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NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    The magnifying power of a telescope is. 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are

    A) 10 cm, 10 cm

    B)        15 cm, 5 cm        

    C)        18 cm, 2 cm        

    D)        11 cm, 9cm

    Correct Answer: C

    Solution :

    Given, \[M=\frac{{{f}_{0}}}{{{f}_{e}}}=9\] and \[{{f}_{0}}+{{f}_{e}}=20\] \[{{f}_{0}}=9{{f}_{e}}\]                 So,          \[9{{f}_{e}}+{{f}_{e}}=20\]                                 \[{{f}_{e}}=2\,cm\] \[\therefore \]  \[{{f}_{0}}=9\times 2\]                                 \[{{f}_{0}}=18\,cm\]


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