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 :
.

Answer

7

Explanation

For reflection from concave mirror ,
`(1)/(v) + (1)/(u) = (1)/(f) " " implies (1)/(v) - (1)/(15) = (-1)/(10)`
`(1)/(v) = (1)/(15) - (1)/(10) = (-1)/(30)`
`therefore v = -30`
magnification `(m_(1)) = -(v)/(u) =-2`
Now for refraction from lens ,
`(1)/(v) - (1)/(u) = (1)/(f) implies (1)/(v) = (1)/(10) - (1)/(20) = (1)/(20)`
`therefore` magnification `(m_(2)) = (v)/(u) =-1`
`therefore M_(1) = m_(1)m_(2) = 2`
Now when the set-up is immersed in liquid , no effect for the image formed by mirror .
we have `(mu_(L) -1) ((1)/(R_(1)) - (1)/(R_(2))) = (1)/(10)`
`implies ((1)/(R_(1)) - (1)/(R_(2))) = (1)/(5)`
when lens is immersed in liquid ,
`(1)/(f_("lens")) = ((mu_(L))/(mu_(S))-1) ((1)/(R_(1)) - (1)/(R_(2)))`
`=(2)/(7) xx (1)/(5) = (2)/(35)`
`therefore (1)/(v)- (1)/(u) = (1)/(f_("liquid"))`
`implies (1)/(v) = (2)/(35) - (1)/(20) = (8-7)/(140) = (1)/(140)`
`therefore` magnification =` -(140)/(20) = -7`
`therefore M_(2) = 2 xx 7 = 14 `
`therefore |(M_(2))/(M_(1))| = 7`