A) 20
B) 30
C) 60
D) 15
Correct Answer: B
Solution :
Given : Focal length of objective lens \[{{f}_{e}}=180\,cm\] Focal length of eye piece \[{{f}_{e}}=6\,cm\] In the normal adjustment, the magnifying power of telescope is given by \[=\frac{{{f}_{0}}}{{{f}_{e}}}=\frac{180}{6}=30\]You need to login to perform this action.
You will be redirected in
3 sec