Physics - Geometrical Optics Question with Solution | TestHub

When an object is placed in front of such a lens,the rays first of all refracted from the convex surface, then refract from the polished plane surface and again refracts from convex surface. If $f_l$ and $f_m$ be the focal lengths of lens (convex surface) and mirror (plane polished surface) respectively, then effective focal length $F$ is given by $\begin{array}{rl}& \frac{1}{F}=\frac{1}{f_l}+\frac{1}{f_m}+\frac{1}{f_l}=\frac{2}{f_l}+\frac{1}{f_m}=\frac{2}{f_l}\left(\because f_m=\frac{R}{2}=\infty \right)\\ & \frac{1}{f_l}=(\mu -1)\left(\frac{1}{R}\right)\\ & \therefore \frac{1}{F}=\frac{2(\mu -1)}{R}\text{ or }F=\frac{R}{2(\mu -1)}\text{ or }R=2F=\frac{R}{(\mu -1)}\end{array}$