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Geometrical optics

  1. Aug 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A needle of length 5.0cm placed at the bottom of a beaker and a converging lens of focal length 20.0cm is held 30.0cm from the needle.The needle is illuminated with a lamp from the bottom.An image of the needle is formed on a screen.The beaker is now filled with water(refractive index=4/3)to a depth of 20cm.What change can be observed in the image?
    If the screen is moved to get a new sharp image,what will the image distance and object be in this new position??

    i squeezed my brain so hard..still don hv any idea at all..becoz theres a space (air)between the lens and the water.When i am using the lens maker's formula,im confusing whether i should assume there r three lens -water,converging lens and water or two or one..
    can somebody give some hints about how to solve it??
     
  2. jcsd
  3. Aug 9, 2008 #2

    Doc Al

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    Staff: Mentor

    Forget about the lens for a moment. From the viewpoint of someone looking down into the beaker (from the position of the lens), will the addition of the water make the apparent position of the needle closer or farther away? By how much?
     
  4. Aug 9, 2008 #3
    oh!!
    the apparent depth is 22.5cm from the lens
    and let u=22.5cm and solve it by lens formula
    is it correct??
     
  5. Aug 9, 2008 #4
    sorry...
    mistake.....real depth shouldnt include the air......u=25cm
    by the way!!thnx!!!!
     
  6. Aug 9, 2008 #5

    Doc Al

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    Staff: Mentor

    How did you get that value? Realize that the water is only 20 cm deep.
     
  7. Aug 9, 2008 #6

    Doc Al

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    Staff: Mentor

    Now you've got it. :smile:
     
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