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 M1. When the set-up is kept in a medium of refractive index 7/6, the magnification becomes M2. The magnitude |M2/M1| is

The first image from reflection by spherical concave mirror can be found from mirror formula:
1/v + 1/u = 1/f.. (i)
For concave surface: u = -15 cm, f = -10 cm
Substituting in equation (i)
1/v + 1/-15 = 1/-10
1/v = 1/10 - 1/15 = 1/30
=> v = 30 cm
the magnification is given as
m = -v/u = -30/-15 = 2
The first image will act as object for lens and it's position can be found using lens formula
1/v - 1/u = 1/f.. (ii)
For concave surface: u = -20 cm, f = 10 cm
Substituting in equation (ii)
1/v - 1/(-20) = 1/10
1/v + 1/20 = 1/10
1/v = 1/10 - 1/20 = 1/20
=> v = 20 cm
the magnification is given as
m = v/u = 20/(-20) = -1
total magnification
M1 = 2 x (-1) = -2
Now the whole set is immersed in liquid of
n = 7/6
the reflection from concave mirror will remain same as their is no change in object position and focal length.
The change in focal length can be found using lens maker formula
1/f = (μ₂ - μ₁/μ₁)(1/R₁ - 1/R₂)
1/f_air = (1.5 - 1/1)(1/R₁ - 1/R₂) .. (iii)
1/f_liquid = ((1.5/(7/6)) - 1/(7/6))(1/R₁ - 1/R₂) .. (iv)
dividing (iii) by (iv)
f_liquid/f_air = 1/(2/7 + 1/6) = 42/(12 + 7) = 42/19
f_liquid/f_air = 1/(1.5/(7/6) - (7/6)) = 1/((9/7) - (7/6)) = 42/(54-49)=42/5 = 7/4
f_liquid = f_air x 7/4 = 70/4 = 17.5 cm
The first image will act as object for lens and it's position can be found using lens formula
1/v - 1/u = 1/f.. (i)
For concave surface: u = -20cm, f = 70/4 cm
Substituting in equation (i)
1/v - 1/(-20) = 4/70
1/v + 1/20 = 4/70
1/v = 2/35 - 1/20 = (40-35)/700 = 5/700 = 1/140
=> v = 140cm
the magnification is given as
m = v/u = 140/-20 = -7
total magnification
M2 = 2 x (-7) = -14
M2/M1 = -14/-2 = 7