An equiconvex lens is cut two halves along (i) XOX' and (ii) YOY'...

Since the lens is equiconvex, the radius of curvature of each half is same.
Using lens maker formula $\frac{1}{f}=(\mu-$1)$\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$Here $R_1=RR_2=-R$ by convection \therefore $\frac{1}{f}=(\mu-$1) $\frac{2}{\mathrm{R}}(\mu-$1) $\frac{1}{\mathrm{R}}=\frac{1}{2 f} \ldots(\mathrm{i})$If we cut the lens along $\mathrm{XOX}^{\prime}$ then two halves of the lens will be having the same radii of curvalur and $$\begin{aligned}& R_1=R, R_2=-R \\ & \therefore \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{2}{R}\right)\\ & \Rightarrow \frac{1}{f^{\prime}}=\frac{2}{2 f}=\frac{1}{f} \end{aligned}$$From (i) it is clear that $f^{\prime}=f$ But when we cut along $\mathrm{YOY}$ ' then$$\begin{aligned}& R_1=R \text{ and }R_2=\infty\\ & \therefore \infty \frac{1}{f^{\prime \prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)\\ & =(\mu-1) \frac{1}{R}=\frac{1}{2 f}\\ & \Rightarrow \quad f^{\prime \prime}=2 f \end{aligned}$$