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Lenses, diminished and magnified

  1. Apr 2, 2005 #1
    This is probably a stupidly easy question, but I'm confused so I need to ask about it. :smile:

    I have this experiment I was doing where I have a light source at one end (which is enclosed in a casing and only lets light get out in the shape of an arrow because of an arrow shaped hole) and a screen at the other. In between is a lens and I move it to get the image of the arrow on the screen in focus. I notice I can get two points of focus, one where the arrow is diminished (smaller) and where it is magnified (larger).

    Also, when I move the screen to 40cm from the light source I only seem to be able to get one in focus image, which I cannot decide is diminished or magnified.

    I'm very confused as to why this happens, any help would be great.

    Thanks! :yuck:
  2. jcsd
  3. Apr 2, 2005 #2

    Andrew Mason

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    Are you sure that you are keeping the lens and object fixed while you move only the screen?

    For a given object distance and focal length, there is only one point at which the image will be in focus. That is determined by the lens equation:

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

    Edit and correction:For a fixed screen to object distance, S, there will be two lens positions where the object is in focus provided S>4f. They will be:

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

    If [itex]S = 4f[/itex] there will be only one solution. So I gather that your screen was 80 cm from the object and the lens has a focal length of 20 cm.

    Last edited: Apr 3, 2005
  4. Apr 3, 2005 #3
    sorry, probabaly didnt explain very well...

    object and screen kept the same then moving the lens. After I have found the two points of focus, move the screen then move the lens again.
  5. Apr 3, 2005 #4


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    Start with

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

    Suppose you've got the lens in one position that gives you an image on the screen. The lens is a distance [itex]o[/itex] from the object, and a distance [itex]i[/itex] from the image. The other position is the one that has the numbers for [itex]i[/itex] and [itex]o[/itex] switched around.

    Exception: if [itex]i[/itex] and [itex]o[/itex] are equal, you don't change anything when you switch them around.
  6. Apr 3, 2005 #5

    Andrew Mason

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    I shouldn't have doubted your experimental results! It seems I was using the midpoint between screen and object as the reference for measuring distances rather than the object location. I have corrected my first answer.

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