Question

A container has two layers of immiscible liquids $1$ and $2$ with refractive indices $1.5$ and $4/3$ respectively. What is apparent position of an object placed at the bottom of the vessel. Thickness of the layers are $3\text{cm}$ and $8\text{cm}$ respectively?

• A. $6\text{cm}\text{below the liquid 2 - air interface.}$
• B. $8\text{cm}\text{below the liquid 2 - air interface.}$
• C. $8\text{cm}\text{below the liquid 1 - liquid 2 interface.}$
• D. $6\text{cm}\text{below the water - air interface.}$

Answer 1

The correct option is B $8\text{cm}\text{below the liquid 2 - air interface.}$

Using Image shift in glass slab formula,
Shift of position of object due to liquids of refractive indices ${\mu }_1,{\mu }_2$ and thickness $d_1,d_2$ is
$d_1(1-\frac{1}{{\mu }_1})+d_2(1-\frac{1}{{\mu }_2})$
Here ${\mu }_1=\frac{3}{2},{\mu }_2=\frac{4}{3}$ and thickness $d_1=3\text{cm},d_2=8\text{cm}$
So shift = $3(1-\frac{2}{3})+8(1-\frac{3}{4})=3\text{cm}$ upwards from bottom of the container.
Hence final position from air-liquid 2 interface is $8+3-3=8\text{cm}$ below it