Physics - Geometrical Optics Question with Solution | TestHub

For bifocal convex lens $\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$ $=\frac{(\mu -1)\times 2}{R}\dots .\left(R_1=R_2=R\right)$ For plane surface ${\mathrm{R}}_2=\infty$ For half plane-convex lens $\begin{array}{rl}& \frac{1}{{\mathrm{f}}^{\prime }}=(\mu -1)\frac{1}{\mathrm{R}}\\ & \frac{1/\mathrm{f}}{1/{\mathrm{f}}^{\prime }}=\frac{(\mu -1)}{\mathrm{R}}\times 2\times \frac{\mathrm{R}}{\mu -1}=2\\ & \frac{{\mathrm{f}}^{\prime }}{\mathrm{f}}=2\\ & {\mathrm{f}}^{\prime }=2\mathrm{f}\end{array}$ As focal length is $F,f^{\prime }=2F$.