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Simple Magnification Problem

  1. May 1, 2009 #1
    GREEN= Correct
    1. The problem statement, all variables and given/known data
    The lens in the figure is used to produce a real image of a candle flame at the 36 cm mark. What is the focal length of the lens?
    focal length= 9


    (My attachments aren't working, however, if you were to draw a line, put a lens in the middle, then a small candle on the left and a magnified, inverted candle on the left, that's the drawing. The length from the object from the small candle to the lens is 12cm at first, and the length from the lens to the larger image of the candle is 36 at first, the changed lengths are mentioned below. I'm so sorry about this).

    Change the location of the object to 11 cm. Determine the location and magnification of the image.

    location= 49.5
    magnification=____
    2. Relevant equations

    h'/h= s'/s= magnification

    In other words: the ratio of height of object to height of image equals the ratio of length to lens for the object to the length to lens of the image, which equals magnification.

    3. The attempt at a solution

    I don't understand why the magnification problem isn't panning out. I'm using the equation above, stating that the image distance over the object distance equals the magnification. So, with the new numbers, it would look like this:

    49.5/11= 4.5

    However, this isn't the answer. Can anyone help me out with this one?

    Thanks in advance!

    ~Phoenix
     

    Attached Files:

    Last edited: May 2, 2009
  2. jcsd
  3. May 1, 2009 #2
    What kind of lens is this? And a figure with ray tracing would be best so we can see your work, since I am a little confused by your explanation.
     
  4. May 2, 2009 #3
    I managed to find a diagram that was basically the same idea (ignoring the already drawn ray lines). Essentially, the left side length is 11cm and the right of the lens is 49.5 cm as I found. All my explanation says is I plugged my numbers into the equation for magnification and it didn't work (neither did the negative of that number). I'm using a universally acknowledged equation.

    I went up and edited my original post to show that new diagram and (just in case) what the variables mean in the equation.

    Can anyone else help me out? Thanks again.
    ~Phoenix
     
  5. May 2, 2009 #4
    Well the image is inverted so the magnification is negative. Be sure you didn't type something wrong, and also you should type the problem out exactly the way it is worded.
     
  6. May 5, 2009 #5
    I was merely using the wrong distance. Thanks though. :)

    Oh, and I just copy and pasted the entire question. That is how it's worded. The image just wouldn't appear.

    ~Phoenix
     
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