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The magnifying power of a telescope in normal adjustment is 24, when the length of the telescope tube 1 meter. The focal length of the eye lens is -

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Question

The magnifying power of a telescope in normal adjustment is 24, when the length of the telescope tube 1 meter. The focal length of the eye lens is

Options

  • 100 cm

  • 24 cm

  • 76 cm

  • 4 cm

MCQ

Solution

4 cm

Explanation:

m = `(fo)/(fe)` = 24, F0 = 24fe 

Length of telescope tube fo + fe = 1 m

2Kfe + fe = 100 cm

25fe = 100 cm

fe = 100 cm / 25 = 4 cm

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