Explanation

(a) Biconvex lens is cut perpendicularly to the principle axis, it will become a plano-convex lens.
Focal length of biconvex lens
$\frac{1}{f} = (n - 1)\left( \frac{1}{R₁} - \frac{1}{R₂} \right)$
$\frac{1}{f} = (n - 1)\frac{2}{R}\ \ \ \ \ \ \ \ (\because R_{1} = R,R_{2} = - R)$
$\Longrightarrow f = \frac{R}{2(n - 1)}\ \ \ \ \ldots.\ldots(i)$
For plano-convex lens
$\frac{1}{f₁} = (n - 1)\left( \frac{1}{R} - \frac{1}{\infty} \right)$
$f^{1} = \frac{R}{(n - 1)}\ \ \ \ \ \ldots\ldots\ldots\ldots.(ii)$
Comparing Eqs. (i) and (ii), we see that focal length becomes double.
Power of lens P $\propto \frac{1}{focal\ length}$
Hence, power will become half.
New power = $\frac{4}{2} = 2\ D$