The focal length of an equiconvex lens is

$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$  ...(i)

For an equiconvex lens, $R_1=R_2=R$

$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2\left(\mu -1\right)}{R}$

Case I:   When the lens is cut along $\mathrm{XOX}\text{'}$ , then each half becomes equiconvex with

$R_1=+R, R_2=-R$

$\therefore \frac{1}{f}=\left(\mu -1\right)\left[\frac{1}{R}-\left(\frac{1}{-R}\right)\right]\Rightarrow \frac{1}{f}\left(\mu -1\right)\left(\frac{1}{R}+\frac{1}{R}\right)$

$= \left(\mu -1\right)\frac{2}{R}=\frac{1}{f^{\prime }} \Rightarrow f^{\prime }=f$

Case II:  When the lens is cut along $\mathrm{YOY}\text{'}$ , then each half becomes plano-convex with

$R_1=+R,R_2=\infty$

$\therefore \frac{1}{f^{\prime \prime }}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$= \left(\mu -1\right) \left(\frac{1}{R}-\frac{1}{\infty }\right)=\left(\frac{\mu -1}{R}\right)=\frac{1}{2f}\Rightarrow f^{\prime \prime }=2f$