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Homework Help: Magnification of telescope

  1. Jun 15, 2006 #1
    Please help!
    # A telescope has an objective of focal length 50 cm and an eye piece of focal length 5 cm. It is focused for distinct vision on a scale 200 cm away from the objective. What is the linear magnification of the telescope numerically? Also what is the magnifying power of the telescope numerically?
    Could someone tell me, if linear magnification of the telescope different from magnifying power of the telescope? I think they both mean the same. I have this doubt because 2 different answers are given in my book for linear magnification and magnifying power of the telescope.
  2. jcsd
  3. Jun 15, 2006 #2


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    Oddly, I've never actually heard the term "linear magnification." The magnification of the telescope is just the focal length of the objective divided by the focal length of the eyepiece. Therefore, this telescope-eyepiece system achieves a magnification of 10x.

    - Warren
  4. Jun 17, 2006 #3


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    I think...
    The linear magnification is the ratio of the length of the image to the length of object.

    This is called in such a way because some times we find the ratio of the area of image to the area of object this is called aerial magnification which is square of the linear magnification.

    The linear magnification hear is not fo/fe because the object (200 cm.) is not at very large distance and the rays, from a point of the object, can not be considered parallel. You have to apply the lens formula to find the object distance and the image distance for the distance of the image from the optical center of objective and so on... This is not simply fo as in the case if the object is at long distance. Try this and you will learn automatically.
  5. Jun 17, 2006 #4
    There is a word linear magnification in physics in terms of optical instruments. It is a unitless number that is the ratio of the height of the image and the height of the object.

    Additionally, I think you are talking about the angular magnification of the telescope for the magnifying power. Angular magnification is different for linear magnification.

    Last edited: Jun 17, 2006
  6. Jun 17, 2006 #5
    According to my Physics 11 textbook, linear magnification is "the ratio of the size of an image to the size of the object". I'm not so sure about the term "magnifying power". WDJr seems to be correct:

    "The angular size of the virtual image seen by the observer will be larger than the actual angular size of the object. The ratio of these two sizes is the magnifying power and is equal to the ratio of the focal lengths of the objective and ocular."
    -- http://www.infoplease.com/ce6/sci/A0861462.html
    Last edited: Jun 17, 2006
  7. Jun 18, 2006 #6
    I solved it in following way:
    Assuming it to be an astronomical telescope and the final image is formed at the least distance of distinct vision(D). Let f(o) & f(e) be the focal length of objective and eye piece. We know that,
    Magnifying power = Angular Magnification = [f(o)/f(e)]{1 + f(e)/D}
    = (50/5){1 + 5/25}
    = 12
    Is it right? But the answer given in my book is 16.
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