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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    The focal length of objective and eye piece of a telescope are\[{{f}_{0}}\] and\[{{f}_{e}}\] respectively. Then its magnification will be:

    A)  \[{{f}_{0}}+{{f}_{e}}\]                                  

    B)  \[{{f}_{0}}\times {{f}_{e}}\]

    C)  \[\frac{{{f}_{0}}}{{{f}_{e}}}\]                     

    D)         \[\frac{{{f}_{e}}}{{{f}_{0}}}\]

    Correct Answer: C

    Solution :

    Magnifying power of telescope for relaxed eye \[M=\frac{{{f}_{0}}}{{{f}_{e}}}=\frac{{{f}_{0}}}{{{f}_{e}}}\] (numerically)


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