Physics - Geometrical Optics Question with Solution | TestHub

The focal length $\left(f_1\right)$ of the plano-cöneex Tens with $n=1.5$ using lens-maker formula $\frac{1}{f_1}=\left(n_1-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]=(1.5-1)\left[\frac{1}{14}-\frac{1}{\infty }\right]=\frac{1}{28}$ The focal length $\left(f_2\right)$ of the plano-convex lens with $n=1.2$ $\frac{1}{f_2}=\left(n_2-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]=(1.2-1)\left[\frac{1}{\infty }-\frac{1}{-14}\right]=\frac{1}{70}$ Focal length F of the combination $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{20}$ Now, applying lens formula for the combination of lens $\begin{array}{ll}& \frac{1}{V}-\frac{1}{U}=\frac{1}{F}\Rightarrow \frac{1}{V}-\frac{1}{-40}=\frac{1}{20}[\text{ Given }\mu =40\mathrm{cm}]\\ \therefore & V=40\mathrm{cm}\end{array}$