An extended object is placed at point O , 10cm in front of a convex...

For convex lens,

$\dfrac{1}{f} = \left(\dfrac{{\mu}_{2}}{{\mu}_{1}} - 1\right)\left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right) ⇒ \dfrac{1}{{f}_{1}} = \left(\dfrac{3}{2} - 1\right)\left(\dfrac{1}{20} - \dfrac{1}{- 20}\right)$

$⇒ {f}_{1} = + 20cm$

Now, image distance from convex,

${v}_{convex} = \dfrac{u{f}_{1}}{u + {f}_{1}} = \dfrac{\left(- 10\right) \times 20}{- 10 + 20}$

$⇒ {v}_{convex} = - 20cm$

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

Therefore, ${u}_{concave} = - 30cm$

For concave lens,

$\dfrac{1}{{f}_{2}} = \left(\dfrac{3}{2} - 1\right)\left(\dfrac{1}{- 20} - \dfrac{1}{20}\right)$

$⇒ {f}_{2} = - 20cm$

Now, distance of image from concave,

${v}_{concave} = \dfrac{\left(- 30\right) \times \left(- 20\right)}{- 30 - 20}$$= - 12cm$

Now, $m = {m}_{convex} \times {m}_{concave}$

$m = \dfrac{\left(- 20\right)}{\left(- 10\right)} \times \dfrac{\left(- 12\right)}{\left(- 30\right)}$$= 0 . 8$