In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to $$\mid R_1 \mid \text{ and } \mid R_2 \mid$$, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is

Taking downward direction as positive:

$$P_1 = \left(\frac{4}{3} - 1\right)\left(\frac{1}{\infty} - \frac{1}{-|R_1|}\right) \implies P_1 = \frac{1}{3|R_1|}$$

$$P_2 = \left(\frac{3}{2} - 1\right)\left(\frac{1}{-|R_1|} - \frac{1}{-|R_2|}\right) \implies P_2 = \frac{1}{2}\left(\frac{1}{|R_2|} - \frac{1}{|R_1|}\right)$$

$$P_3 = \left(\frac{4}{3} - 1\right)\left(\frac{1}{-|R_2|} - \frac{1}{\infty}\right) \implies P_3 = -\frac{1}{3|R_2|}$$

Equivalent power:

$$P_{\text{eq}} = P_1 + P_2 + P_3 \implies P_{\text{eq}} = \frac{1}{3|R_1|} + \frac{1}{2|R_2|} - \frac{1}{2|R_1|} - \frac{1}{3|R_2|}$$

$$P_{\text{eq}} = -\frac{1}{6|R_1|} + \frac{1}{6|R_2|} \implies P_{\text{eq}} = -\frac{1}{6}\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)$$