elrodnumber3
elrodnumber3 elrodnumber3
  • 10-08-2018
  • Physics
contestada

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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Asdfjkadfjkakjf Asdfjkadfjkakjf
  • 11-08-2018

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


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zlskfrum zlskfrum
  • 11-08-2018

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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