Three immiscible transparent liquids with refractive indices `3//2,4//3` and `6//5` are arranged one on top of another. The depth of the liquid are `3 cm, 4 cm` and `6 cm` respectively. The apparent depth of the vessel is

Options

A. `10 cm`
B. `9 cm`
C. `8 cm`
D. `7 cm`

Explanation

To find the apparent depth of a vessel containing three immiscible transparent liquids with different refractive indices, we can use the formula for apparent depth in a medium with varying refractive indices. The formula states that the apparent depth \( d \) is given by:

\[ d = \frac{d_1}{\mu_1} + \frac{d_2}{\mu_2} + \frac{d_3}{\mu_3} \]

where:
- \( d_1, d_2, d_3 \) are the actual depths of the liquids,
- \( \mu_1, \mu_2, \mu_3 \) are the refractive indices of the liquids.

### Step-by-Step Solution:

1. **Identify the depths and refractive indices:**
- For the first liquid:
- Depth \( d_1 = 3 \, \text{cm} \)
- Refractive index \( \mu_1 = \frac{3}{2} \)
- For the second liquid:
- Depth \( d_2 = 4 \, \text{cm} \)
- Refractive index \( \mu_2 = \frac{4}{3} \)
- For the third liquid:
- Depth \( d_3 = 6 \, \text{cm} \)
- Refractive index \( \mu_3 = \frac{6}{5} \)

2. **Substitute the values into the formula:**
\[ d = \frac{3}{\frac{3}{2}} + \frac{4}{\frac{4}{3}} + \frac{6}{\frac{6}{5}} \]

3. **Calculate each term:**
- For the first term:
\[ \frac{3}{\frac{3}{2}} = 3 \cdot \frac{2}{3} = 2 \, \text{cm} \]
- For the second term:
\[ \frac{4}{\frac{4}{3}} = 4 \cdot \frac{3}{4} = 3 \, \text{cm} \]
- For the third term:
\[ \frac{6}{\frac{6}{5}} = 6 \cdot \frac{5}{6} = 5 \, \text{cm} \]

4. **Sum the results:**
\[ d = 2 + 3 + 5 = 10 \, \text{cm} \]

5. **Conclusion:**
The apparent depth of the vessel is \( 10 \, \text{cm} \).

### Final Answer:
The apparent depth of the vessel is \( 10 \, \text{cm} \).