Three immiscibles transparent liquids with erefractive 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 solve the problem of finding the apparent depth of three immiscible transparent liquids with given refractive indices and depths, we can follow these steps:

### Step 1: Identify the given data
- Refractive indices:
- Liquid 1: \( \mu_1 = \frac{3}{2} \)
- Liquid 2: \( \mu_2 = \frac{4}{3} \)
- Liquid 3: \( \mu_3 = \frac{6}{5} \)
- Depths:
- Liquid 1: \( d_1 = 3 \, \text{cm} \)
- Liquid 2: \( d_2 = 4 \, \text{cm} \)
- Liquid 3: \( d_3 = 6 \, \text{cm} \)

### Step 2: Use the formula for apparent depth
The relationship between real depth and apparent depth is given by:
\[ \mu = \frac{d}{h'} \]
Where:
- \( \mu \) is the refractive index,
- \( d \) is the real depth,
- \( h' \) is the apparent depth.

From this, we can rearrange the formula to find the apparent depth:
\[ h' = \frac{d}{\mu} \]

### Step 3: Calculate the apparent depths for each liquid
1. For Liquid 1:
\[ h_1' = \frac{d_1}{\mu_1} = \frac{3 \, \text{cm}}{\frac{3}{2}} = \frac{3 \times 2}{3} = 2 \, \text{cm} \]

2. For Liquid 2:
\[ h_2' = \frac{d_2}{\mu_2} = \frac{4 \, \text{cm}}{\frac{4}{3}} = \frac{4 \times 3}{4} = 3 \, \text{cm} \]

3. For Liquid 3:
\[ h_3' = \frac{d_3}{\mu_3} = \frac{6 \, \text{cm}}{\frac{6}{5}} = \frac{6 \times 5}{6} = 5 \, \text{cm} \]

### Step 4: Calculate the total apparent depth
Now, we can find the total apparent depth \( h' \) by summing the apparent depths of all three liquids:
\[ h' = h_1' + h_2' + h_3' = 2 \, \text{cm} + 3 \, \text{cm} + 5 \, \text{cm} = 10 \, \text{cm} \]

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

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