Three immiscible liquids of densities `d_1 gt d_2 gt d_3` and refractive indices `mu_1 gt mu_2 gt 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.

Options

A. `h/6(1/(mu_(1))+1/(mu_(2))+1/(mu_(3)))`
B. `h/6(1/(mu_(1))-1/(mu_(2))-1/(mu_(3)))`
C. `h/3(1/(mu_(1))-1/(mu_(2))-1/(mu_(3)))`
D. `h/3(1/(mu_(1))+1/(mu_(2))+1/(mu_(3)))`

Explanation

To find the apparent depth of the dot at the bottom of the beaker containing three immiscible liquids, we can follow these steps:

### Step 1: Understand the Setup
We have three immiscible liquids stacked in a beaker, each with a height of \( \frac{h}{3} \). The refractive indices of the liquids are given as \( \mu_1 > \mu_2 > \mu_3 \).

### Step 2: Define Real Depth
The real depth of the dot at the bottom of the beaker is the total height of the three liquid columns:
\[ \text{Real Depth} = \frac{h}{3} + \frac{h}{3} + \frac{h}{3} = h \]

### Step 3: Calculate Apparent Depth for Each Liquid
Using the formula for apparent depth, which relates the real depth and the refractive index:
\[ \text{Apparent Depth} = \frac{\text{Real Depth}}{\mu} \]

For each liquid, we can calculate the apparent depth as follows:

1. **For the first liquid (height \( \frac{h}{3} \), refractive index \( \mu_1 \)):**
\[ x_1 = \frac{\frac{h}{3}}{\mu_1} = \frac{h}{3\mu_1} \]

2. **For the second liquid (height \( \frac{h}{3} \), refractive index \( \mu_2 \)):**
\[ x_2 = \frac{\frac{h}{3}}{\mu_2} = \frac{h}{3\mu_2} \]

3. **For the third liquid (height \( \frac{h}{3} \), refractive index \( \mu_3 \)):**
\[ x_3 = \frac{\frac{h}{3}}{\mu_3} = \frac{h}{3\mu_3} \]

### Step 4: Sum the Apparent Depths
The total apparent depth \( x \) when viewed from the air through all three liquids is the sum of the individual apparent depths:
\[ x = x_1 + x_2 + x_3 \]
Substituting the values we calculated:
\[ x = \frac{h}{3\mu_1} + \frac{h}{3\mu_2} + \frac{h}{3\mu_3} \]

### Step 5: Factor Out Common Terms
We can factor out \( \frac{h}{3} \) from the expression:
\[ x = \frac{h}{3} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} \right) \]

### Final Answer
Thus, the apparent depth of the dot is:
\[ \boxed{x = \frac{h}{3} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} \right)} \]