Questions
The magnifying power of an astronomical telescope for normal vision is 7 and its tube length is 1.6 m. Then focal lengths of eye piece and objective are
detailed solution
Correct option is B
for normal vision M=f0fe and L=f0+fe∴L=fe(1+M)⇒fe=L1+M=160(1+7)cm=20 cmand f0=7×20 cm=140 cmTalk to our academic expert!
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