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AMU Medical AMU Solved Paper-1995

  • question_answer
    In a telescope, the distance between the objective and the eye piece is 24.0 cm. The focal length of the objective is 18.0 cm. Its magnifying power is

    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 \[\frac{{{f}_{o}}}{{{f}_{e}}}=\frac{18}{6}=3\]


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