A thin equiconvex lens of glass of refractive index `mu=3//2` & of focal length `0.3m`in air is sealed into an opening at one end of a tank filled with water `(mu=4//3)` .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 the mirror is `0.8m`A small object is placed outside the tank in front of the lens at a distance of `0.9`m form the lens along its axis .Find the position (relative to the lens)of the image of the object fromed by the stsyem.

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.