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Physics question

  • 20-09-2012 05:03PM
    #1
    FudgeBrace
    Registered Users, Registered Users 2 Posts: 216 ✭✭


    Hi, could someone please show me how to do this question please.?
    Q : where must an object be placed so that an image one third the height of the object is formed in a convex mirror of focal length 20cm?


Comments

  • #2
    mulciber
    Registered Users, Registered Users 2 Posts: 200 ✭✭mulciber


    Step 1. Start with the formula -1/f= 1/u - 1/v (Convex mirror formula)

    Step 2. Magnification= v/u so therefore 1/3=v/u
    Therefore: v=1/3u

    Step 3. Sub in 1/3u for v in your equation: -1/f= 1/u -1/(1/3)u or -1/f= 1/u -3/u

    Step 4. Sub in 20 for f, solve for u and you should get an answer of 40cm.

    I hope this helps. =)


  • #3
    therambler
    Registered Users, Registered Users 2 Posts: 64 ✭✭therambler


    magnification=v/u

    m=1/3

    v/u=1/3

    u=3v

    REAL IMAGE

    1/u + 1/v = 1/f ==> 1/3v + 1/v =1/f

    4/3v=1/f

    4/3v=1/20

    3v=80

    3v=u

    80=u

    VIRTUAL IMAGE

    1/u-1/v=1/f

    1/3v -1/v = 1/f

    2/3v = 1/20

    3v=40
    u=40

    For the required real image the object must be placed 80cm from the lens and for the virtual image, the object must be 40cm from the lens.


  • #4
    FudgeBrace
    Registered Users, Registered Users 2 Posts: 216 ✭✭FudgeBrace


    thanks guys that was great help!!


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