An equi-convex lens has power P. It is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be

When the lens is cut along its principal axis, the focal length of the two halves will remain same because the radius of curvature of both the surfaces are still same. So, the power also remains same as $\mathrm{P}=\frac{1}{\mathrm{f}}$