JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer One plano-convex and one plano-concave lens of same radius of curvature 'R' but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is \[{{\mu }_{1}}\]and that of 2 is \[{{\mu }_{2}}\], then the focal length of the combination is : [JEE Main 10-4-2019 Morning]

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

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

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

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

    Correct Answer: D

    Solution :

    For 1st lens\[\frac{1}{{{f}_{1}}}=\left( \frac{{{\mu }_{1}}-1}{1} \right)\left( \frac{1}{\infty }-\frac{1}{-R} \right)=\frac{{{\mu }_{1}}-1}{R}\] for 2nd lens\[\frac{1}{{{f}_{2}}}=\left( \frac{{{\mu }_{2}}-1}{1} \right)\left( \frac{1}{-R}-0 \right)=-\frac{{{\mu }_{2}}-1}{R}\] \[\frac{1}{{{f}_{eq}}}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] \[\frac{1}{{{f}_{eq}}}=\frac{R}{{{\mu }_{1}}-1}+\frac{R}{-({{\mu }_{2}}-1)}\Rightarrow \frac{1}{{{f}_{eq}}}=\frac{R}{{{\mu }_{1}}-{{\mu }_{2}}}\] Hence \[{{f}_{eq}}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\]


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