The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will

As we know the relations between $\mathrm{f},\mathrm{\mu },{\mathrm{R}}_1$ and R₂ is known as lens maker's formula :
$\begin{array}{l}\frac{1}{\mathrm{f}}=(\mathrm{\mu }-1)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)\\ {\mathrm{R}}_1=\mathrm{\infty },{\mathrm{R}}_2=-\mathrm{R}\\ \mathrm{f}=\frac{\mathrm{R}}{(\mathrm{\mu }-1)}\end{array}$
Given that'
R=20 cm,
$\mathrm{\mu }=1.5$
Put the values;
$\mathrm{f}=\frac{\mathrm{R}}{\mathrm{\mu }-1}=\frac{20}{15-1}=40\mathrm{cm}$
As f>0,  it means converging nature of the lens.
So, lens act as a convex lens irrespective of the side on which the object lies.