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 :
.

Options

A. 5
B. 6
C. 7
D. 8

Answer

C

Explanation

Image formation by concave mirror is unaffected by presence of medium as for mirror, `(1)/(v)+(1)/(u)=(1)/(f)`
Here, `u = -15 cm, f = -10 cm`
Let `v = v_m`. From `(1)/(v)=(1)/(f)-(1)/(u) , (1)/(v_m)=(1)/(-10)+(1)/(15)` ,
`v_m = -30 cm`
For the lens in air, `R_1 = R, R_2 = -R`
`(1)/(f)=(mu - 1)((1)/(R_1) - (1)/(R_2))` or `(1)/(10)=((3)/(2) - ) ((2)/( R))`
or `(1)/(5)=(2)/( R)`
In medium of refractive index `7//6`,
`(1)/(f_2)=((3 //2)/(7//6) -1)((2)/(R)) = (2)/(7) xx (1)/(5) =(2)/(35)`
Fpr lens in air, `u = -20 cm, f = + 10 cm`
`(1)/(v_a)-(1)/((-20))=(1)/(10)` or `v_a = 20 cm`
For lens in medium, `(1)/(v'_m) -(1)/((-20)) = (2)/(35)`
or `v'_m = 140 cm`
:. `(M_2)/(M_1)=(v'_m)/(v_a) = (140)/(20) = 7`.