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Distance of the final image in a compound microscope

  1. Oct 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A compound microscope has an objective of focal length 12 mm and an eyepiece of focal length 50 mm. The lenses have a separation of 90 mm and an object of height 0.3 mm is placed 15 mm from the objective. Calculate: (a) the distance of the intermediate image from the objective; (b) the distance of the final image from (i) the eyepiece, (ii) the objective; (c) the size of the intermediate image; (d) the angular magnification.

    Answers: (a) 60 mm; (b) (i) 75 mm, (ii) 15 mm; (c) 1.2 mm; (d) 33.3

    2. The attempt at a solution
    (a) (1 / u) + (1 / v) = (1 / f)
    u = 15 mm, f = 12 mm. Plug in and v = 60 mm

    (c) (h / ho) = (v / u)
    ho = 0.3 mm, v = 60 mm, u = 15 mm. Plug in and h = 1.2 mm

    (d) M = β / α
    α = ho / D = 0.3 / 250 = 1.2 * 10-3
    β = h / (90 mm - 30 mm) = 0.04
    M = 33.3

    But (b)? I used this graph from the book for my question.
    145e5c540df9.jpg

    And according to this graph the distance of the final image (final virtual image) from the eyepiece is D and is 250 mm. And the distance from FVI to the objective should be more than 15 mm. Maybe I don't quite understand what shall I find. Really stuck on this one. Any help please?
     
  2. jcsd
  3. Oct 21, 2015 #2

    ehild

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    You need to find the position of the final image, made by the eyepiece lens. The object is the first image. What is the object distance?
     
  4. Oct 22, 2015 #3
    By object you mean "intermediate real image"?

    Is it something like this:
    From (a) we have v = 60 mm for the obj. lens. So u = 30 for the eyepiece lens.

    1 / 30 + 1 / v = 1 / 50
    v = - 75 mm. So the distance from the eyepiece is 75 mm and the image is virtual.

    Because S = 90 mm, 90 - 75 = 15 mm, distance from the objective.

    Is this right? And why is it -75 but in the answer it's just 75? And is the 90-75 way a correct way to find 15 mm? And how come it is not negative?

    Sorry for so many questions, a bit confused on this part.
     
  5. Oct 22, 2015 #4

    BvU

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    You're doing fine. The -75 is correct, and it's a position on a coordinate axis with the objective at the origin.
    The book answer is |-75| mm = 75 mm because they ask for a distance.
    And yes, the virtual image is at position (+90) + (-75) wrt the objective (*), so at | +15| = 15 mm distance from the objective.

    (*) relative position wrt objective = relative position of ocular wrt objective + relative position of virtual image wrt ocular

    --
     
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