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Q. There is a telescope of magnifying power $5$ and tube length $60\, cm$ . What is the focal length of its eye piece?

NTA AbhyasNTA Abhyas 2022

Solution:

$m=\frac{f_{0}}{f_{e}}$
$5=\frac{f_{0}}{f_{e}}$
$f_{0}=5f_{e}$
$f_{0}+f_{e}=60$
$6f_{e}=60$
$f_{e}=10$ .