The radii of curvature of both the surfaces of a convex lens of focal length ' $f$ ' and focal power '…
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 $\quad P^{\prime}=\frac{1}{f^{\prime}}$
$\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{f}}{\mathrm{f}^{\prime}}=\frac{1}{2}$
$\mathrm{P}^{\prime}=\frac{\mathrm{P}}{2}$