varying object distances on image formation
  • Investigate the applications of lenses in optical devices
USEFUL FOR

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

mike2007
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A positive lens has a focal length of 12cm. An object is located a distance of 3 cm from the lens.
How far is the lens from the image?
Is the image real or virtual, erect or inverted?

My attempt
f = +12cm
o = +3cm
i =?
1/o + 1/I = 1/f
1/i = 1/f – 1/o
= 1/12cm – 1/3cm
= 1/12 – 4/12
= - (3/12)
i = -4cm on the same side of the object.

Since the object lies in within the focal point, the image is virtual and erect.
 
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Always draw the ray diagram before trying to do the maths.
 
Looks good to me.
 

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