A vessel is quarter filled with a liquid of refractive index `mu`. The remaining parts of the vessel is filled with an immiscible liquid of refractive index `3mu//2`. The apparent depth of the vessel is 50% of the actual depth. The value of `mu` is

Options

A. 1.4
B. 1.5
C. 1.8
D. 1.67

Explanation

To solve the problem, we need to find the value of the refractive index \( \mu \) for the liquid in the vessel. Let's break down the steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The vessel is filled with two liquids. The first liquid has a refractive index \( \mu \) and occupies \( \frac{1}{4} \) of the height \( H \) of the vessel. Therefore, the height of this liquid is \( \frac{H}{4} \).
- The second liquid has a refractive index \( \frac{3\mu}{2} \) and occupies the remaining \( \frac{3}{4} \) of the height of the vessel. Thus, the height of this liquid is \( \frac{3H}{4} \).

2. **Apparent Depth Calculation**:
- The apparent depth \( d_a \) is given as 50% of the actual depth \( H \). Therefore, we have:
\[ d_a = \frac{H}{2} \]

3. **Using the Formula for Apparent Depth**:
- The formula for the apparent depth when dealing with two different refractive indices is:
\[ d_a = \frac{d_1}{\mu_1} + \frac{d_2}{\mu_2} \]
- Here, \( d_1 = \frac{H}{4} \) (depth of the first liquid) and \( d_2 = \frac{3H}{4} \) (depth of the second liquid).
- The refractive indices are \( \mu_1 = \mu \) and \( \mu_2 = \frac{3\mu}{2} \).

4. **Substituting Values into the Formula**:
- Substitute the known values into the formula:
\[ \frac{H}{2} = \frac{\frac{H}{4}}{\mu} + \frac{\frac{3H}{4}}{\frac{3\mu}{2}} \]

5. **Simplifying the Equation**:
- Simplifying the right-hand side:
\[ \frac{H}{2} = \frac{H}{4\mu} + \frac{2H}{4} \]
- This simplifies to:
\[ \frac{H}{2} = \frac{H}{4\mu} + \frac{H}{2} \]

6. **Eliminating \( H \)**:
- Since \( H \) is common in all terms, we can cancel it out (assuming \( H \neq 0 \)):
\[ \frac{1}{2} = \frac{1}{4\mu} + \frac{1}{2} \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{1}{2} - \frac{1}{2} = \frac{1}{4\mu} \]
- This leads to:
\[ 0 = \frac{1}{4\mu} \]
- This is incorrect; we need to adjust our approach.

8. **Correcting the Equation**:
- Going back to:
\[ \frac{H}{2} = \frac{H}{4\mu} + \frac{3H}{8} \]
- Multiply through by \( 8\mu \) to eliminate the fractions:
\[ 4\mu = 2 + 3\mu \]

9. **Solving for \( \mu \)**:
- Rearranging gives:
\[ 4\mu - 3\mu = 2 \]
\[ \mu = 2 \]

### Final Answer:
The value of \( \mu \) is \( 2 \).