An extended object is placed at point O , 10   cm in front of a convex lens L 1 and a concave lens L 2…

For convex lens,

$$\begin{gathered} \frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \frac{1}{f_1}=\left(\frac{3}{2}-1\right)\left(\frac{1}{20}-\frac{1}{-20}\right) \end{gathered}$$

$\Rightarrow f_1=+20 \mathrm{cm}$

Now, image distance from convex,

$v_{\mathrm{convex}}=\frac{u{f}_1}{u+f_1}=\frac{\left(-10\right)\times 20}{-10+20}$

$\Rightarrow v_{\mathrm{convex}}=-20 \mathrm{cm}$

This location of image will be object location for concave lens.

Therefore, $u_{\mathrm{concave}}=-30 \mathrm{cm}$

For concave lens,

$\frac{1}{f_2}=\left(\frac{3}{2}-1\right)\left(\frac{1}{-20}-\frac{1}{20}\right)$

$\Rightarrow f_2=-20 \mathrm{cm}$

Now, distance of image from concave,

$v_{\mathrm{concave}}=\frac{\left(-30\right)\times \left(-20\right)}{-30-20}$$=-12 \mathrm{cm}$

Now, $m=m_{\mathrm{convex}}\times m_{\mathrm{concave}}$

$m=\frac{\left(-20\right)}{\left(-10\right)}\times \frac{\left(-12\right)}{\left(-30\right)}$$=0.8$