Physics - Geometrical Optics Question with Solution | TestHub

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, $R_1=R,R_2=\infty$ $\therefore \frac{1}{{\mathrm{f}}^{\prime }}=(\mu -1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty }\right)$ $\frac{1}{{\mathrm{f}}^{\prime }}=\frac{1}{\mathrm{R}}(\mu -1)$ $\therefore \frac{{\mathrm{f}}^{\prime }}{\mathrm{f}}=2$ ${\mathrm{f}}^{\prime }=2\mathrm{f}$ $P=\frac{1}{f}$ and $P^{\prime }=\frac{1}{f^{\prime }}$ $\frac{{\mathrm{P}}^{\prime }}{\mathrm{P}}=\frac{\mathrm{f}}{{\mathrm{f}}^{\prime }}=\frac{1}{2}$ ${\mathrm{P}}^{\prime }=\frac{\mathrm{P}}{2}$