Three immiscible liquids of densities d1 d2 d3 and refractive indices μ1 μ2 μ3 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.

Question

Three immiscible liquids of densities d1 > d2 > d3 and refractive indices μ1 > μ2 > μ3 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) (h/6)((1/μ1) +(1/μ2)+(1/μ3)) (B) (h/6)((1/μ1) -(1/μ2) - (1/μ3)) (C) (h/3)((1/μ1) -(1/μ2) - (1/μ3)) (D) (h/3)((1/μ1) +(1/μ2)+(1/μ3))

Tardigrade Admin · Accepted Answer

Apparent depth of the dot = (h/3μ1)+(h/3μ2)+(h/3μ3) = (h/3)[(1/μ1)+(1/μ2)+(1/μ3)]

Q. Three immiscible liquids of densities $d_{1} > d_{2} > d_{3}$ and refractive indices $μ_{1} > μ_{2} > μ_{3}$ 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.

$\frac{h}{6} (\frac{1}{μ _{1}} + \frac{1}{μ _{2}} + \frac{1}{μ _{3}})$

$\frac{h}{6} (\frac{1}{μ _{1}} - \frac{1}{μ _{2}} - \frac{1}{μ _{3}})$

$\frac{h}{3} (\frac{1}{μ _{1}} - \frac{1}{μ _{2}} - \frac{1}{μ _{3}})$

$\frac{h}{3} (\frac{1}{μ _{1}} + \frac{1}{μ _{2}} + \frac{1}{μ _{3}})$

Apparent depth of the dot
$= \frac{h}{3 μ _{1}} + \frac{h}{3 μ _{2}} + \frac{h}{3 μ _{3}}$
$= \frac{h}{3} [\frac{1}{μ _{1}} + \frac{1}{μ _{2}} + \frac{1}{μ _{3}}]$

Q. Three immiscible liquids of densities d1​>d2​>d3​ and refractive indices μ1​>μ2​>μ3​ 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.

6h​(μ1​1​+μ2​1​+μ3​1​)

6h​(μ1​1​−μ2​1​−μ3​1​)

3h​(μ1​1​−μ2​1​−μ3​1​)

3h​(μ1​1​+μ2​1​+μ3​1​)