Question

An object is located at the bottom of a tank containing two immiscible liquids and is seen vertically from above. The lower and upper liquids are of depths `h_1` and `h_2` respectively and of refractive indices `mu_1` and `mu_2` respectively. Locate the position of the image of the object O from the surface.

Answer 1

Shift by (1): `h_1(1-1//mu_1)`
Shift by (2): `h_2(1-1//mu_2)`
Total shift `OI=h_1(1-1//mu_1)+h_2(1-1//mu_2)`
Distance of final image from P
`PI=PO-OI`
`=(h_1+h_2)-[h_1(1-1//mu_1)+h_2(1-1//mu_2)]`
`=(h_1)/(mu_1)+(h_2)/(mu_2)`