A) 3.0
B) 3/2
C) 2/3
D) 0.3
Correct Answer: A
Solution :
: In a telescope, distance between objective\[({{f}_{o}})\]and eyepiece\[({{f}_{e}})={{f}_{o}}+{{f}_{e}}\] \[\therefore \] \[{{f}_{o}}+{{f}_{e}}=24\] or \[18+{{f}_{e}}=24\] \[\therefore \] \[{{f}_{e}}=6\,cm\] Magnifying power \[\frac{{{f}_{o}}}{{{f}_{e}}}=\frac{18}{6}=3\]You need to login to perform this action.
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