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Angular Magnification and the Magnifying Glass

  1. Apr 6, 2006 #1
    I'm having problems with the following problem:

    A person who has a near point of 25.0 cm is looking with unaided eyes at an object that is located at the near point. The object has an angular size of 0.012 rad. Then, holding a magnifying glass (f= 10.0 cm) next to her eye, she views the image of this object , the image being located at the near point. What is the angular size of the image?

    The answer is suppose to be 0.042 rad. However I'm getting a different answer. I used the formula M = (1/f - 1/di)N

    I assumed that N = 25.0 cm
    do = 25.0cm
    M = 0.012 rad

    For the first part and in the second part
    f = 10.0 cm
    di = -25.0cm

    However when I plug in my numbers I get an answer of 3.5. What am I doing wrong?
     
  2. jcsd
  3. Apr 7, 2006 #2

    andrevdh

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    It can be solved with just the basic formulas. First work out the height of the object from the given angle (0.012 rad) and distance (25 cm). Then get the object distance with the thin lens formula for the magnifying glass. Use the magnification of this setup to get the image height. From this one can calculate the angle that the image makes at the eye. The angles are small so one can approximate the radian angle calculations with height/(distance from eye).
     
  4. Apr 8, 2006 #3

    andrevdh

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    The magnifation [itex]M=3.5[/itex] that you get is the ratio of the height of the image of the object to the real height of the object. You can work out the real height [itex]h[/itex] of the object with the [itex]0.012\ rad[/itex] angle. This and the magnification enables you to calculate the height of the image [itex]h\prime[/itex]
     
    Last edited: Apr 8, 2006
  5. Apr 19, 2011 #4
    Since magnification is 3.5, thus the image size is 0.3cm x 3.5 = 1.05 cm.
    θ=〖tan〗^(-1) 1.05/25=2.405°=0.042rad
     
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