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