A biconvex thick lens is constructed with glass $\left(\mathrm{\mu }=1.50\right)$. Each of the surfaces has a radius of 10 cm and the thickness at the middle is 5 cm. Locate the image of an object placed far away from the lens is

For first refraction at A
$\begin{array}{l}\frac{{\mathrm{\mu }}_2}{\mathrm{v}}-\frac{{\mathrm{\mu }}_1}{\mathrm{u}}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\mathrm{R}}\therefore \mathrm{u}=\mathrm{\alpha }\\ \frac{1.5}{\mathrm{v}}-0=\frac{0.5}{10}\Rightarrow \frac{1.5}{\mathrm{v}}=\frac{0.5}{10}\Rightarrow \mathrm{v}=30\mathrm{cm}\end{array}$
For second refraction at B
$\begin{array}{l}\mathrm{u}=(30-5)=25\mathrm{cm}\\ {\mathrm{R}}_2=-10\\ \frac{{\mathrm{\mu }}_2}{\mathrm{v}}-\frac{{\mathrm{\mu }}_1}{\mathrm{u}}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\mathrm{R}}\Rightarrow \frac{1}{\mathrm{v}}-\frac{1.5}{25}=\frac{1-1.5}{-10}\end{array}$
$\begin{array}{l}\Rightarrow \frac{1}{\mathrm{v}}-\frac{1.5}{2.5}=\frac{0.5}{10}\Rightarrow \frac{1}{\mathrm{v}}=\frac{1.25+1.5}{25}\\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{2.75}{25}\Rightarrow \frac{1}{\mathrm{v}}=\frac{25}{2.75}=9.1\mathrm{cm}\end{array}$