Focal length of plano-convex lens be $f_1$ $\;\frac{1}{f_1}\;=(1.5-1) \left(\;\frac{1}{15}-\;\frac{1}{\infty}\right)$ $\text{Or,}\quad \;\frac{1}{f_1}\;=\;\frac{1}{30}$ $\therefore \ f_1\;=30 \text{ cm}$ Focal length of plano-concave lens be $f_2$ $\;\frac{1}{f_2}\;=(1.5-1) \left(\;\frac{1}{\infty}-\;\frac{1}{15}\right)$ $\therefore \;\; f_2\;=-30 \text{ cm}$ since, parallel rays are incident on the plano convex lens it will form an image at a distance $30 \text{ cm}$ from the lens (since, focal length is $30 \text{ cm}$ ). This image will act as an object for the plano concave lens. Object distance $=30-20=10 \text{ cm}$ Applying lens formula, $\;\frac{1}{v_2}-\;\frac{1}{u_2}\;=\;\frac{1}{f_2}$ $\;\text{ Or, }\;\; \;\frac{1}{v_2}-\;\frac{1}{10}\;=-\;\frac{1}{30}$ So, $v_2=15\text{ cm}$