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Elementary lens system, microscope with negative ocular

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  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Surgeons often use a type of "binocular microscope" [that's a word by word translation]. It consists of a 50mm lens close to the patient and a -15mm right in front of the doctor's eye.

    The distance between the patient and the 50 mm lens is 400mm and the negative lens is placed so that the image and object have the same distance from the doctor's eye (for correct depth perception).

    What should the distance between the lenses be?


    2. Relevant equations
    [tex]\frac{1}{f} = \frac{1}{s} + \frac{1}{s'}[/tex]


    3. The attempt at a solution
    I know how to solve, quite easily in fact. The mathematics isn't mind boggling but where I'm having trouble is visualizing it. I assumed it would work like a regular microscope because of no reason but "I don't know how it would work with a negative rather than positive eyepiece".

    I sat trying to ray trace it to come to some conclusion of my own but didn't get anywhere. Can't wrap my head around how to go about it when there's a lens system with a negative lens.
    This is where I'd like some help, my thinking (probably faulty) was that something like this could occur. Is this way off the realm of actual physics?
    http://img163.imageshack.us/img163/3164/j51b.png [Broken]

    In any case, without just know the fact that the image from the lens closest to the patient should be between the two lenses and works as an object for the second lens how do I deduce it?

    Thank you

    Edit: Changed img tags to not mess with forum layout
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 21, 2013 #2
    Excellent diagram. From it you can see that the image from the first lens is actually behind the second lens, so it is actually a virtual object for the negative lens. It is not between the two lenses as for a normal microscope.
     
  4. Aug 21, 2013 #3
    Thanks but diagram is faulty for the exercise, unfortunately, good to know I wasn't completely in fairyland though. I think I figured out where my thinking was wrong, as the exam question said "ray trace to show" I assumed I should be able to essentially ray trace it without doing any maths on the side because it would be "obvious if I just knew the rules".

    Turns out that doesn't quite work as well as I thought it should.
     
  5. Aug 21, 2013 #4
    Did they want you to draw it to scale? Is that where you went wrong? Problem is where to place the 2nd lens. Your diagram succesfully indicates the relationships between the lenses, the object and images formed. It can be a good guideline to a more detailed drawing or second attempt. Sometimes they expect you to "read their minds" (joke).
     
  6. Aug 21, 2013 #5
    Don't think it necessarily had to be to scale, but it had to be in the right ballpark (ie image should be between the lenses to serve as an object for lens 2). What tripped me over was that the ocular was a negative lens. I knew how it would work with a positive ocular, but with a negative it could be radically different. Galilean telescopes for example. Suppose I should've just done it mathematically form the get go and then ray traced it based on certain information, would have saved some headache.

    Been approaching lens systems the wrong wa. Last time I completely ignore one course just because it's open book!
     
  7. Aug 22, 2013 #6
    The image needen't be between the lenses to serve as an object for the negative lens, as your diagram shows the image is actually behind the negative lens at the point where the rays of the 50 mm lens converges on the optical axis. All it means is that it then serves as a virtual object for the negative lens. The object distance is then taken to be a negative value since it is a virtual object.
     
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