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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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