Let us first locate the image of S formed by the lens L. Here u = 12 cm and f = 15 cm. We have,

                     $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

or,                 $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

                       $=\frac{1}{15\text{ cm}}-\frac{1}{12\text{ cm}}$

or,                $v=-60\text{ cm.}$

The negative sign shows that the image is formed to the left of the lens as suggested in the figure. The image I₁ acts as the source for the mirror. The mirror forms an image I₂ of the source I₁. This image I₂ then acts as the source for the lens and the final beam comes out parallel to the principal axis. Clearly I₂ must be at the focus of the lens. We have,

                    I₁ I₂ = I₁L + LI₂ = 60 cm + 15 cm = 75 cm.

Suppose the distance of the mirror from I₂ is x cm. For the reflection from the mirror,

           u = MI₁ = - (75 + x)cm, v = - x cm and f = - 20 cm.

Using                      $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$,

                              $\frac{1}{\text{x}}+\frac{1}{75+\text{x}}=\frac{1}{20}$

or,                        $\frac{75+2\text{x}}{\left(75+\text{x}\right)\text{x}}=\frac{1}{20}$

or,                        ${\text{x}}^2+35\text{x}-1500=0$

or,                        $\text{x}=\frac{-35\pm \sqrt{35\times 35+4\times 1500}}{2}$.

This gives x = 25 or - 60.

As the negative sign has no physical meaning, only positive sign should be taken. Taking x = 25, the separation between the lens and the mirror is (15 + 25) cm = 40 cm.

Hence k = 4