A fish is rising up veritcally inside a pond with velocity `4 m//s,` and notices a bird, which is dividing downward and its velocity appear to be `16 m//s` (to fish). What is the real velocity of the diving bird, if refractive index of water is `4//3?`

Explanation

To solve the problem, we need to find the real velocity of the diving bird as observed from the fish's perspective, taking into account the refractive index of water.

### Step-by-Step Solution:

1. **Identify the given values:**
- Velocity of the fish (upward): \( v_f = 4 \, \text{m/s} \)
- Apparent velocity of the bird (downward, as seen by the fish): \( v_{ab} = 16 \, \text{m/s} \)
- Refractive index of water: \( n = \frac{4}{3} \)

2. **Understand the relationship between apparent velocity and real velocity:**
The apparent velocity of the bird as seen by the fish is related to the real velocity of the bird (\( v_b \)) and the velocity of the fish (\( v_f \)) by the formula:
\[ v_{ab} = v_b - v_f \cdot n \]
where \( n \) is the refractive index.

3. **Rearrange the equation to solve for the real velocity of the bird:**
\[ v_b = v_{ab} + v_f \cdot n \]

4. **Substitute the known values into the equation:**
\[ v_b = 16 \, \text{m/s} + 4 \, \text{m/s} \cdot \frac{4}{3} \]

5. **Calculate the second term:**
\[ 4 \cdot \frac{4}{3} = \frac{16}{3} \approx 5.33 \, \text{m/s} \]

6. **Add the two velocities:**
\[ v_b = 16 + \frac{16}{3} = \frac{48}{3} + \frac{16}{3} = \frac{64}{3} \approx 21.33 \, \text{m/s} \]

7. **Conclusion:**
The real velocity of the diving bird is approximately \( 21.33 \, \text{m/s} \).