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

[Darth Frodo]

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 = \frac{Image}{Object}

M = \frac{V}{U}

M = \frac{1}{4}

V = \frac{U}{4}


\frac{1}{f} = \frac{1}{U} + \frac{1}{V}

\frac{1}{50} = \frac{1}{U} + \frac{4}{U} (From above)

\frac{1}{50} = \frac{5}{U}

U = 250 cm

U = 0.25 cm


Is this right?

 

[technician]

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