A symmetric double convex lens is cut in two equal parts by a plane...

Biconvex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens. Focal length of biconvex lens $\begin{gathered} \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \frac{1}{f}=(n-1) \frac{2}{R} \quad\left(\because R_1=R, R_2=-R\right) \end{gathered} \Rightarrow \quad f=\frac{R}{2(n-1)}\ldots(i)$ For plano-convex lens $\begin{gathered} \frac{1}{f_1}=(n-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\ f_1=\frac{R}{(n-1)}\ldots(ii) \end{gathered}$ Comparing Eqs. (i) and (ii), we see that focal length becomes double. As power of lens $P \propto \frac{1}{\text{ focal length }}$ Hence, power will become half. $\text{ New power }=\frac{4}{2}=2$$\mathrm{D}$