Question

Three immiscible liquid of densities $d_1$ > $d_2$ > $d_3$ and refractive indices ${\mu }_1$ > ${\mu }_2$ > ${\mu }_3$ are put in a beaker. The height of each liquid column is $\frac{h}{3}$. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

• A. $\frac{h}{6}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2}+\frac{1}{{\mu }_3})$
• B. $\frac{h}{6}(\frac{1}{{\mu }_1}-\frac{1}{{\mu }_2}-\frac{1}{{\mu }_3})$
• C. $\frac{h}{3}(\frac{1}{{\mu }_1}-\frac{1}{{\mu }_2}+\frac{1}{{\mu }_3})$
• D. $\frac{h}{3}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2}+\frac{1}{{\mu }_3})$

Answer 1

Answer: (D)

$\frac{h}{3}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2}+\frac{1}{{\mu }_3})$

For Liquid 1,

Apparent depth$=\frac{h}{3{\mu }_1}$

For Liquid 2,

Apparent depth$=\frac{h}{3{\mu }_2}$

For Liquid 3,

Apparent depth$=\frac{h}{{\mu }_3}$

Total Apparent Depth$=\frac{h}{3}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2}+\frac{1}{{\mu }_3})$