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 Fig. 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 this set up is kept in a medium of refractive index `7//6`, the magnification becomes `M_2`. The magnitude `((M_2)/(M_1))` is :
.

Explanation

Applying mirror formula
`(1)/(v)+(1)/(u)=(1)/(f)`
`(1)/(v)=(1)/(f)-(1)/(u)=(1)/(-10)+(1)/(15)`
`(1)/(v)=(-15+10)/(150)=(-5)/(150)=(-1)/(30)`
`:'` `v=-30cm`
For convex lens `u=|2f_l|`
Therefore image will have a magnification of l. when the set-up is kept in a medium The focal length of the lens will change
`((1)/(f_l))/((1)/(f_l^(`)))=(((n_l)/(n_s)-1))/(((n_l)/(n_s^(`))-1))implies(f_l^(`))/(10)=[[(1.5)/(1)-1]]/([1.5/((7)/(6))-1])`
`impliesf_l^`=17.5cm`
Applying lens formula `(1)/(v)-(1)/(u)=(1)/(f_l^`)`
`(1)/(v)-(1)/(-20)=(1)/(17.5)impliesv=15=40cm`
`M_l^``=`Magnification by lens `=(v)/(u)=(140)/(-20)=-7`
Now `[[M_2)/(M_1)]=` `|(MmirrorxxM_l^`)/(MmirrorxxM_l)|=7`