How Does Lens Focal Length Affect Image Size on a Wall?

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SUMMARY

The discussion focuses on the relationship between lens focal length and image size on a wall, specifically using a lens with a focal length of 50 cm. The calculations demonstrate that a magnification of 1/4 results in an object distance (U) of 250 cm, confirming the lens formula application. The conversation also highlights the distinction between linear magnification and area magnification, emphasizing that linear magnification affects lengths, while area magnification scales the area by the square of the linear magnification factor.

PREREQUISITES
  • Understanding of lens formulas, specifically the thin lens equation.
  • Knowledge of magnification concepts, including linear and area magnification.
  • Familiarity with basic optics principles.
  • Ability to perform algebraic manipulations involving fractions and ratios.
NEXT STEPS
  • Research the thin lens equation in detail, focusing on its applications in optics.
  • Explore the concept of area magnification and its mathematical implications.
  • Learn about different types of lenses and their focal lengths.
  • Investigate practical applications of lens magnification in photography and imaging systems.
USEFUL FOR

Optics students, photographers, and anyone interested in understanding the principles of lens magnification and image projection.

Darth Frodo
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Problem 1

A flat screen TV is place on a wall in a room. A lens of focal length 50cm is placed between the television and the opposite wall so that a sharp image with one quarter of that of the area of the television is produced on the opposite wall.

Answer:

Magnification = [itex]\frac{Image}{Object}[/itex]

M = [itex]\frac{V}{U}[/itex]

M = [itex]\frac{1}{4}[/itex]

V = [itex]\frac{U}{4}[/itex]


[itex]\frac{1}{f}[/itex] = [itex]\frac{1}{U}[/itex] + [itex]\frac{1}{V}[/itex]

[itex]\frac{1}{50}[/itex] = [itex]\frac{1}{U}[/itex] + [itex]\frac{4}{U}[/itex] (From above)

[itex]\frac{1}{50}[/itex] = [itex]\frac{5}{U}[/itex]

U = 250 cm

U = 0.25 cm


Is this right?
 
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The magnification given by m=v/u is a LINEAR magnification... this is how much lengths are magnified. If m = 2 it means lengths are 2x greater... what will that mean for AREA magnification?
Hope this helps
 

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