JEE Main 2025 Jan 22 Shift 2, Physics Q1: Thin Lens and Mirrors

For the original biconvex lens (both surfaces of radius $R$): $\frac{1}{f}=(\mu -1)\frac{2}{R}=P=4D$.

The horizontal cut AB along the principal axis halves the aperture but does not change either radius of curvature. The optical power is independent of aperture, so power remains $4D$.

The vertical cut CD perpendicular to the principal axis converts the biconvex lens into a plano-convex piece: one surface has radius $R$, the other becomes flat ($R=\infty$). Hence:

$\frac{1}{f^{\prime }}=(\mu -1)\frac{1}{R}=\frac{P}{2}=2D$.

Each of the four quarter pieces is a plano-convex half, so its power is $2D$.