physics
posted by andrew .
An object is placed 75.0 cm from a screen.
(a)Where should a converging lens of focal length 8.0 cm be placed to form an image on the screen?
shorter distance __cm from the screen
farther distance __cm from the screen
(b)Find the magnification of the lens.
magnification if placed at the shorter distance ___
magnification if placed at the farther distance ___

Let do = object distance
di = image distance
di + do = 75
1/di + 1/do = 1/f = 1/8
1/(75  do) + 1/do = 1/8
Solve for do.
do + (75  do) = (1/8)(do)(75do)
75 = (1/8)do*(75do)
600 = do*(75do)
do^2 75do + 600 = 0
do = (1/2)[75 +/sqrt(56252400)]
= 37.5 +/28.4
= 65.9 or 9.1 cm
di = 9.1 cm from lens when do = 65.9
di = 65.9 cm from lens when do = 9.1
(magnification = 7.24)