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Image formed by a plane-concave lens and a concave mirror

  1. Sep 9, 2015 #1
    1. The problem statement, all variables and given/known data
    An observer looks at an object of height 2 cm through a plane-convex lens of radius 20 cm and refraction index 1.5. There's a concave mirror of radius 30 cm 40 cm behind it, and the object is at the same distance of both the lens and the mirror. Find the images formed by the optical system and draw the diagram.
    Solutions:
    Image 1: s'=-13,3 cm; y'=-1.33 cm
    Image 2: s'=40 cm; y'=-12 cm

    2. Relevant equations

    [itex]1/s + 1/s=(n_2/n_1 -1)(1/R_1-1/R_2)[/itex]
    [itex]1/s + 1/s'=1/f[/itex]



    3. The attempt at a solution

    For the first image: [itex]\frac{1}{-20} + \frac{1}{-s'}=\frac{1}{2}(\frac{1}{\infty}-\frac{1}{-20}) \implies s'=-13,33 cm[/itex]
    This is correct (one question: why do I have to use "-s'"? Shouldn't the result give the correct sign using s'?).
    But I can't get the diagram to agree with this result. Could anyone make one?

    And then, for the second image, using the first image as an object:
    s=26,67, f=R/2=15 cm. s'=34 cm
    Using the first object:
    s=20 cm, f=15 cm. s'=60 cm.
    In this case, both possible values are wrong. I think the problem derives from my incapacity to picture the diagram. Could it be that I have to account for the divergence of light caused by the first lens to calculate the position of the image created by the mirror? And how do I do that?
     
  2. jcsd
  3. Sep 11, 2015 #2

    Simon Bridge

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    Try sketching the diagram. There are three images.
    I can never remember the sign conventions so I rely on sketching to get the placement right. If the diagram disagrees with the calculation, then the calculation is usualky in error.
     
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