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September 18, 2014

September 18, 2014

Posted by **andrew** on Wednesday, May 23, 2012 at 5:01am.

(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 ___

- physics -
**drwls**, Wednesday, May 23, 2012 at 6:53amLet 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)

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