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VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    The power of a thin convex lens \[\left( _{a}{{n}_{g}}=1.5 \right)\] is 5.0 D. When it is placed in a liquid of refractive index \[_{a}{{n}_{l}},\] then it behaves as a concave lens of focal length 100 cm. The refractive index of the liquid\[_{a}{{n}_{l}},\]will be

    A)  5/3             

    B)  4/3

    C)  \[\sqrt{3}\]             

    D)  5/4  

    Correct Answer: A

    Solution :

    Here,    \[5=(1.5-1)\left( \frac{2}{R} \right)\] If a lens (made of glass) of refractive index \[{{\mu }_{g}}\] is immersed in a liquid of refractive index a \[{{\mu }_{l}},\]then its focal length in liquid \[{{f}_{1}}\] is given by \[\frac{1}{{{f}_{1}}}={{(}_{l}}{{\mu }_{g}}-1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] \[-1=\left( \frac{1.5}{n}-1 \right)\left( \frac{2}{R} \right)\] Dividing, \[-5=\frac{0.5n}{1.5-n}\] or \[-7.5+5n=0.5n\]c or \[-7.5=-4.5n\] or \[n=\frac{75}{45}=\frac{5}{3}\]


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