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Lens question

  1. Feb 21, 2005 #1
    One of our lab assignments was to use a lens to focus an arrow-shaped beam of light onto a sheet of paper. We were then told to predict what would happen if we covered up the top/bottom/left/right half of the lens. I predicted that covering the top portion would cause the bottom part of the image to disappear, covering the bottom portion would cause the top part of the image to disappear, etc. As it turned out however, covering up any half of the lens only caused the image to appear about half as bright, but it was still completely whole. Can somebody help me understand what exactly is going on here? I have been attempting to draw ray diagrams to visualize it, but to no avail. Thanks.
     
  2. jcsd
  3. Feb 21, 2005 #2
    Rays that hit any part of the lense contribute to the image. For example, when you cover the top, You still have rays hitting the bottom part forming the image. Low intensity due to the less number of rays. Drawing a ray diagram will make you understand this. Are you able to draw ray diagrams where the rays hitting the lense are in arbitrory direction?
     
  4. Feb 21, 2005 #3
    Thanks for the response. My difficulty in understand this is obviously due to my inabilty to draw ray diagrams.

    The only ray diagrams I am familiar with drawing are those with the three principle rays. I suppose I should learn to draw ones that are more complex, but at this point I cannot.
     
  5. Feb 21, 2005 #4

    GCT

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    Well, a point on the object is associated with light spreading out in all directions and they will all converge at a point to form a real image point. You should understand now.
     
  6. Feb 21, 2005 #5
    I have uploaded a ray diagram to the following site.

    http://img.photobucket.com/albums/v645/city0818/Raydiagram.jpg

    you may have to copy and paste the url.


    This diagram shows few rays that hit the bottom half of the lense.

    1. First ray is familiar. Goes undeflacted.

    2. Second one comes through the focul point. Deflacted ray goes parallel to the main axis.

    Above two is enough to determine the image. For your information I am writing how to find the delacted beam for an arbitrory beam.

    3. For a ray that is in an arbitrory direction, (3rd one) draw a line parallel to the insident beam through the lense (red line). The deflacted ray goes through the intersecting point of this parallel line with a line through the focul point (see the diagram).

    Hope this helps
     
  7. Feb 21, 2005 #6
    Excellent! Thank you for the info, that does definitely help...a lot!
     
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