Question

Three immisible liquids of derivatives d1>d2>d3 and refractive index ¶1>¶2>¶3 are put.the height of liquid column is h/3. A dot is made at the bottom of the Beaker f normal vision. Find the apparent depth of the dot

Answer 1

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Answer 2

The apparent depth of the dot is $\frac{h}{3}\left[\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}+\frac{1}{\mu_{3}}\right]$

Explanation:

Let the real depth of the dot undo liquid of density d₁ be h/3

Let the apparent depth be x₁

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

$x_{1}=\frac{h}{3 \mu_{1}}$

Similarly, the apparent depth of the dot on the second liquid is:

$x_{2}=\frac{h}{3 \mu_{2}}$

Similarly, the apparent depth of the dot on the third liquid is:

$x_{3}=\frac{h}{3 \mu_{3}}$

Now, the depth is:

$x=x_{1}+x_{2}+x_{3}$

On substituting the values, we get,

$x=\frac{h}{3 \mu_{1}}+\frac{h}{3 \mu_{2}}+\frac{h}{3 \mu_{3}}$

$\therefore x=\frac{h}{3}\left[\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}+\frac{1}{\mu_{3}}\right]$