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Question on Optics

  1. Dec 31, 2008 #1
    Hi. "A biconvеx lеns has a focal length of 100mm. an object is placеd at distancеs of 150 mm and 75 mm from teh lens. Sketch a ray diagram to find the position of thе respective images. confirm these positiones via calculatinons.

    This is the diagram I have drawn:
    http://www.freewebs.com/questionsuk/diagram.JPG

    Does this seem ok?

    However, what equations would be used to confirm the position?

    Any help is appreciated.
     
  2. jcsd
  3. Dec 31, 2008 #2

    dlgoff

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    Give this wikipedia page a read and see if it helps. In paticular the part on the Lensmaker's equation.
     
  4. Dec 31, 2008 #3
    Thanks for your help, although how does that equation consider the position of the object from the lens?
     
  5. Dec 31, 2008 #4

    dlgoff

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  6. Jan 1, 2009 #5
    Test
     
    Last edited: Jan 2, 2009
  7. Jan 1, 2009 #6
    Test
     
    Last edited: Jan 2, 2009
  8. Jan 1, 2009 #7

    Doc Al

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    Not really. To learn how to draw a ray diagram to calculate the image distance, read this: Ray Diagrams for Lenses.

    Think of the thin lens equation as 1/f = 1/do+ 1/di. Here do is the object distance, the distance between the object and the lens, and di is the image distance, the distance between the image and the lens. Use the equation to solve for di.

    Read this: Thin Lens Equation

    You are supposed to figure out the location of the image for two cases: one where do = 75 mm and one where do = 150 mm. (The big difference is that in one case the object is inside the focal length, while in the other it is outside that length.)
     
  9. Jan 1, 2009 #8
    Thanks for the reply.

    Just a question about the diagram, does this seem more correct?

    http://www.freewebs.com/questionsuk/diagram.JPG (you might need to refresh)

    Test
     
    Last edited: Jan 2, 2009
  10. Jan 1, 2009 #9

    Doc Al

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    Yes, much better. But to get a numerical result you must draw it carefully and to scale.

    What does the lens equation tell you?
     
  11. Jan 1, 2009 #10
    Thanks for the reply. For the 150 mm object I worked the image out to be 250 mm from the lens.

    For the 75 mm however I've done

    1 / 100 = 1 / 75 + 1 / v

    = 0.003

    = 333 mm (I think it's a negative as it is behind the focal length)
     
  12. Jan 1, 2009 #11

    Doc Al

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    Redo that calculation.

    Redo that more carefully. Your equation is correct, just solve for the unknown.
     
  13. Jan 1, 2009 #12

    How comes? I did 1 / f = 1 / u + 1 / v

    1 / 100 = 1 / 150 + 1 / v2

    0.01 = 0.006 + 1 / v2

    0.01 - 0.006 = 0.004 which is equivalent to 1 / v 2 so

    v2 = 0.004

    = 250 mm ?

    v2 by the way is not v x 2

    Thanks.
     
  14. Jan 1, 2009 #13

    Doc Al

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    Don't round off. That should be 1/150 = 0.006666...

    Even better, do this:

    1/100 - 1/150 = 3/300 - 2/300 = 1/300 = 1/v2.
     
  15. Jan 1, 2009 #14
    Thanks! For both I got 300 mm although the object at 75 mm is a virtual image.
     
  16. Jan 1, 2009 #15

    Doc Al

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    Good!
     
  17. Jan 1, 2009 #16
    Also, it has given the index as 1.6 and has asked to calculate the radii of the curvature of the lens.

    I've used this equation:

    Radii of curvature R = 1 / u + 1 / v = 1 / f (n1 – 1) (1 / R1 – 1 / R2)

    Now, I assume it's asking for one radius so I've not included the 1 / R2 in that equation.

    To make it simpler the equation is: 1 / f (n1 – 1) (1 / R1 – 1 / R2)

    I did: 0.01 = 0.6 [1 / R1] although there may also need to be a -1 / R2 there as well.

    I then made R1 to be 6mm.


    Any replies are appreciated!
     
  18. Jan 1, 2009 #17

    Doc Al

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    A more reasonable assumption would be to assume that the biconvex lens is symmetrical--both sides have the same radius. (And be more careful with your arithmetic.)
     
  19. Jan 1, 2009 #18
    Ah yes thanks very much!

    One last thing, it says "calculate the longitudinal and transverse magnification of the image if the object is 10 mm high and 5 mm long".

    For the longitudinal it has given fo x fi / xo^2


    fo is the focal length of the object
    fi is the focal length of the image
    xo is the distance from the object to the focal length

    For the 150 mm object I got -4 and the 75 mm object 16

    However, where does the objects dimensions (10 mm high and 5 mm long) come into this? Perhaps it just comes into the transverse equation which is

    fo / xo = xi / fi = -yi / yo = xo xi = fo fi

    ??
     
  20. Jan 1, 2009 #19
    Actually, the yo and yi represent the lengths of the object and its respective image. Not sure about the 5mm long though.
     
  21. Jan 1, 2009 #20

    Doc Al

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    Only the longitudinal magnification will depend on the object size.
     
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