In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to $|R_1|$ and $|R_2|$, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is
Answer: (D) $-\dfrac{1}{6}\left(\dfrac{1}{|R_1|}-\dfrac{1}{|R_2|}\right)$
For thin elements in contact the power is the sum of the powers of all refracting surfaces, $\dfrac{\mu_2-\mu_1}{R}$. The flat water surfaces contribute nothing, so only the two curved surfaces of the glass matter.
Take light travelling downward. Both curved surfaces are concave upward, so their centres of curvature lie above them (on the incoming side): $R = -|R|$.
Upper surface, water $\to$ glass:
$$P_1 = \frac{\frac{3}{2}-\frac{4}{3}}{-|R_1|} = -\frac{1}{6|R_1|}$$
Lower surface, glass $\to$ water:
$$P_2 = \frac{\frac{4}{3}-\frac{3}{2}}{-|R_2|} = +\frac{1}{6|R_2|}$$
$$P = P_1 + P_2 = -\frac{1}{6}\left(\frac{1}{|R_1|}-\frac{1}{|R_2|}\right)$$
Solution by Sreeraj P, M.Sc Physics