2. Solved Papers
  3. NEET

NEET NEET 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 \[\frac{1}{{{f}_{1}}}=\frac{({{\mu }_{1}}-1)}{R}-\frac{({{\mu }_{2}}-1)}{R}\] \[=\frac{[{{\mu }_{1}}-1-{{\mu }_{2}}+1]}{R}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\]


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