Two thin lenses, both of 10 cm focal length—one convex and other concave, are placed 5 cm apart. An object is placed 20 cm in front of the convex lens. Find the nature and position of the final image.

$\Rightarrow$                 ${\mathrm{v}}_1 = 0 +20 \mathrm{cm}$

The convex lens produces converging rays trying to meet at I₁, at a distance of 20 cm from the convex lens, i.e., 15 cm behind the concave lens.
I₁ will serve as a virtual object for the concave lens. 

For refraction at the concave lens, we have
 
Object distance, $\mathrm{u} = 20 - 5 = 15 \mathrm{cm}$
Focal length, $\mathrm{f} = -10 \mathrm{cm}$ 

As per sign convention
                      $$\begin{gathered} \mathrm{u} = -15 \\ \mathrm{f} = -10 \end{gathered}$$ 

Now using lens formula, 
                     $\frac{1}{\mathrm{f}} = \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$ 

                    $\frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}$
                          $$\begin{gathered} =-\frac{1}{10}-\frac{1}{15} \\ =\frac{-3-2}{30} =\frac{-6}{30} \end{gathered}$$  

                     $\frac{1}{\mathrm{v}} =-\frac{1}{5}$
                       $\mathrm{v} = -5 \mathrm{cm}$ 

i.e., this image is in the side of object at a distance of 5 cm to the right of concave lens and at a distance of 10 cm(5 + 5) from convex lens.  

Now,
u = + 15 cm; v =?, f = –10cm

Using lens formula, we have 

                 $\frac{1}{\mathrm{v}}-\frac{1}{15}=-\frac{1}{10}$
$\Rightarrow$                        $\mathrm{v} = -30 \mathrm{cm}$ 

Hence, the final image is virtual and is located at 30 cm to the left of the concave lens.