Two thin similar watch glass pieces are joined together,front to front ,with near convex portion silvered and the combination of glass pieces is placed at a distance `a=60cm` from a screen.A small object is placed normal to the optical axis of the combination such that its image is two time the object. Now water is filled in between the glass. water`mu=4//3`,calculated the distance through,which the object must be displayed so that a sharp image is again formed on the screen.

Answer

15 cm towards the combination

Explanation

Initially the system will act as simple concave mirror given

v = -60, m=-2 `implies u = -30`
`implies (1)/(v) + (1)/(u) = (1)/(f) implies (1)/(-60) - (1)/(30) = (1)/(f)`
f = -20 cm `implies` R = 40 cm
In second case : lens + mirror system
Then `P_("meq") = P_(L) + P_(L) + P_(M)`
`P_("meq")` = equivalent power of the mirror
`P_(L)` = power of the lens
`P_(M)` = power of the mirror
`P_("meq") = (1)/(f_(L)) + (1)/(f_(L)) + (-1)/(f_(M)) because (1)/(f_(L)) = ((1)/(3)) (2)/(R)`
`(1)/(f_(L)) = ((1)/(3)) (2)/(R) = (2 xx 2)/(3R) + ((-2)/(-R)) ((3)/(3))`
`P_("meq") = (10)/(3R) = (-1)/(f_(m)) implies f_("meq") = (-3R)/(10)`
`f_("eqm")= -12 cm`
Now if object is placed at distance = x then
`(1)/(v) + (1)/(u) = (1)/(f_("eq")), (because v = -60)`
`(1)/(-60) + (1)/(u) = (-1)/(12) implies (1)/(u) = (1)/(60) - (1)/(12) = (-4)/(50)`
`implies u = -15 cm`
So object displaced = 15 cm