A fish rising vertically to the surface of water in a lake uniformly at the rate of `3ms^(-1)` observes a bird diving vertically towards the water at a rate of `9ms^(-1)` vertically above it. If the refractive index of water is 4/3 the actual velocity of the dive of the bird is (in `ms^(-1)`)

Options

A. 6
B. 4.5
C. 1.5
D. 2

Answer

To solve the problem, we need to find the actual velocity of the bird diving towards the water, given the apparent velocity observed by the fish and the refractive index of water.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Apparent velocity of the bird (as seen by the fish), \( v_{apparent} = 9 \, \text{m/s} \)
- Velocity of the fish rising, \( v_{fish} = 3 \, \text{m/s} \)
- Refractive index of water, \( \mu = \frac{4}{3} \)

2. **Understand the Relationship:**
The relationship between the apparent velocity, actual velocity, and the velocities involved can be expressed using the equation:
\[ v_{apparent} = v_{fish} + \mu \cdot v_{actual} \]
Here, \( v_{actual} \) is the actual velocity of the bird diving.

3. **Substitute the Known Values:**
Plugging in the known values into the equation:
\[ 9 = 3 + \frac{4}{3} \cdot v_{actual} \]

4. **Rearranging the Equation:**
To isolate \( v_{actual} \), first subtract \( 3 \) from both sides:
\[ 9 - 3 = \frac{4}{3} \cdot v_{actual} \]
\[ 6 = \frac{4}{3} \cdot v_{actual} \]

5. **Solve for \( v_{actual} \):**
Multiply both sides by \( \frac{3}{4} \) to solve for \( v_{actual} \):
\[ v_{actual} = 6 \cdot \frac{3}{4} \]
\[ v_{actual} = \frac{18}{4} = 4.5 \, \text{m/s} \]

6. **Conclusion:**
The actual velocity of the bird diving towards the water is \( 4.5 \, \text{m/s} \).