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Expression of an object and its image onto a screen with two lens?

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Formulate an expression that gives the focal length of a corrective lens, to be used in conjunction with a fixed lens. You may assume that both lenses are on the same plane, i.e. that they are very close together. The corrective lens should allow an object at a specified distance from the lenses to produce an image at a specified distance from the lenses. You may assume that the corrective lens is “weaker” than the other lens, i.e. |f0| < |fc|.

    2. Relevant equations

    Assuming they are both thin lenses:

    Thin lens equation:

    1/do+1/di = 1/f

    (1/object distance + 1/image distance = 1/focal length)


    3. The attempt at a solution

    So I have to derive this equation myself and I'm really lost on where to start. I tried to draw out a good picture of the situation but am still confused. Do I want to start with the thin lens equation as my starting out point?
     

    Attached Files:

    Last edited: Oct 2, 2013
  2. jcsd
  3. Oct 3, 2013 #2
    I think you should assume that the objective is a converging lens and the corrective lens is a diverging lens. The diverging lens can then form a virtual image of the object at a slightly offset position for the objective thereby shifting the final image slightly to the "correct" position. The two lenses are in close proximity. So the problem suggests that you assume that they are in the same location.
     
    Last edited: Oct 3, 2013
  4. Oct 17, 2013 #3
    Hi, thanks so much for your response! I got the derivation wrong before so hoping to try again with figuring this out.

    So I'll use fo for the objective lens (convex) and fc for the corrective lens (concave).

    I eventually need to redrew my ray diagram, but if I have an object and the objective lens a distance (do) from it with the image a distance of (di) from the objective lens, I believe I would get:

    1/do + 1/di = 1/fo

    Using this, I can find where image one (di) will be:

    1/do - 1/fo = 1/di

    For the second image, I can have the distance between the objective lens and the corrective lens be a distance (x) from each other and that the second image forms at a distance (di2) from the corrective lens:

    1/di2 = 1/fc - 1/x

    So using the equations:

    1/do - 1/fo = 1/di

    1/di2 = 1/fc -1/x

    I can predict the image distances (di1 and di2) of the object. Does this sound correct?
     
  5. Oct 17, 2013 #4
    I really wish I didn't throw my derivation away... anyway.

    Your second equation is not correct.
    At the point where you start talking about x things really get unclear.
    The lenses are virtually on top of each other so you should assume that they are at the same position.
    You can then continue by using the image distance of the objective lens as the object distance for the corrective lens. This object distance should be made negative - here I am assuming that the original object is outside of the focal point of the objective lens - though since it is on the "wrong" side of the corrective lens.
     
    Last edited: Oct 17, 2013
  6. Oct 17, 2013 #5
    I wasn't sure what you meant by your first statement but I apologize if you felt your time was wasted...:/

    Thank you for looking at it again. I redrew my diagram to try to show both lenses in closer proximity to one another. I'm also making so that fo and fc are equal to one another.

    So if fo = fc, then would my equation just be:

    1/do + 1/di = 1/fo

    From there then, I get for image distance:

    1/do - 1/fo = 1/di

    New diagram is attached.

    Thank you again.
     

    Attached Files:

  7. Oct 17, 2013 #6
    No. It is just I did the formula derivation on paper but threw it away when you didn't come back, so now I have to do it again. I also did it the other way round, that is started with the corrective lens but, no problem, lets try it the other way round.

    fc is not equal to fo, they are different lenses. The original problem statement also states that the corrective lens have a longer (negative) focal length than the objective (positive) lens.

    Continuing with your derivation - this image distance now becomes the object distance for the corrective lens. You should just make it negative since the image is on the right-hand side of the corrective lens, if we assume that the original object is outside of the focal length of the objective lens. Keep in mind that the lenses are on top of each other, so there are no extra distance to take into account.
     
  8. Oct 17, 2013 #7
    Oh I see. Sorry about not responding back earlier on, I think I thought I had figured it out but did it wrong after getting feedback from it earlier this week. I'm here to stay now though until I have it solved! Thank you again.

    So I think I'm still lost on this. I do see now that fo needs to be less then fc, so fo < fc. I'm wondering if I will apply the thin lens equation for both lens and then put them together but making it so that fo is less then fc?

    1/do + 1/di = fo-fc ?

    So then, solving for di:

    1/do + 1/fc - 1/fo = 1/di?
     
  9. Oct 18, 2013 #8
    You were on the right track before. Use the thin lens formula to obtain the image distance of the objective lens. The negative of this becomes the object distance for the corrective lens.
     
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