A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4 D then the power of a part of the divided lens is

$$P = (\mu - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = \frac{2(\mu - 1)}{R} = 4\text{ D}$$

$$P_{\text{horizontal half}} = P = 4\text{ D}$$

Cutting perpendicular to the principal axis (plane CD):

$$R_1 = R, \quad R_2 = \infty \implies P_{\text{vertical half}} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{\mu - 1}{R} = \frac{P}{2}$$

For a quarter part (divided by both planes AB and CD):

$$P' = \frac{P}{2} \implies P' = \frac{4\text{ D}}{2} = 2\text{ D}$$