The radii of curvature of both the surfaces of a convex lens of focal length ' \(f\) ' and focal power ' \(\mathrm{P}\) ' are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens is

From Lens maker's formula,
\(\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
As the radius of both the surfaces is the same,
\(\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)\)
\(\frac{1}{\mathrm{f}}=\frac{2}{\mathrm{R}}(\mu-1)\)
When one surface is plane, \(R_1=R, R_2=\infty\)
\(\therefore \quad \frac{1}{\mathrm{f}^{\prime}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)\)
\(\frac{1}{\mathrm{f}^{\prime}}=\frac{1}{\mathrm{R}}(\mu-1)\)
\(\therefore \quad \frac{\mathrm{f}^{\prime}}{\mathrm{f}}=2\)
\(\mathrm{f}^{\prime}=2 \mathrm{f}\)
\(P=\frac{1}{f} \quad\) and \(\quad P^{\prime}=\frac{1}{f^{\prime}}\)
\(\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{f}}{\mathrm{f}^{\prime}}=\frac{1}{2}\)
\(\mathrm{P}^{\prime}=\frac{\mathrm{P}}{2}\)