Question

Three immiscible liquids 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 $h/3$. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

• A. $\frac{h}{6}\left(\frac{1}{\mu_{1}} +\frac{1}{\mu_{2}}+\frac{1}{\mu_{3}}\right)$
• B. $\frac{h}{6}\left(\frac{1}{\mu_{1}} -\frac{1}{\mu_{2}} - \frac{1}{\mu_{3}}\right)$
• C. $\frac{h}{3}\left(\frac{1}{\mu_{1}} -\frac{1}{\mu_{2}} - \frac{1}{\mu_{3}}\right)$
• D. $\frac{h}{3}\left(\frac{1}{\mu_{1}} +\frac{1}{\mu_{2}}+\frac{1}{\mu_{3}}\right)$

Answer 1

Answer: (D)

$\frac{h}{3}\left(\frac{1}{\mu_{1}} +\frac{1}{\mu_{2}}+\frac{1}{\mu_{3}}\right)$

Answer 2

Apparent depth of the dot $= \frac{h}{3\mu_{1}}+\frac{h}{3\mu_{2}}+\frac{h}{3\mu_{3}}$ $= \frac{h}{3}\left[\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}+\frac{1}{\mu_{3}}\right]$