JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Geometrical Optics

The cut is made along a plane perpendicular to the principal axis, so the biconvex lens separates into two plano-convex pieces, each retaining one curved face and acquiring one flat face. For the original symmetric biconvex lens, $\frac{1}{f}=(\mu -1)\left(\frac{2}{R}\right)=\frac{1}{10}{\text{cm}}^{-1}$. Each plano-convex piece has only a single curved surface, so $\frac{1}{f^{\prime }}=(\mu -1)\left(\frac{1}{R}\right)$, which is exactly half of $\frac{1}{f}$. Hence $f^{\prime }=20$ cm, and the power of each piece is $P^{\prime }=\frac{1}{0.20}=5$ D.