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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Two plano-concave lenses (1 and 2) of) glass of refractive index 1.5 have radii of curvature 25 cm and 20 cm. They are placed in contact with their curved surfaces towards each other and the space between them is filled with liquid of refractive index\[\frac{4}{3}\]. Then the combination is

    A)  convex of focal length 70 cm

    B)  concave of focal length 70 cm

    C)  concave of focal length 66.6 cm

    D)  convex of focal length 66.6 cm

    E)  concave of focal length 72.5 cm

    Correct Answer: C

    Solution :

    The focal length \[\frac{1}{{{f}_{1}}}=(\mu -1)\frac{1}{-R}\] \[\frac{1}{{{f}_{1}}}=(1.5-1)\frac{1}{-25}\] \[{{f}_{1}}=-50\,cm\] The focal length                 \[\frac{1}{{{f}_{2}}}=(1.5-1)\times \frac{1}{-20}\]                 \[\frac{1}{{{f}_{2}}}=0.5\times \frac{1}{-20}m\]                 \[{{f}_{2}}=-\,40\,cm\] The focal length of bi-convex lens                 \[\frac{1}{{{f}_{3}}}={{(}_{1}}{{n}_{2}}-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\]                 \[\frac{1}{{{f}_{3}}}=\left( \frac{4}{3}-1 \right)\left( \frac{1}{20}+\frac{1}{25} \right)\]                 \[\frac{1}{{{f}_{3}}}=\frac{1}{3}\times \left( \frac{5+4}{100} \right)\]                 \[\frac{1}{{{f}_{3}}}=\frac{1}{3}\times \frac{9}{100}\]                 \[\frac{1}{{{f}_{3}}}=\frac{3}{100}\]                 \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}+\frac{1}{{{f}_{3}}}\]                 \[=-\frac{1}{50}-\frac{1}{40}+\frac{3}{100}\]                 \[\frac{1}{F}=\frac{-200}{3}\]      \[F=-\,66.6\text{ }cm\] Negative sign represent the concave lens.


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