A symmetric double convex lens is sliced into two equal parts by a...

The power of a symmetric double convex lens is given by $P = (\mu - 1)\left(\dfrac{1}{R} - \dfrac{1}{-R}\right) = \dfrac{2(\mu - 1)}{R}$.

Given $P = 4 \text{ D}$.

When the lens is sliced into two equal parts by a plane perpendicular to its principal axis, each part becomes a plano-convex lens with radii of curvature $R$ and $\infty$.

The power of one such cut lens is $P' = (\mu - 1)\left(\dfrac{1}{R} - \dfrac{1}{\infty}\right) = \dfrac{\mu - 1}{R}$.

Comparing the two expressions, we get $P' = \dfrac{P}{2}$.

Substituting the value of $P$, $P' = \dfrac{4}{2} = 2 \text{ D}$.