We know from Lens maker's formula

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

Initially, Here R₁ = R and R₂ = –R by convection.

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

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

If we cut the lens along XOX' then the two halves of the lens will be having the same radii of curvature and so, focal length f' = f .

But when we cut it along YOY' then, we will have

R₁ = R but R₂ = $\infty$

$\therefore$ ${1 \over {f''}} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over \infty }} \right)$ = $\left( {\mu - 1} \right){1 \over R}$ = ${1 \over {2f}}$

$\Rightarrow$ f'' = 2f