An object is placed 75.0 cm from a screen.

Question

(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 ___

Answer 1

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)(75-do)
75 = (1/8)do*(75-do)
600 = do*(75-do)
do^2 -75do + 600 = 0
do = (1/2)[75 +/-sqrt(5625-2400)]
= 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)