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3 May, 21:56

compute the value of the magnifying power of a 150-inch fl8 telescope with an eyepiece of one-half-inch focal length

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  1. 3 May, 22:02
    0
    assume fl8 is f/8

    150" at f/8 means the focal length = 150*8 = 1200 inches

    divide by eyepiece of 1/2", the magnifying power = 1200 / (1/2) = 2400X
  2. 3 May, 22:21
    0
    150 inches at f/8 gives a focal length of 1200 inches.

    Divide 1200 inches by 1/2 inch eye piece to get the magnification of 2400x.
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