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physics

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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 ___

  • physics - ,

    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)

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