A bi-convex lens is formed with two thin plano-convex lenses as...

According to lens maker's formula $\frac{1}{f}=(n-1)$\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$For the first lens :$n=1.5, R_1=+14 \mathrm{~cm}, R_2=\infty \therefore \quad \frac{1}{f_1}=(1.5-1)$\left(\frac{1}{14}-\frac{1}{\infty}\right)=\frac{0.5}{14}$ For the second lens : $n=1.2, R_1=\infty, R_2=-14 \mathrm{~cm}$ $\therefore \quad \frac{1}{f_2}=(1.2-1)$\left(\frac{1}{\infty}-\frac{1}{-14}\right)=\frac{0.2}{14}$The focal length of the bi-convex lens is$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{0.5}{14}+\frac{0.2}{14}=\frac{0.7}{14}=\frac{1}{20}$According to thin lens formula;$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-40}=\frac{1}{20}$or$v=40 \mathrm{~cm}