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AIIMS AIIMS Solved Paper-2006

  • question_answer
    A wire mesh consisting of very small squares is viewed at a distance of 8 cm through a magnifying converging lens of focal length 10 cm, kept close to the eye. The magnification produced by the lens is:

    A) 5             

    B)       8                             

    C)       10                          

    D)       20

    Correct Answer: A

    Solution :

                    Lens formula is given by \[\frac{1}{f}=\frac{1}{\upsilon }-\frac{1}{u}\]                          ?..(i) where f is focal length of lens, v is image distance and u is object distance. Given, f= 10 cm (as lens is converging) \[u=-8\text{ }cm\] (as object is placed on left side of the lens) Substituting these values in Eq. (i), we get \[\frac{1}{10}=\frac{1}{v}-\frac{1}{-8}\] \[\Rightarrow \] \[\frac{1}{v}=\frac{1}{10}-\frac{1}{8}\] \[\Rightarrow \]  \[\frac{1}{v}=\frac{8-10}{80}\] \[v=\frac{80}{-2}=-40cm\] Hence, magnification produced by the lens \[m=\frac{v}{u}=\frac{-40}{-8}=5\]


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