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MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • question_answer
    An astronomical telescope has an objective of focal length 100 cm and magnifying power of the distance between the two lenses in normal adjustment will be

    A)  106 cm                

    B)  102 cm

    C)  92 cm                  

    D)  78 cm

    Correct Answer: B

    Solution :

                    \[M=-\frac{{{f}_{o}}}{{{f}_{e}}},\]where\[{{f}_{o}}\]is focal length of object and\[{{f}_{e}}\]is focal length of eyepiecs \[{{f}_{e}}=\frac{{{f}_{o}}}{|M|}=\frac{100}{50}=2\,cm\] Length of the telescope \[={{f}_{1}}={{f}_{e}}=100+2=102\,cm\]


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