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 

There is no effect on magnification of mirror due to change in medium
For mirror,  $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}+\frac{1}{15}$ 
So image due to mirror will form at a distance of 30 cm from the mirror in front of it.
So distance of object from the lens $=50-30=20cm$  
In first case, the object is at a distance of $2f$ from lens. So magnification will be 1
In second case, $\frac{1}{f\text{'}}=(\mu \text{'}-1)x$  
But $\frac{1}{f}=(\mu -1)x\Rightarrow f\text{'}=\left(\frac{\mu -1}{\mu \text{'}-1}\right)f=\frac{70}{4}cm$ 
So magnification $=\frac{f\text{'}}{f\text{'}+u}=7$  
So ratio $\frac{M_2}{M_1}=\frac{7\times \text{\hspace{0.17em}\hspace{0.17em}Magnification\hspace{0.17em}of\hspace{0.17em}mirror}}{1\times \text{\hspace{0.17em}\hspace{0.17em}Magnification\hspace{0.17em}of\hspace{0.17em}mirror}}=7$