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Graphically interpreting data regarding magnification with a convex lens

  1. Dec 16, 2007 #1
    1. The problem statement, all variables and given/known data

    How do I find the relationship of magnification to the graph of "Object Distance vs. Image Distance" or the graph of "Inverse Object Distance vs. Inverse Image Distance" for data points collected from a light box and a convex lens?

    Basically for the lab I used a light box to project an image through a convex lens onto a piece of paper. Then, I changed the distance of the light box, which was the "object" in this case and adjusted the distance of the paper to focus the "image".

    Here is the lab data that I collected:

    D obj D img image size 1/D obj 1/D img |magnification|
    27 101 19 0.03704 0.00990 3.75
    40 45 5 0.025 0.02222 1.25
    35 53 6.5 0.02857 0.01887 1.63
    45 39 5.5 0.02222 0.02564 0.875
    50 35 3 0.02 0.02857 0.7
    55 32 2.5 0.01818 0.03125 0.582
    60 30 2 0.01667 0.03333 0.5

    The variables were object distance (x) and image distance (y). I had to plot the graph of
    "Object Distance vs. Image Distance" and then the graph of "Inverse Object Distance vs. Inverse Image Distance". Using regression techniques, I found the equation of the "inverse" graph to be y = -1.125x + .052 and the equation of the original graph to be y = (-1.125/x + 0.052)^(-1). I am very certain that both of these equations accurately reflect the data that I collected.

    Another possibly relevant calculation was the focal distance which I found with the thin lens equation, which was about 20 - 21 cm.

    My problem came when I tried to relate the magnification to the graphs of the data, which is required for the lab analysis (my attempted method is below at part 3).

    2. Relevant equations

    Thin Lens Equation (probably irrelevant)
    Magnification = - ((image distance)/(object distance))
    slope = change in y / change in x

    3. The attempt at a solution

    I looked at the graph of the rational equation which I first plotted (y = (-1.125/x + 0.052)^(-1)) and figured that since my y axis was image distance, and my x axis was object distance (both in cm). I figured that slope is rise/run which would be image distance/object distance. I also noted that Magnification = - ((image distance)/(object distance)), so the negative of the slope must be the magnification! Because the equation was a rational equation (and therefore curved), I took the derivative of the equation, so I could have a function which would yield the slope at any given point of the curve, and took the negative of the derivative to find the magnification.

    I figured: y' = -M(x) (where M(x) would yield the magnification at a given point)
    therefore: M(x) = -y'

    and found through the Chain Rule: M(x) = -y' = 1.125/((x^2)(-1.125/x + 0.052)^2)

    I checked the derivative at a couple of points on my TI 84 calculator and it seems to be correct. However, the equation does not yield the magnification. For example:

    when Magnification = 3.75, image distance = 96cm (according to my equation), and object distance = 27 cm

    but when calculated with M(x), the magnification should be about 14. This is where I got stuck.

    I would appreciate any help somebody could give me. (this is my first use of this site, so please tell me if I posted this incorrectly as well!) Thank you very much!

    -- Dave
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 16, 2007 #2

    rl.bhat

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    Homework Helper

    I took the derivative of the equation, so I could have a function which would yield the slope at any given point of the curve, and took the negative of the derivative to find the magnification.
    This does not give the magnification. But it indicates the square of the magnification.
    Consider the lens formula.1/f = 1/u + 1/v ( without any sign convention). Take the derivative with respect to u. And see what expression you get.
     
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