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Converging lens

  1. Apr 16, 2005 #1
    A bright object is placed on one sid eof a converging lens of focal length f, and a white screen for viewing the image is on the opposite side. The distance [tex]d_T = d_i + d_0 [/tex] between the object and the screen is kept fixed, but the lens can be moved.

    Show that if [tex]d_t > 4f[/tex] , there will be two positions where the lens can be placed and a sharp image can be produced on the screen.
    And if [tex]d_t < 4f [/tex], no lens position where a shakrp image is formed.
    Also determine a formula for the distance b/w the two lens position,and the ratio of the image sizes.

    the attachement is the work that i was trying figure out how to do it. but i dont seem to know how to tackle this problem
     

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  2. jcsd
  3. Apr 16, 2005 #2

    GCT

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    You'll need to do some rearranging of equations and plugging in, try finding dt as a function of f.
     
  4. Apr 17, 2005 #3

    Andrew Mason

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    From the lens equation:

    [tex]\frac{1}{f} = \frac{1}{i} + \frac{1}{o} [/tex]

    since S = i + o (S = object to screen distance):

    [tex]\frac{1}{f} = \frac{1}{i} + \frac{1}{S-i} [/tex]

    This gives you a quadratic equation in terms of i. Solve that using the quadratic formula and you should get two solutions for i:

    [tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex]

    AM
     
    Last edited: Apr 17, 2005
  5. Apr 17, 2005 #4
    after [tex]\frac{1}{f} = \frac{1}{i} + \frac{1}{S-i} [/tex] , I have [tex] f = \frac{si - i^2}{s} [/tex] , and then i set the equation equals to zero: [tex] i^2 + fs - si = 0 [/tex], but i dont get how you can solve for i from [tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex].
     
  6. Apr 17, 2005 #5

    Andrew Mason

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    Well, you can't solve for i unless you know S. But that is not what the question asks.

    What is the condition for i to be real? What is the condition for i to have 2 real values?

    AM
     
  7. Apr 17, 2005 #6
    for i to be real and have two solutions, [tex]\sqrt{S^2 - 4Sf}[/tex] must be greater than 0.
    So, [tex] s^2 - 4sf > 0 [/tex]

    [tex] s^2 > 4sf [/tex]
    [tex] s > 4f [/tex] and we set [tex] s = i + o [/tex] before, so we get the answer for the first two questiosn.

    But I don't know how you can get [tex]i = \frac{S \pm \sqrt{S^2 - 4Sf}}{2}[/tex], from [tex] i^2 + fs - si = 0 [/tex]
     
  8. Apr 17, 2005 #7

    Andrew Mason

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    That is just the quadratic formula. The general quadratic equation:

    [tex]ax^2 + bx + c = 0[/tex]

    has solutions:

    [tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

    In this case, a = 1, b = -S, c = sf

    AM
     
  9. Apr 18, 2005 #8
    AHAHHAHHAHAHA. I feel myself really stupid now....can't even find the coeiffient term of a quadratic equation!
     
  10. Apr 19, 2005 #9

    Andrew Mason

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    That happens during exam time! Ease up .. it was a bit of a tricky question.

    AM
     
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