2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    A thin convex lens made from crown glass\[\left[ \mu =\frac{3}{2} \right]\]has focal length f. When it is measured in two different liquids having refractive indices\[\frac{4}{3}\]and\[\frac{5}{3},\]it has the focal lengths f1 and f2 respectively. The correct relation between the focal lengths is :   JEE Main  Solved  Paper-2014

    A) \[{{f}_{2}}>f\]and\[{{f}_{1}}\] becomes negative

    B) \[{{f}_{1}}\]and\[{{f}_{2}}\] both become negative

    C) \[{{f}_{1}}={{f}_{2}}<f\]

    D) \[{{f}_{1}}>f\]and\[{{f}_{2}}\]becomes negative

    Correct Answer: D

    Solution :

    \[\frac{1}{f}=\left( \frac{3}{2}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]\[\frac{1}{f}=\frac{1}{2}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]        ?.(1) \[\frac{1}{{{f}_{1}}}=\left( \frac{\frac{3}{2}}{\frac{4}{3}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]\[\frac{1}{{{f}_{1}}}=\frac{1}{8}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]?.(2) \[\frac{1}{{{f}_{2}}}=\left( \frac{\frac{3}{2}}{\frac{5}{3}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]\[\frac{1}{{{f}_{2}}}=-\frac{1}{10}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]..(3) From (1)\[f=\frac{2}{\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)}\] From (2) \[{{f}_{1}}=\frac{8}{\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)}\] From (3)\[{{f}_{2}}=-\frac{10}{\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner