Question

Three immiscible liquids of densities and refractive indies $\mu _ { 1 } > \mu _ { 2 } > \mu _ { 3 }$ 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. $\frac { \mathrm { h } } { 6 } \left( \frac { 1 } { \mu _ { 1 } } + \frac { 1 } { \mu _ { 2 } } + \frac { 1 } { \mu _ { 3 } } \right)$
• B. $\frac { \mathrm { h } } { 6 } \left( \frac { 1 } { \mu _ { 1 } } - \frac { 1 } { \mu _ { 2 } } - \frac { 1 } { \mu _ { 3 } } \right)$
• C. $\frac { \mathrm { h } } { 3 } \left( \frac { 1 } { \mu _ { 1 } } - \frac { 1 } { \mu _ { 2 } } - \frac { 1 } { \mu _ { 3 } } \right)$
• D. $\frac { \mathrm { h } } { 3 } \left( \frac { 1 } { \mu _ { 1 } } + \frac { 1 } { \mu _ { 2 } } + \frac { 1 } { \mu _ { 3 } } \right)$

Answer 1

Answer: (D)

$\frac { \mathrm { h } } { 3 } \left( \frac { 1 } { \mu _ { 1 } } + \frac { 1 } { \mu _ { 2 } } + \frac { 1 } { \mu _ { 3 } } \right)$

Answer 2

: Let be apparent depth of the dot when seen from air.

$\therefore \mu _ { 1 } = \frac { \mathrm { h } / 3 } { \mathrm { x } _ { 1 } }$

(Here, h/3 is real depth of the dot under liquid of density $\mathrm { d } _ { 1 }$)

Similarly, apparent depths of the dot when seen from air through two other liquids are

$\mathrm { x } _ { 2 } = \frac { \mathrm { h } } { 3 \mu _ { 2 } }$ and

$\therefore$ Apparent depth of the dot

$= \frac { 1 } { 3 \mu } + \frac { 1 } { 3 \mu _ { 2 } } + \frac { \mathrm { h } } { 3 \mu _ { 3 } } = \frac { \mathrm { h } } { 3 } \left[ \frac { 1 } { \mu _ { 1 } } + \frac { 1 } { \mu _ { 2 } } + \frac { 1 } { \mu _ { 3 } } \right]$