JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    If for a spherical mirror object distance, \[u=(50.1\pm 0.5)\] and image distance\[v=(20.1\pm 0.2)\], then focal length of the spherical mirror will be

    A) \[(14.3\pm 0.1)cm\]      

    B) \[(14.3\pm 0.5)cm\]

    C) \[(30.1\pm 0.1)cm\]      

    D)  \[(25.3\pm 0.5)cm\]

    Correct Answer: A

    Solution :

    We have\[\frac{1}{f}=\frac{1}{u}+\frac{1}{v}=\frac{v+u}{uv}\]; \[\therefore \]  \[f=\frac{uv}{v+u}=\frac{(50.1)(20.1)}{(50.1+20.1)}\]                 \[=14.3\,\,cm\] From     \[\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\]                 \[\frac{-\Delta f}{{{f}^{2}}}=\frac{-\Delta u}{{{u}^{2}}}+\frac{-\Delta u}{{{v}^{2}}}\] \[\Rightarrow \]               \[\Delta f=\Delta u{{\left( \frac{f}{u} \right)}^{2}}+\Delta v{{\left( \frac{f}{v} \right)}^{2}}\]                 \[=0.5{{\left( \frac{14.3}{50.1} \right)}^{2}}+0.2{{\left( \frac{14.3}{20.1} \right)}^{2}}\]                 \[=0.04+0.10=0.14\,\,cm\] \[\therefore \]  \[f=(14.3\pm 0.1)cm\]


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