JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Geometrical Optics

For a thin biconvex lens with both faces of radius of curvature R, the lens maker's formula gives
$\frac{1}{f}=(\mu -1)\left(\frac{1}{R}+\frac{1}{R}\right)=(\mu -1)\frac{2}{R}$
so the power of the whole lens is $P=(\mu -1)\frac{2}{R}=4\text{ D}$.

The plane CD (perpendicular to the optical axis, through the centre) splits the lens along its thickness into a front piece and a back piece. Each of these is no longer biconvex but plano-convex, with only one curved surface of radius R and a flat face at the cut:
$\frac{1}{f^{\prime }}=(\mu -1)\frac{1}{R}$
which is exactly half of the biconvex expression, so its power is $P^{\prime }=\frac{P}{2}$.

The plane AB (containing the axis) further splits each plano-convex piece by aperture only — this changes the size of the piece but not the radius of curvature of its curved surface, so the power is unaffected by this second cut.

Hence each of the four pieces has power
$P^{\prime }=\frac{4}{2}=2\text{ D}$