A vesse of depth x is half filled with oil of refractive index `mu_(1)` and the other half is filled with water of refrative index `mu_(2)`. The apparent depth of the vessel when viewed above is

Options

A. `(x(mu_(1)+mu_(2)))/(2mu_(1)mu_(2))`
B. `(xmu_(1)mu_(2))/(2(mu_(1)+mu_(2)))`
C. `(xmu_(1)mu_(2))/((mu_(1)+mu_(2)))`
D. `(2x(mu_(1)+mu_(2)))/(mu_(1)mu_(2))`

Explanation

To solve the problem of finding the apparent depth of a vessel that is half-filled with oil and half-filled with water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Depths**:
The total depth of the vessel is given as \( x \). Since the vessel is half-filled with oil and half-filled with water, the depth of each liquid is:
\[ \text{Depth of oil} = \frac{x}{2} \]
\[ \text{Depth of water} = \frac{x}{2} \]

2. **Calculate Apparent Depth for Oil**:
The formula for apparent depth when viewed from above is given by:
\[ d_1 = \frac{\text{Real depth}}{\text{Refractive index}} = \frac{\frac{x}{2}}{\mu_1} = \frac{x}{2\mu_1} \]

3. **Calculate Apparent Depth for Water**:
Similarly, for water, the apparent depth is:
\[ d_2 = \frac{\text{Real depth}}{\text{Refractive index}} = \frac{\frac{x}{2}}{\mu_2} = \frac{x}{2\mu_2} \]

4. **Total Apparent Depth**:
The total apparent depth \( d \) when viewed from the top is the sum of the apparent depths of both liquids:
\[ d = d_1 + d_2 = \frac{x}{2\mu_1} + \frac{x}{2\mu_2} \]

5. **Combine the Terms**:
To combine the terms, we can factor out \( \frac{x}{2} \):
\[ d = \frac{x}{2} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} \right) \]

6. **Final Expression**:
We can express the final result as:
\[ d = \frac{x}{2} \cdot \frac{\mu_1 + \mu_2}{\mu_1 \mu_2} \]

### Final Answer:
Thus, the apparent depth of the vessel when viewed from above is:
\[ d = \frac{x(\mu_1 + \mu_2)}{2\mu_1\mu_2} \]