Biconvex lens is cut perpendicularly to the principle axis, it will become a plano-convex lens.

Focal length of biconvex lens

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R₁}-\frac{1}{R₂}\right)$

$\frac{1}{f}=\left(n-1\right)\frac{2}{R} (\because R_1=R,R_2=-R)$

$⟹f=\frac{R}{2\left(n-1\right)} \dots .\dots \left(i\right)$

For plano-convex lens

$\frac{1}{f₁}=\left(n-1\right)\left(\frac{1}{R}-\frac{1}{\infty }\right)$

$f^1=\frac{R}{\left(n-1\right)} \dots \dots \dots \dots .\left(ii\right)$

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 = $\frac{4}{2}=2\mathrm{ }\mathrm{D}$