A tank contains three layers of immiscible liquids. The first layer is of water with refractive index `4//3` and thickness 8cm. The second layer is of oil with refractive index `3//2` and thickness 9cm while the third layer is of glycerine with refractive index 2 and thickness 4cm. Find the apparent depth of the bottom of the container.

Explanation

Method of interfaces:
A ray of light from the object undergoes refraction at three interfaces: (1) Water-oil, (2) Oil-glycerine, and (3) Glycerine-air. The coordinate system for each of the interfaces is shown in the figure below.
Water-oil interface:
`d_(1)=-8cm, mu_(1)=(4)/(3), mu_(2)=1.5 `
`(mu_(1))/(d_(1))=(mu_(2))/(d_(2))` or `d_(2)=(mu_(2))/(mu_(1)) d_(1)` , we get `d_(2)=-9cm`
Oil-glycerine interface:
`d_(1)=-(9+9)=-18cm, mu_(1)=1.5, mu_(2)=2`
As `(mu_(1))/(d_(1))=(mu_(2))/(d_(2))` or `d_(2)=(mu_(2))/(mu_(1))d_(1)`
`d_(2)=-24cm`
Glycerine-air interface
`d_(1)=-(4+24)=-28cm, mu_(1)=2, mu_(2)=1`
As `(mu_(1))/(d_(1))=(mu_(2))/(d_(2))` or `d_(2)=(mu_(2))/(mu_(1))d_(1)` , we get `d_(2)=-14cm`
Thus, the final image is 14 cm below the glycerine-air interface.
Method 2: Method of shifting
Net shifting
`s=d_(1)(1-(1)/(mu_(1)))+d_(2)(1-(1)/(mu_(2)))+d_(3)(1-(1)/(mu_(3)))`
`=8(1-(1)/(4//3))+9(1-(1)/(3//2))+4(1-(1)/(2))`
`=7cm`
The shifting will be in the direction of ray travelling, i.e., upwards.
Hence, apparent depth `h^(')=(21-7)=14cm` .