A diverting lens of focal length $20cm$ and converging mirror of focal length $10cm$ are placed coaxially at a separation of $5cm$. Where should an object be placed so that a real image is found at the object itself?

Let the object to placed at a distance $x$ from the lens further away from the mirror.

For the concave lens (1st refraction)

$u=-x,f=-20cm$

From lens formula,

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

$\Rightarrow \frac{1}{v}=\frac{1}{-20}+\frac{1}{-x}$

$\Rightarrow v=-\left(\frac{20x}{x+20}\right)$

So, the virtual image due to first refraction lies on the same side as that of object. $\left(A^{\prime }{B}^{\prime }\right)$

This image becomes the object for the concave mirror.

For the mirror,

$u=-(5+\frac{20x}{x+20})=-\left(\frac{25x+100}{x+20}\right)$

$f=-10cm$

From mirror equation,

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

$\Rightarrow \frac{1}{v}=\frac{1}{-10}+\frac{x+20}{25x+100}$

$\Rightarrow v=\frac{50(x+4)}{3x-20}$

So, this image is formed towards left of the mirror.

Again for second refraction in concave lens,

$u=-[5-\frac{50(x+4)}{3x-20}]$ (assuming that image of mirror is formed between the lens and mirro)

$v=+x$ (since, the final image is produced on the object)

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

$\Rightarrow \frac{1}{x}+\frac{1}{5}-\frac{50(x+4)}{3x-20}=\frac{1}{-20}$

$\Rightarrow x=60cm$

The project should be placed at a distance $60cm$ from the lens further away from the mirror.

So that the final image is formed on itself.