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QJEEM22Phb27c: JEE Main 2022 Physics - Class 12 JEEM22Phb27c Solution: The power of a lens (biconvex) is in particular medium.…

The power of a lens (biconvex) is 1.25 m−11.25 \mathrm{~m}^{-1} in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm20 \mathrm{~cm} and 40 cm40 \mathrm{~cm} respectively. The refractive index of surrounding medium:

Multiple Choice Options

  • A 1
  • B 97\frac{9}{7}
  • C 32\frac{3}{2}
  • D 43\frac{4}{3}

Step-by-Step Verified Solution

∵\because 1f=(μ2μ1−1)(1R1−1R2){1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

⇒1.25100=(1.5μ1−1)(120+140)\Rightarrow {{1.25} \over {100}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {20}} + {1 \over {40}}} \right)

⇒180=(1.5μ1−1)×(4+2)80\Rightarrow {1 \over {80}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right) \times {{(4 + 2)} \over {80}}

⇒1.5μ1−1=16\Rightarrow {{1.5} \over {{\mu _1}}} - 1 = {1 \over 6}

⇒1.5μ1=76\Rightarrow {{1.5} \over {{\mu _1}}} = {7 \over 6}

⇒μ1=1.5×67=97\Rightarrow {\mu _1} = {{1.5 \times 6} \over 7} = {9 \over 7}

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