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  $\frac{7}{6},$  the magnification becomes  $M_2$. The magnitude $\left|\frac{M_2}{M_1}\right|$  is 

Applying mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}or\frac{1}{v}=\frac{1}{f}-\frac{1}{w}=\frac{1}{-10}+\frac{1}{15}$
 $\therefore \frac{1}{v}=\frac{-15+10}{150}=\frac{-5}{150}=\frac{-1}{30}\therefore v=-30cm$
And magnification,  $m_1=-\frac{v}{u}=-\frac{-30}{-15}=-2$

Now for refraction  from lens,  $u=-(50-30)=-20cm$
 $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}\Rightarrow \frac{1}{v}-\frac{1}{-20}=\frac{1}{10}$
$\frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}\therefore v=20cm$ 
Magnification,  $m_2=\frac{v}{u}=\frac{20}{-20}=-1$
Magnification produced by the combination, 
 $M_1=m_1\times m_2=(-2)\times (-1)=2$
Again, when  system  is kept in a medium  of refractive index 7/6.
There is no change for mirror in this case, 
For lens,  $\frac{1}{f_l\text{'}}=(\frac{{\mu }_l}{{\mu }_s}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
 $\frac{1}{f_l\text{'}}=(\frac{3/2}{7/6}-1)(\frac{1}{R_1}-\frac{1}{R_2})\Rightarrow \frac{1}{f_l\text{'}}=\frac{2}{7}(\frac{1}{R_1}-\frac{1}{R_2})$
Also, when lens was in  air  $\frac{1}{f_l}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{10}\therefore (\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{5}$ 
Using this result in eqn. (i), we get 
 $\frac{1}{f_1\text{'}}=\frac{2}{7}\times \frac{1}{5}\therefore f_l\text{'}=\frac{35}{2}cm$
Again using lens formula,  $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_l\text{'}}$
$\frac{1}{v}-\frac{1}{-20}=\frac{2}{35}\Rightarrow \frac{1}{v}=\frac{2}{35}-\frac{1}{20}=\frac{1}{140}\therefore v=140cm$ 
Magnification,  $\frac{1}{v}-\frac{1}{-20}=\frac{2}{35}\Rightarrow \frac{1}{v}=\frac{2}{35}-\frac{1}{20}=\frac{1}{140}\therefore v=140cm$
Magnification produced by the combination, 
$M_2=m_1\times m_2\text{'}=(-2)\times (-7)=14$

$\therefore \left|\frac{M_2}{M_1}\right|=\frac{14}{2}=7$