An optical arrangement consists of two concave mirrors M _ 1 and M...

For reflection from ${M}_{2}$:

Applying mirror formula, we get

$\dfrac{1}{v} + \dfrac{1}{\left(- 10\right)} = \dfrac{1}{\left(- 12\right)}$

$⇒ \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{12}$

$⇒ v = + 60cm$ (for ${I}_{1}$)

For refraction from $L$:

Applying lens equation, we get

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

$⇒ \dfrac{1}{v} - \dfrac{1}{\left(- 80\right)} = \dfrac{1}{10}$

$⇒ v = + \dfrac{80}{7}$ (for ${I}_{2}$)

This image should be at focus of ${M}_{1}$

$∴ \dfrac{20}{2} + \dfrac{80}{7} = \dfrac{n}{7}$

$⇒ n = 150$

Also,

If ${I}_{2}$ is formed at pole of ${M}_{1}$

then $\dfrac{n}{7} = \dfrac{80}{7}$

$⇒ n = 80$

And further if ${I}_{2}$ is formed at centre of curvature of ${M}_{1}$ then

$\dfrac{n}{7} = \dfrac{80}{7} + 20$

$∴ n = 220$

So value of $n$ can be either $80\text{or}150\text{or}220$.