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WB JEE Medical WB JEE Medical Solved Paper-2013

  • question_answer
    A plano-convex lens fits exactly into a plano-concave   lens.   Their   plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices \[{{\mu }_{1}}\]and \[{{\mu }_{2}}\]and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

    A)  \[\frac{R}{2({{\mu }_{1}}+{{\mu }_{2}})}\]

    B)  \[\frac{R}{2({{\mu }_{1}}-{{\mu }_{2}})}\]

    C)  \[\frac{R}{({{\mu }_{1}}-{{\mu }_{2}})}\]

    D)  \[\frac{2R}{({{\mu }_{2}}-{{\mu }_{1}})}\]

    Correct Answer: C

    Solution :

     Focal length of the combination \[\frac{1}{f}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] ?(i) We have \[{{f}_{1}}=\frac{R}{({{\mu }_{1}}-1)}\]and \[{{f}_{2}}=\frac{R}{({{\mu }_{2}}-1)}\] or \[\frac{1}{{{f}_{1}}}=\frac{R}{({{\mu }_{1}}-1)}\]or \[\frac{1}{{{f}_{2}}}=-\frac{R}{({{\mu }_{2}}-1)}\] Putting these values in Eq. (i), we get \[\frac{1}{f}=\frac{({{\mu }_{1}}-1)}{R}-\frac{({{\mu }_{2}}-1)}{R}.\] \[=\frac{[{{\mu }_{1}}-1-{{\mu }_{2}}+1]}{R}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\]


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