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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    The focal lengths of the objective and of the eye-piece of a compound microscope are\[{{f}_{o}}\]and\[{{f}_{e}}\]respectively. If L is the tube length and D, the least distance of distinct vision, then its angular magnification, when the image is formed at infinity, is

    A)  \[\left( 1-\frac{L}{{{f}_{o}}} \right)\left( \frac{D}{{{f}_{e}}} \right)\]

    B)         \[\left( 1+\frac{L}{{{f}_{o}}} \right)\left( \frac{D}{{{f}_{e}}} \right)\]

    C)  \[\frac{L}{{{f}_{o}}}\left( 1-\frac{D}{{{f}_{e}}} \right)\] 

    D)         \[\frac{L}{{{f}_{o}}}\left( 1+\frac{D}{{{f}_{e}}} \right)\]

    E)  \[\frac{L}{{{f}_{o}}}\left( \frac{D}{{{f}_{e}}} \right)\]

    Correct Answer: E

    Solution :

    In compound microscope, the final image is formed at infinite. Magnification \[{{M}_{\infty }}=-\frac{{{v}_{o}}}{{{u}_{o}}}\left( \frac{D}{{{f}_{e}}} \right)\] Or           \[{{m}_{\infty }}=\frac{L}{{{f}_{o}}}\left( \frac{D}{{{f}_{e}}} \right)\]       


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