Consider a concave mirror and a convex lens (refractive index=1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index=1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification  $M_1.$  When the set-up is kept in a medium of refractive index  $7/6$  the magnification becomes $M_2.$   The magnitude $\left|\frac{M_2}{M_1}\right|$  is.

 $\frac{1}{v}=\frac{1}{15}-\frac{1}{10}=\frac{-1}{30}\therefore v=-30$
Magnification  $\left(m_1\right)=-\frac{v}{u}=-2Now$
For refraction from lens, 

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}\therefore$
 Magnification $\left(m_2\right)=\frac{v}{u}=-1$    
$\therefore M_1=m_1{m}_2=2$ Now   when the set up is immersed in liquid, no effect for the image formed   by mirror. We have 
 $({\mu }_L-1)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{10}$
$\Rightarrow (\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{5}$  When lens is immersed in liquid,
 $\frac{1}{f_{lens}}=(\frac{{\mu }_L}{{\mu }_S}-1)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{2}{7}\times \frac{1}{5}=$$\therefore \frac{1}{v}-\frac{1}{u}=\frac{1}{f_{Liquid}}$
 
    $\Rightarrow \frac{1}{v}=\frac{2}{35}-\frac{1}{20}=\frac{8-7}{140}=\frac{1}{140}$
Magnification    $=-\frac{140}{20}=-7$   $\therefore M_2=2\times 7=14\therefore \left|\frac{M_2}{M_1}\right|=7$