Verified Solution • 2025–26 Session moderate

QJEEM22Ph7e26: JEE Main 2022 Physics - Class 12 JEEM22Ph7e26 Solution: A convex lens has power P. It is cut into two halves along its…

A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figures). Choose the incorrect option for the reported pieces.

JEE Main 2022 (Online) 27th June Evening Shift Physics - Geometrical Optics Question 101 English

Multiple Choice Options

  • A Power of L1=P2{L_1} = {P \over 2}
  • B Power of L2=P2{L_2} = {P \over 2}
  • C Power of L3=P2{L_3} = {P \over 2}
  • D Power of L1 = P

Step-by-Step Verified Solution

We know P=1f=(μ−1)(1R1−1R2)P = {1 \over f} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

L1:1f1=(μ−1)(1R1−1R2)=P1=(μ−1)(2R)=P{L_1}:{1 \over {{f_1}}} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) = {P_1} = (\mu - 1)\left( {{2 \over R}} \right) = P

L2:1f2=(μ−1)(1R1)=P2=μ−1R{L_2}:{1 \over {{f_2}}} = (\mu - 1)\left( {{1 \over {{R_1}}}} \right) = {P_2} = {{\mu - 1} \over R}

L3:1f3=(μ−1)(−1R2)=P3=μ−1R{L_3}:{1 \over {{f_3}}} = (\mu - 1)\left( { - {1 \over {{R_2}}}} \right) = {P_3} = {{\mu - 1} \over R}

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