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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    An infinitely long rod lies along the axis of concave mirror of focal length\[f\]. The near end of the rod is at a distance\[x>f\]from the mirror. Then the length of the image of the rod is:

    A)  \[\frac{{{f}^{2}}}{x+f}\]               

    B)         \[\frac{{{f}^{2}}}{x}\]                   

    C)  \[\frac{xf}{x-f}\]             

    D)         \[\frac{xf}{x+f}\]

    E)  \[\frac{{{f}^{2}}}{x-f}\]

    Correct Answer: E

    Solution :

    \[\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\] and      \[m=\frac{v}{u}=-\frac{I}{O}\] Using above relation, the length of image                 \[I=\frac{{{f}^{2}}}{x-f}\]


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