Two convex lenses of focal length 20cm each are placed coaxially...

Let ${L}_{1}$ be the first lens and ${L}_{2}$ be the second lens.

The first refraction is in ${L}_{1}$.

Using lens formula,

${u}_{1} = \infty , {f}_{1} = 20cm, \dfrac{1}{{v}_{1}} - 0 = \dfrac{1}{{f}_{1}} ⇒ {v}_{1} = 20cm$

Hence, the distance between the second image and the ${L}_{2}$ would be \left(60 - 20\right) = 40cm$.</p><p>For the second lens${L}_{2}$,</p><p>${u}_{2} = - 40cm, {f}_{2} = 20cm, \dfrac{1}{{v}_{2}} - \dfrac{1}{{u}_{2}} = \dfrac{1}{{f}_{2}} ⇒ \dfrac{1}{{v}_{2}} + \dfrac{1}{40} = \dfrac{1}{20} ⇒ \dfrac{1}{{v}_{2}} = \dfrac{1}{40} ⇒ {v}_{2} = 40cm$</p><p>So, the distance of image from first lens is</p><p>$x = 40cm + 60cm$</p><p>$= 100cm$