Question

Two immiscible transparent liquids of refractive indices $1.5$ and $\frac{4}{3}$ are arranged as shown in figure. The apparent depth of mark at the bottom from top surface is

• A. $10cm$
• B. $7cm$
• C. $\frac{43}{3}cm$
• D. $12cm$

Answer 1

The correct option is B $7cm$

shift observed due to them is $t_1(1-\frac{1}{{\mu }_1})+t_2(1-\frac{1}{{\mu }_2})$

$6(1-\frac{2}{3})+4(1-\frac{3}{4})=3$

observed depth is $10-3=7$

option $B$ is correct