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) - 1)((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)`
For 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`