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QJEEM19Ph9a17: JEE Main 2019 Physics - Class 12 JEEM19Ph9a17 Solution: One plano-convex and one plano-concave lens of same radius of…

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 μ\mu 1 and that of 2 is μ\mu 2, then the focal length of the combination is : JEE Main 2019 (Online) 10th April Morning Slot Physics - Geometrical Optics Question 162 English

Multiple Choice Options

  • A 2Rμ1−μ2{2R \over { {{\mu _1} - {\mu _2}}}}
  • B R2−(μ1−μ2){R \over {2 - \left( {{\mu _1} - {\mu _2}} \right)}}
  • C Rμ1−μ2{R \over { {{\mu _1} - {\mu _2}}}}
  • D R2(μ1−μ2){R \over {2 \left( {{\mu _1} - {\mu _2}} \right)}}

Step-by-Step Verified Solution

For 1st lens 1f1=(μ1−11)(1∞−1−R)=μ1−1R{1 \over {{f_1}}} = \left( {{{{\mu _1} - 1} \over 1}} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right) = {{{\mu _1} - 1} \over R}

For 2nd lens 1f2=(μ2−11)(1−R)−0=μ2−1R{1 \over {{f_2}}} = \left( {{{{\mu _2} - 1} \over 1}} \right)\left( {{1 \over { - R}}} \right) - 0 = {{{\mu _2} - 1} \over R}

1feq=1f1+1f2{1 \over {{f_{eq}}}} = {1 \over {{f_1}}} + {1 \over {{f_2}}}

1feq=μ1−1R+−(μ2−1)R{1 \over {{f_{eq}}}} = {{{\mu _1} - 1} \over R} + {{ - \left( {{\mu _2} - 1} \right)} \over R}

⇒feq=Rμ1−μ2\Rightarrow {f_{eq}} = {R \over {{\mu _1} - {\mu _2}}}

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