An equiconvex lens is cut into two halves along (i) XOX' and (ii) YOY' as shown in the figure. Let…

For an equiconvex lens with radius R on both sides: $\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R} + \dfrac{1}{R}\right) = \dfrac{2(\mu - 1)}{R}$
Case (i), cut along XOX' (the principal axis): each half is still bounded by both curved surfaces of radius R; only the aperture is halved. So each half has the same focal length: $f' = f$.
Case (ii), cut along YOY': each half is a plano-convex lens with one curved surface of radius R and one plane surface ($R_2 = \infty$).
$\dfrac{1}{f''} = (\mu - 1)\left(\dfrac{1}{R} - \dfrac{1}{\infty}\right) = \dfrac{\mu - 1}{R} = \dfrac{1}{2f}$
$\Rightarrow f'' = 2f$
So $f' = f$ and $f'' = 2f$.