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System of lenses with a light source at first focal point

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  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data

    So I have this system of lenses with this problem:

    First lens: focal length ##f_1=50 mm##
    Second lens : focal length ##f_2=400 mm##
    Distance between them ##d=600mm##

    A small light source is placed at the focal point for the first lens. At the focal point for the second lens we have an image of the light source, what is the magnification?

    2. Relevant equations

    Lens formula: ##\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}##

    3. The attempt at a solution

    Since we have the objekt at the first focal point the image of that lens is the infinity. This implies that the objekt for the second lens is the infinity and that would give us the image at the second focal point to the right of the second lens. Is this correct? How would I continue to find the magnification of this?
     
  2. jcsd
  3. Nov 9, 2015 #2

    BvU

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    How about a drawing ?
     
  4. Nov 9, 2015 #3
    I have done a ray diagram, but it is on paper. Is there anyway you could lead me on anyway?
     
  5. Nov 9, 2015 #4

    BvU

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    Picture "object at f" here might help...
    Rays from "foot" of arrow are easy to draw for this 2 lens system. Now the rays for the tip of the arrow ...
     
  6. Nov 9, 2015 #5
    But how do I know the height of the arrow? Also, if the height is to big I can't draw it at all, right?
     
  7. Nov 9, 2015 #6

    BvU

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    You don't need to know the height of the arrow. All you want is the magnification factor. So take something useful and make the drawing !

    Two of these rules are very useful for finding rays from tip of arrow to second lens...
     
  8. Nov 9, 2015 #7
    These I all know but its hard to draw something when the first focal length is 50 mm then followed by a gap of 600 mm. I want to calculate it before I draw it to see where I end up. I already have a "feeling" for where I want to end up but I want to calculate it, that is the problem.
     
  9. Nov 9, 2015 #8

    BvU

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    If you make a sketchy drawing, you'll see what it is you want to calculate :smile: !
     
  10. Nov 9, 2015 #9
    From my sketchy drawing it feels like I should end up with the magnification ##M \approx 6## but that doesn't help me, haha.
     
  11. Nov 9, 2015 #10

    BvU

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    See any nice triangles you can use effectively ?

    [edit] I did make the sketch. It makes the answer so obvious that I don't even want to post that picture: PF rules don't allow helpers to do your homework for you. Only help.

    [more edit] from post #4: Rays from "foot" of arrow are easy to draw for this 2 lens system.
    What do you see ?
     
    Last edited: Nov 9, 2015
  12. Nov 10, 2015 #11
    Ok! It's not a homework, just an excerise, but i understand. At least now I am on my way.

    Is it enough to draw rays form the foot and the top of the arrow? I should look for some angles, right?
     
  13. Nov 10, 2015 #12

    BvU

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    Remember: the object is at the focal point of the first lens. That says something about all the rays that come from any point on the object.
    Can you post at least some sketch ?
     
  14. Nov 10, 2015 #13
    I'm at the university at the moment, I might be able to take a photograph with my cell phone.

    I abit ashamed that i cant manage this haha

    Edit: this is the best i can do at this moment

    http://sv.tinypic.com/r/2a6j9ma/9
     
  15. Nov 10, 2015 #14
    I think that i solved it now.. I got the answer 8 and that seems to be right. I Will post My solution when I get home, maybe you can check it and it might help others in the future. :)
     
  16. Nov 10, 2015 #15

    BvU

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    There is no need for shame. My PhD didn't prevent me from looking at this exercise in wonder either. Just hang in there !

    Your picture is just fine, I am completely reassured.
    (And trust that the picture from the rays emerging from the foot of the arrow is correct too. It helped you to find the image location on the axis -- and perhaps you noticed that the distance between seceond lens and image plane does not depend on distance between lenses :wink: )​

    So you see that the rays from the top of the arrow after the first lens have a property: they are parallel. You already know where they focus, so in fact one ray is already enough to establish the image height. There were three rules to make these drawings. One of them isn't usable :smile:

    I eagerly hope for an aha moment now :rolleyes: !

    --
     
  17. Nov 10, 2015 #16

    BvU

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    It's not the destination, it's the journey that makes life so valuable ...
     
  18. Nov 10, 2015 #17
    Haha yes, but its not good for my self conscious to struggle with this when i have no problem at all with much heavier physics and math.. Anyway, thanks for your patience :)
     
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