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

`90 cm` from the lens towards right

Explanation

Form lens Maker's formula,
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
we have `(1)/(0.3)=((30)/(2)-1)((1)/(R)-(1)/(-R))`
`(Here R_(1)=R_(2)=-R) therefore R=0.4 m`
Now applying `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)` at air galss surface, we get
`((3)/(2))/v_(1)-(1)/(-(0.9))=((3)/(2)-1)/(0.3) therefore v_(1)=2.7 m`
`i.e,.` first iamge `_(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)=1.2m`
`i.e,.` second iamge `_(2)` is formed at `1.2 m` from the lens or `0.4 m` from the plane mirror. This will act as virtual object for mirror. Therefore, third real iamge `_(3)` will be formed at a distance of `0.4 m` in front of the mirror after reflection form it. Now this iamge 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)=-0.54 m`
`i.e.,` fourth iamge is `0.9 m` relative to the lens (rightward) or the image is formed `0.1 m` behind the mirror .