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MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • question_answer
    The radii of curvature of the two surfaces of a lens are 20 cm and 30 cm and the reparative index of the material of the lens is 1.5. If the lens is concave-convex, then the focal length of the lens is

    A)  24 cm                  

    B)  10 cm  

    C)  15 cm                  

    D)  120 cm

    Correct Answer: C

    Solution :

                    The focal length of the tens \[\frac{1}{f}=(\mu -1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] \[\frac{1}{f}=(1.5-1)\left( \frac{1}{20}-\frac{1}{30} \right)\] \[\frac{1}{f}=0.5\left( \frac{30-20}{600} \right)\] \[\frac{1}{f}=\frac{1}{2}\times \frac{10}{600}=\frac{1}{120}cm\] or            \[f=120cm\]


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