Placement of an object and magnification

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SUMMARY

The discussion centers on the placement of a 0.05 m focal length lens to produce a sharp image on a screen located 1 m away. The lens maker equation, 1/s + 1/s' = 1/f, was applied to determine that the lens should be positioned 0.05 m from the screen. The corresponding magnification calculated using m = -s'/s resulted in -0.05, indicating a reduced inverted image. The solution confirms the correct application of optical principles to achieve the desired image formation.

PREREQUISITES
  • Understanding of the lens maker equation
  • Familiarity with magnification calculations
  • Basic knowledge of optics and image formation
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the impact of varying focal lengths on image formation
  • Explore the principles of convex and concave lenses
  • Learn about real vs. virtual images in optics
  • Investigate the application of the thin lens formula in complex optical systems
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Students studying optics, physics educators, and anyone interested in understanding lens behavior and image formation principles.

Violagirl
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Homework Statement



An object is 1 m from a screen. At what points may a 0.05-m focal length lens be placed so as to produce a sharp image on the screen? What are the corresponding magnifications?

Homework Equations



Lens maker equation:

1/s + 1/s' = 1/f

Magnification equation:

m = -s'/s = -f/s


The Attempt at a Solution



For the first question:

1/s' = 1/f -1/s = 1/0.05 m-1 = -.95 m.

1 m - .95 m = 0.05 m away from the screen.

Magnification:

m = -0.05 m / 1 m = -0.05

Is this right?
 
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Use the fact that the sum of the object and image distances is 100 cm.
 

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