A thin equiconvex lens made of glass of refractive index `3//2` and of focal length `0.3m` in air is sealed into an opening at one end of a tank filled with water `(mu=(3)/(2))`. On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis as shown in Figure. The separation between the lens and mirror is `0.8m. ` A small object is placed outside the tank in front of the lens at a distance oa `0.9m` from the lens along its axis. Find the position (relative to lens) of the image of the object formed by the system.

Answer

0.9 m from the lens (rightwards)

Explanation

From the lens maker's formula
`(1)/(f) = (mu-1)((1)/(R_(1)) - (1)/(R_(2)))`
we have `(1)/(0.3) = ((3)/(2)-1)((1)/(R) - (1)/(-R))`
(Here `R_(1) = R "and" R_(2) = -R`) `" " therefore R = 0.3`
Now applying `(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2)- mu_(1))/(R) ` at air glass surface , we get `(3//2)/(v_(1))- (1)/(-(09)) = (3//2 - 1)/(0.3)`
`therefore v_(1) = 2.7 "m"`
i.e., first image `I_(1)` will be formed at 2.7 m from the lens. This will act as the virtual object for glass water surface.
Therefore, applying `(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)` at glass water surface.
we have `(4//3)/(v_(2)) - (3//2)/(2.7) = (4//3 - 3//2)/(-0.3)`
`therefore v_(2) = 1.2 m`
i.e, second image `I_(2)` is formed at 1.2 m from the lens or 0.4 m from the plane mirror. This will act as a virtual object for mirror. Therefore , third real image `I_(3)` will be formed at a distance of 0.4 m in front of the mirror after reflection from it. Now this will work as a real object for water glass interface.
Hence applying `(mu_(2))/(v)- (mu_(1))/(u) = (mu_(2)- mu_(1))/(R)`
we get `(3//2)/(v_(4)) - (4//3)/(-(0.8 - 0.4)) = (3//2-4//3)/(0.3)`
`therefore v_(4) = 0.54m` i.e. fourth image is formed to the right of the lens at a distance of 0.54 m from it. Now finally applying the same formula for glass air surface.
`(1)/(v_(2)) - (3//2)/(0.54) - (1-3//2)/(-0.3) " " therefore v_(theta) = 0.9 m`
i.e., position of final image is 0.9 m relative to the lens frightwards or the image is formed 0.1 m behind the mirror.